C/C++ Users Group Library 1996 July/C-C++ Users Group Library July 1996.iso/vol_400/442_01/ptrtut02.txt

Text File | 1996-02-15 | 87.7 KB | 2,472 lines

PTRTUT02.TXT   12/19/95

           A TUTORIAL ON POINTERS AND ARRAYS IN C
                     by Ted Jensen
                      Version 0.2
      This material is hereby placed in the public domain.

                    TABLE OF CONTENTS

        Preface

        Introduction

        Chapter 1:  What is a Pointer?

        Chapter 2:  Pointer Types and Arrays.

        Chapter 3:  Pointers and Strings

        Chapter 4:  More on Strings

        Chapter 5:  Pointers and Structures

        Chapter 6:  More on Strings and Arrays of Strings

        Chapter 7:  More on Multi-Dimensional Arrays

        Chapter 8:  Pointers to Arrays

        Chapter 9:  Pointers and Dynamic Allocation of Memory

        Chapter 10: Pointers to Functions

        Epilog

->============================================================
PREFACE

    This document is intended to introduce pointers to beginning
programmers in the C programming language.  Over several years of
reading and contributing to various conferences on C including
those on the FidoNet and UseNet, I have noted a large number of
newcomers to C appear to have a difficult time in grasping the
fundamentals of pointers.  I therefore undertook the task of
trying to explain them in plain language with lots of examples.

    The first version of this document was placed in the public
domain, as is this one.  It was picked up by Bob Stout who
included it as a file called PTR-HELP.TXT in his widely
distributed collection of SNIPPETS.  Since that release, I have
added a significant amount of material and made some minor
corrections in the original work.

Acknowledgements:

    There are so many people who have unknowingly contributed to
this work because of the questions they have posed in the FidoNet
C Echo, or the UseNet Newsgroup comp.lang.c, or several other
conferences in other networks, that it would be impossible to
list them all.  Special thanks go to Bob Stout who was kind
enough to include the first version of this material in his
SNIPPETS file.

About the Author:

    Ted Jensen is a retired Electronics Engineer who worked as a
hardware designer or manager of hardware designers in the field
of magnetic recording.  Programming has been a hobby of his off
and on since 1968 when he learned how to keypunch cards for
submission to be run on a mainframe.  (The mainframe had 64K of
magnetic core memory!).

Use of this Material:

    Everything contained herein is hereby released to the Public
Domain.  Any person may copy or distribute this material in any
manner they wish.  The only thing I ask is that if this material
is used as a teaching aid in a class, I would appreciate it if it
were distributed in its entirety, i.e. including all chapters,
the preface and the introduction.  I would also appreciate it if
under such circumstances the instructor of such a class would
drop me a note at one of the addresses below informing me of
this.  I have written this with the hope that it will be useful
to others and since I'm not asking any financial remuneration,
the only way I know that I have at least partially reached that
goal is via feedback from those who find this material useful.

    By the way, you needn't be an instructor or teacher to
contact me.  I would appreciate a note from _anyone_ who finds
the material useful, or who has constructive criticism to offer.
I'm also willing to answer questions submitted by mail.

Ted Jensen                        [e-mail]
P.O. Box 324                      1-415-365-8452
Redwood City, CA 94064
Dec. 1995

->============================================================
INTRODUCTION

    If one is to be proficient in the writing of code in the C
programming language, one must have a thorough working knowledge
of how to use pointers.  Unfortunately, C pointers appear to
represent a stumbling block to newcomers, particularly those
coming from other computer languages such as Fortran, Pascal or
Basic.

    To aid those newcomers in the understanding of pointers I have
written the following material.  To get the maximum benefit from
this material, I feel it is important that the user be able to
run the code in the various listings contained in the article.  I
have attempted, therefore, to keep all code ANSI compliant so
that it will work with any ANSI compliant compiler.  And I have
tried to carefully block the code within the text so that with
the help of an ASCII text editor one can copy a given block of
code to a new file and compile it on their system.  I recommend
that readers do this as it will help in understanding the
material.



->==================================================================
CHAPTER 1: What is a pointer?

    One of the things beginners in C find most difficult to
understand is the concept of pointers.  The purpose of this
document is to provide an introduction to pointers and their use
to these beginners.

    I have found that often the main reason beginners have a
problem with pointers is that they have a weak or minimal feeling
for variables, (as they are used in C).  Thus we start with a
discussion of C variables in general.

    A variable in a program is something with a name, the value
of which can vary.  The way the compiler and linker handles this
is that it assigns a specific block of memory within the computer
to hold the value of that variable.  The size of that block
depends on the range over which the variable is allowed to vary.
For example, on PC's the size of an integer variable is 2 bytes,
and that of a long integer is 4 bytes.  In C the size of a
variable type such as an integer need not be the same on all
types of machines.

    When we declare a variable we inform the compiler of two
things, the name of the variable and the type of the variable.
For example, we declare a variable of type integer with the name
k by writing:

    int k;

    On seeing the "int" part of this statement the compiler sets
aside 2 bytes (on a PC) of memory to hold the value of the
integer.  It also sets up a symbol table. And in that table it
adds the symbol k and the relative address in memory where those
2 bytes were set aside.

    Thus, later if we write:

    k = 2;

at run time we expect that the value 2 will be placed in that
memory location reserved for the storage of the value of k.  In C
we refer to a variable such as the integer k as an "object".

    In a sense there are two "values" associated with the object
k, one being the value of the integer stored there (2 in the
above example) and the other being the "value" of the memory
location where it is stored, i.e. the address of k.  Some texts
refer to these two values with the nomenclature rvalue (right
value, pronounced "are value") and lvalue (left value, pronounced
"el value") respectively.

    In some languages, the lvalue is the value permitted on the
left side of the assignment operator '=' (i.e. the address where
the result of evaluation of the right side ends up).  The rvalue
is that which is on the right side of the assignment statement,
the '2' above.  Rvalues cannot be used on the left side
of the assignment statement.  Thus:    2 = k;   is illegal.

    Actually, the above definition of "lvalue" is somewhat
modified for C.  According to K&R-2 (page 197):  [1]

    "An _object_ is a named region of storage; an _lvalue_ is an
     expression referring to an object."

However, at this point, the definition originally cited above is
sufficient.  As we become more familiar with pointers we will go
into more detail on this.

    Okay, now consider:

    int j, k;
    k = 2;
    j = 7;    <-- line 1
    k = j;    <-- line 2

    In the above, the compiler interprets the j in line 1 as the
address of the variable j (its lvalue) and creates code to copy
the value 7 to that address.  In line 2, however, the j is
interpreted as its rvalue (since it is on the right hand side of
the assignment operator '=').  That is, here the j refers to the
value _stored_ at the memory location set aside for j, in this
case 7.  So, the 7 is copied to the address designated by the
lvalue of k.

    In all of these examples, we are using 2 byte integers so all
copying of rvalues from one storage location to the other is done
by copying 2 bytes.  Had we been using long integers, we would be
copying 4 bytes.

    Now, let's say that we have a reason for wanting a variable
designed to hold an lvalue (an address).  The size required to
hold such a value depends on the system.  On older desk top
computers with 64K of memory total, the address of any point in
memory can be contained in 2 bytes.  Computers with more memory
would require more bytes to hold an address.  Some computers,
such as the IBM PC might require special handling to hold a
segment and offset under certain circumstances.  The actual size
required is not too important so long as we have a way of
informing the compiler that what we want to store is an address.

    Such a variable is called a "pointer variable" (for reasons
which hopefully will become clearer a little later).  In C when
we define a pointer variable we do so by preceding its name with
an asterisk.  In C we also give our pointer a type which, in this
case, refers to the type of data stored at the address we will be
storing in our pointer.  For example, consider the variable
declaration:

    int *ptr;

    ptr is the _name_ of our variable (just as 'k' was the name
of our integer variable).  The '*' informs the compiler that we
want a pointer variable, i.e. to set aside however many bytes is
required to store an address in memory.  The "int" says that we
intend to use our pointer variable to store the address of an
integer. Such a pointer is said to "point to" an integer.
However, note that when we wrote  "int k;" we did not give k a value.
If this definition was made outside of any function many compilers
will initialize it to zero.  Similarly, ptr has no value, that is
we haven't stored an address in it in the above declaration.  In
this case, again if the declaration is outside of any function,
it is initialized to a value #defined by your compiler as NULL. It
is called a NULL pointer.  While in most cases NULL is #defined
as zero, it need not be.  That is, different compilers handle
this differently.  Also while zero is an integer, NULL
need not be.  However, the value that NULL actually has
internally is of little consequence to the programmer since at
the source code level   NULL == 0  is guaranteed to evaluate to
true regardless of the internal value of NULL.

    But, back to using our new variable ptr.  Suppose now that we
want to store in ptr the address of our integer variable k.  To
do this we use the unary '&' operator and write:

    ptr = &k;

    What the '&' operator does is retrieve the lvalue (address)
of k, even though k is on the right hand side of the assignment
operator '=', and copies that to the contents of our pointer ptr.
Now, ptr is said to "point to" k.  Bear with us now, there is
only one more operator we need to discuss.

    The "dereferencing operator" is the asterisk and it is used
as follows:

    *ptr = 7;

will copy 7 to the address pointed to by ptr.  Thus if ptr
"points to" (contains the address of) k, the above statement will
set the value of k to 7.  That is, when we use the '*' this way
we are referring to the value of that which ptr is pointing
to, not the value of the pointer itself.

    Similarly, we could write:

    printf("%d\n",*ptr);

to print to the screen the integer value stored at the address
pointed to by "ptr".

    One way to see how all this stuff fits together would be to
run the following program and then review the code and the output
carefully.

------------ Program 1.1 ---------------------------------

/* Program 1.1 from PTRTUT02.TXT   12/14/95 */

#include <stdio.h>

int j, k;
int *ptr;


int main(void)
{
   j = 1;
   k = 2;
   ptr = &k;
   printf("\n");
   printf("j has the value %d and is stored at %p\n",j,&j);
   printf("k has the value %d and is stored at %p\n",k,&k);
   printf("ptr has the value %p and is stored at %p\n",ptr,&ptr);
   printf("The value of the integer pointed to by ptr is %d\n",
           *ptr);
   return 0;
}
---------------------------------------
To review:

    A variable is declared by giving it a type and a name (e.g.
     int k;)

    A pointer variable is declared by giving it a type and a name
     (e.g. int *ptr) where the asterisk tells the compiler that
     the variable named ptr is a pointer variable and the type
     tells the compiler what type the pointer is to point to
     (integer in this case).

    Once a variable is declared, we can get its address by
     preceding its name with the unary '&' operator, as in &k.

    We can "dereference" a pointer, i.e. refer to the value of
     that which it points to, by using the unary '*' operator as
     in *ptr.

    An "lvalue" of a variable is the value of its address, i.e.
     where it is stored in memory.  The "rvalue" of a variable is
     the value stored in that variable (at that address).

References in Chapter 1:

    [1]  "The C Programming Language"   2nd Edition
          B. Kernighan and D. Ritchie
          Prentice Hall
          ISBN 0-13-110362-8

->==================================================================
CHAPTER 2: Pointer types and Arrays

    Okay, let's move on.  Let us consider why we need to identify
the "type" of variable that a pointer points to, as in:

        int *ptr;

    One reason for doing this is so that later, once ptr "points
to" something, if we write:

        *ptr = 2;

the compiler will know how many bytes to copy into that memory
location pointed to by ptr.  If ptr was declared as pointing to an
integer, 2 bytes would be copied, if a long, 4 bytes would be
copied.  Similarly for floats and doubles the appropriate number
will be copied.  But, defining the type that the pointer points
to permits a number of other interesting ways a compiler can
interpret code.  For example, consider a block in memory
consisting if ten integers in a row.  That is, 20 bytes of memory
are set aside to hold 10 integers.

    Now, let's say we point our integer pointer ptr at the first
of these integers.  Furthermore lets say that integer is located
at memory location 100 (decimal).  What happens when we write:

    ptr + 1;

    Because the compiler "knows" this is a pointer (i.e. its
value is an address) and that it points to an integer (its
current address, 100, is the address of an integer), it adds 2 to
ptr instead of 1, so the pointer "points to" the _next_
_integer_, at memory location 102.  Similarly, were the ptr
declared as a pointer to a long, it would add 4 to it instead of
1.  The same goes for other data types such as floats, doubles,
or even user defined data types such as structures.  This is
obviously not the same kind of "addition" that we normally think
of.  In C it is referred to as addition using "pointer
arithmetic", a term which we will come back to later.

    Similarly, since ++ptr and ptr++ are both equivalent to
ptr + 1 (though the point in the program when ptr is incremented
may be different), incrementing a pointer using the unary ++
operator, either pre- or post-, increments the address it stores
by the amount sizeof(type) where "type" is the type of the object
pointed to. (i.e. 2 for an integer, 4 for a long,
etc.).

    Since a block of 10 integers located contiguously in memory
is, by definition, an array of integers, this brings up an
interesting relationship between arrays and pointers.

    Consider the following:

    int my_array[] = {1,23,17,4,-5,100};

    Here we have an array containing 6 integers.  We refer to
each of these integers by means of a subscript to my_array, i.e.
using my_array[0] through my_array[5].  But, we could
alternatively access them via a pointer as follows:

    int *ptr;

    ptr = &my_array[0];       /* point our pointer at the first
                                 integer in our array */

    And then we could print out our array either using the array
notation or by dereferencing our pointer.  The following code
illustrates this:
-----------  Program 2.1  -----------------------------------
/* Program 2.1 from PTRTUT02.TXT   12/14/95 */


#include <stdio.h>

int my_array[] = {1,23,17,4,-5,100};
int *ptr;

int main(void)
{
    int i;
    ptr = &my_array[0];     /* point our pointer to the first
                                      element of the array */
    printf("\n\n");
    for(i = 0; i < 6; i++)
    {
      printf("my_array[%d] = %d   ",i,my_array[i]);   /*<-- A */
      printf("ptr + %d = %d\n",i, *(ptr + i));        /*<-- B */
    }
    return 0;
}
----------------------------------------------------
   Compile and run the above program and carefully note lines A
and B and that the program prints out the same values in either
case.  Also observe how we dereferenced our pointer in line B,
i.e. we first added i to it and then dereferenced the new
pointer. Change line B to read:

     printf("ptr + %d = %d\n",i, *ptr++);

and run it again... then change it to:

     printf("ptr + %d = %d\n",i, *(++ptr));

and try once more.  Each time try and predict the outcome and
carefully look at the actual outcome.

    In C, the standard states that wherever we might use
&var_name[0] we can replace that with var_name, thus in our code
where we wrote:

        ptr = &my_array[0];

    we can write:

        ptr = my_array;     to achieve the same result.

    This leads many texts to state that the name of an array is a
pointer.  While this is true, I prefer to mentally think "the
name of the array is the address of first element in the array".
Many beginners (including myself when I was learning) have a
tendency to become confused by thinking of it as a pointer.
For example, while we can write ptr = my_array; we cannot write

    my_array = ptr;

    The reason is that the while ptr is a variable, my_array is a
constant.  That is, the location at which the first element of
my_array will be stored cannot be changed once my_array[] has
been declared.

    Earlier when discussing the term "lvalue" I cited K&R-2 where
it stated:

    "An _object_ is a named region of storage; an _lvalue_ is an
     expression referring to an object".

This raises an interesting problem.  Since my_array is a named
region of storage, why is "my_array" in the above assignment
statement not an lvalue?  To resolve this problem, some refer to
"my_array" as an "unmodifiable lvalue".

Modify the example program above by changing

    ptr = &my_array[0];     to     ptr = my_array;

and run it again to verify the results are identical.

    Now, let's delve a little further into the difference between
the names "ptr" and "my_array" as used above.  Some writers will
refer to an array's name as a _constant_ pointer.  What do we
mean by that?  Well, to understand the term "constant" in this
sense, let's go back to our definition of the term "variable".
When we declare a variable we set aside a spot in memory to hold
the value of the appropriate type.  Once that is done the name of
the variable can be interpreted in one of two ways.  When used on
the left side of the assignment operator, the compiler interprets
it as the memory location to which to move that value resulting
from evaluation of the right side of the assignment operator.
But, when used on the right side of the assignment operator, the
name of a variable is interpreted to mean the contents stored at
that memory address set aside to hold the value of that variable.

    With that in mind, let's now consider the simplest of
constants, as in:

    int i, k;
    i = 2;

    Here, while "i" is a variable and then occupies space in the
data portion of memory, "2" is a constant and, as such, instead
of setting aside memory in the data segment, it is imbedded
directly in the code segment of memory.  That is, while writing
something like k = i;  tells the compiler to create code which at
run time will look at memory location &i to determine the value
to be moved to k, code created by  i = 2;  simply puts the '2' in
the code and there is no referencing of the data segment.  That
is, both k and i are objects, but 2 is not an object.

    Similarly, in the above, since "my_array" is a constant, once
the compiler establishes where the array itself is to be stored,
it "knows" the address of my_array[0] and on seeing:

    ptr = my_array;

it simply uses this address as a constant in the code segment and
there is no referencing of the data segment beyond that.

    Well, that's a lot of technical stuff to digest and I don't
expect a beginner to understand all of it on first reading.  With
time and experimentation you will want to come back and re-read
the first 2 chapters.  But for now, let's move on to the
relationship between pointers, character arrays, and strings.

->==================================================================
CHAPTER 3:  Pointers and Strings

    The study of strings is useful to further tie in the
relationship between pointers and arrays.  It also makes it easy
to illustrate how some of the standard C string functions can be
implemented. Finally it illustrates how and when pointers can and
should be passed to functions.

    In C, strings are arrays of characters.  This is not
necessarily true in other languages.  In BASIC, Pascal, Fortran
and various other languages, a string has its own data type.  But
in C it does not.  In C a string is an array of characters
terminated with a binary zero character (written as '\0'). To
start off our discussion we will write some code which, while
preferred for illustrative purposes, you would probably never
write in an actual program.  Consider, for example:

    char my_string[40];

    my_string[0] = 'T';
    my_string[1] = 'e';
    my_string[2] = 'd':
    my_string[3] = '\0';

    While one would never build a string like this, the end
result is a string in that it is an array of characters
_terminated_with_a_nul_character_.  By definition, in C, a string
is an array of characters terminated with the nul character. Be
aware that "nul" is _not_ the same as "NULL".  The nul refers to a zero
as is defined by the escape sequence '\0'.  That is it occupies
one byte of memory.  The NULL, on the other hand, is the value of
an uninitialized pointer and pointers require more than one byte
of storage.  NULL is #defined in a header file in your C
compiler, nul may not be #defined at all.

    Since writing the above code would be very time consuming, C
permits two alternate ways of achieving the same thing.  First,
one might write:

    char my_string[40] = {'T', 'e', 'd', '\0',};

    But this also takes more typing than is convenient.  So, C
permits:

    char my_string[40] = "Ted";

    When the double quotes are used, instead of the single quotes
as was done in the previous examples, the nul character ( '\0' )
is automatically appended to the end of the string.

    In all of the above cases, the same thing happens.  The
compiler sets aside an contiguous block of memory 40 bytes long
to hold characters and initialized it such that the first 4
characters are Ted\0.

    Now, consider the following program:

------------------program 3.1-------------------------------------
/* Program 3.1 from PTRTUT02.TXT   12/14/95 */


#include <stdio.h>

char strA[80] = "A string to be used for demonstration purposes";
char strB[80];

int main(void)
{
   char *pA;     /* a pointer to type character */
   char *pB;     /* another pointer to type character */
   puts(strA);   /* show string A */
   pA = strA;    /* point pA at string A */
   puts(pA);     /* show what pA is pointing to */
   pB = strB;    /* point pB at string B */
   putchar('\n');       /* move down one line on the screen */
   while(*pA != '\0')   /* line A (see text) */
   {
     *pB++ = *pA++;     /* line B (see text) */
   }
   *pB = '\0';          /* line C (see text) */
   puts(strB);          /* show strB on screen */
   return 0;
}
--------- end program 3.1 -------------------------------------

    In the above we start out by defining two character arrays of
80 characters each.  Since these are globally defined, they are
initialized to all '\0's first.  Then, strA has the first 42
characters initialized to the string in quotes.

    Now, moving into the code, we declare two character pointers
and show the string on the screen.  We then "point" the pointer pA
at strA.  That is, by means of the assignment statement we copy
the address of strA[0] into our variable pA.  We now use puts()
to show that which is pointed to by pA on the screen.  Consider
here that the function prototype for puts() is:

    int puts(const char *s);

    For the moment, ignore the "const".  The parameter passed to
puts is a pointer, that is the _value_ of a pointer (since all
parameters in C are passed by value), and the value of a pointer
is the address to which it points, or, simply, an address.  Thus
when we write:

    puts(strA);        as we have seen, we are passing the

address of strA[0].  Similarly, when we write:

    puts(pA);          we are passing the same address, since

we have set pA = strA;

    Given that, follow the code down to the while() statement on
line A.  Line A states:

    While the character pointed to by pA (i.e. *pA) is not a nul
character (i.e. the terminating '\0'), do the following:

    line B states:  copy the character pointed to by pA to the
space pointed to by pB, then increment pA so it points to the
next character and pB so it points to the next space.

    When we have copied the last character, pA now points to the
terminating nul character and the loop ends. However, we have not
copied the nul character.  And, by definition a string in C
_must_ be nul terminated.  So, we add the nul character with line
C.

    It is very educational to run this program with your debugger
while watching strA, strB, pA and pB and single stepping through
the program.  It is even more educational if instead of simply
defining strB[] as has been done above, initialize it also with
something like:

 strB[80] = "12345678901234567890123456789012345678901234567890"

where the number of digits used is greater than the length of
strA and then repeat the single stepping procedure while watching
the above variables.  Give these things a try!

    Getting back to the prototype for puts() for a moment, the
"const" used as a parameter modifier informs the user that the
function will not modify the string pointed to by s, i.e. it will
treat that string as a constant.

    Of course, what the above program illustrates is a simple way
of copying a string.  After playing with the above until you have
a good understanding of what is happening, we can proceed to
creating our own replacement for the standard strcpy() that comes
with C.  It might look like:

   char *my_strcpy(char *destination, char *source)
   {
        char *p = destination;
        while (*source != '\0')
        {
           *p++ = *source++;
        }
        *p = '\0';
        return destination;
   }

    In this case, I have followed the practice used in the
standard routine of returning a pointer to the destination.

    Again, the function is designed to accept the values of two
character pointers, i.e. addresses, and thus in the previous
program we could write:

int main(void)
{
  my_strcpy(strB, strA);
  puts(strB);
}

    I have deviated slightly from the form used in standard C
which would have the prototype:

    char *my_strcpy(char *destination, const char *source);

    Here the "const" modifier is used to assure the user that the
function will not modify the contents pointed to by the source
pointer.  You can prove this by modifying the function above, and
its prototype, to include the "const" modifier as shown.  Then,
within the function you can add a statement which attempts to
change the contents of that which is pointed to by source, such
as:

    *source = 'X';

which would normally change the first character of the string to
an X.  The const modifier should cause your compiler to catch
this as an error.  Try it and see.

    Now, let's consider some of the things the above examples
have shown us.  First off, consider the fact that *ptr++ is to be
interpreted as returning the value pointed to by ptr and then
incrementing the pointer value.  This has to
do with the precedence of the operators.  Were we to write
(*ptr)++ we would increment, not the pointer, but that which the
pointer points to!  i.e. if used on the first character of the
above example string the 'T' would be incremented to a 'U'.  You
can write some simple example code to illustrate this.

    Recall again that a string is nothing more than an array
of characters, with the last character being a '\0'.  What we
have done above is deal with copying an array.  It happens to be
an array of characters but the technique could be applied to an
array of integers, doubles, etc.  In those cases, however, we
would not be dealing with strings and hence the end of the array
would not be marked with a special value like the nul character.
We could implement a version that relied on a special value to
identify the end. For example, we could copy an array of positive
integers by marking the end with a negative integer.  On the
other hand, it is more usual that when we write a function to
copy an array of items other than strings we pass the function
the number of items to be copied as well as the address of the
array, e.g. something like the following prototype might
indicate:

    void int_copy(int *ptrA, int *ptrB, int nbr);

where nbr is the number of integers to be copied.  You might want
to play with this idea and create an array of integers and see if
you can write the function int_copy() and make it work.

    This permits using functions to manipulate large arrays. For
example, if we have an array of 5000 integers that we want to
manipulate with a function, we need only pass to that function
the address of the array (and any auxiliary information such as
nbr above, depending on what we are doing). The array itself does
_not_ get passed, i.e. the whole array is not copied and put on
the stack before calling the function, only its address is sent.

    This is different from passing, say an integer, to a
function.  When we pass an integer we make a copy of the integer,
i.e. get its value and put it on the stack.  Within the function
any manipulation of the value passed can in no way effect the
original integer.  But, with arrays and pointers we can pass the
address of the variable and hence manipulate the values of the
original variables.

->==================================================================
CHAPTER 4: More on Strings

    Well, we have progressed quite a way in a short time!  Let's
back up a little and look at what was done in Chapter 3 on
copying of strings but in a different light.  Consider the
following function:

   char *my_strcpy(char dest[], char source[])
   {
        int i = 0;

        while (source[i] != '\0')
        {
           dest[i] = source[i];
           i++;
        }
        dest[i] = '\0';
        return dest;
   }

    Recall that strings are arrays of characters.  Here we have
chosen to use array notation instead of pointer notation to do
the actual copying.  The results are the same, i.e. the string
gets copied using this notation just as accurately as it did
before.  This raises some interesting points which we will
discuss.

    Since parameters are passed by value, in both the passing of
a character pointer or the name of the array as above, what
actually gets passed is the address of the first element of each
array.  Thus, the numerical value of the parameter passed is the
same whether we use a character pointer or an array name as a
parameter.  This would tend to imply that somehow:

        source[i]  is the same as  *(p+i);

    In fact, this is true, i.e wherever one writes   a[i]  it can
be replaced with  *(a + i) without any problems.  In fact, the
compiler will create the same code in either case.  Thus we see
that pointer arithmetic is the same thing as array indexing.
Either syntax produces the same result.

    This is NOT saying that pointers and arrays are the same
thing, they are not.  We are only saying that to identify a given
element of an array we have the choice of two syntaxes, one using
array indexing and the other using pointer arithmetic, which
yield identical results.

    Now, looking at this last expression, part of it..  (a + i)
is a simple addition using the + operator and the rules of c
state that such an expression is commutative.  That is   (a + i)
is identical to (i + a).  Thus we could write *(i + a) just as
easily as *(a + i).

    But *(i + a) could have come from i[a] !  From all of this
comes the curious truth that if:

    char a[20];
    int i;

    writing    a[3] = 'x';   is the same as writing

               3[a] = 'x';

    Try it!  Set up an array of characters, integers or longs,
etc. and assigned the 3rd or 4th element a value using the
conventional approach and then print out that value to be sure
you have that working.  Then reverse the array notation as I have
done above.  A good compiler will not balk and the results will
be identical.  A curiosity... nothing more!

    Now, looking at our function above, when we write:

        dest[i] = source[i];

due to the fact that array indexing and pointer arithmetic yield
identical results, we can write this as:

        *(dest + i) = *(source + i);

    But, this takes 2 additions for each value taken on by i.
Additions, generally speaking, take more time than
incrementations (such as those done using the ++ operator as in
i++).  This may not be true in modern optimizing compilers, but
one can never be sure.  Thus, the pointer version may be a bit
faster than the array version.

    Another way to speed up the pointer version would be to
change:

    while (*source != '\0')     to simply    while (*source)

since the value within the parenthesis will go to zero (FALSE) at
the same time in either case.

    At this point you might want to experiment a bit with writing
some of your own programs using pointers.  Manipulating strings
is a good place to experiment.  You might want to write your own
versions of such standard functions as:

            strlen();
            strcat();
            strchr();

and any others you might have on your system.

    We will come back to strings and their manipulation through
pointers in a future chapter.  For now, let's move on and discuss
structures for a bit.

->==================================================================
CHAPTER 5: Pointers and Structures

    As you may know, we can declare the form of a block of data
containing different data types by means of a structure
declaration.  For example, a personnel file might contain
structures which look something like:

  struct tag{
       char lname[20];        /* last name */
       char fname[20];        /* first name */
       int age;               /* age */
       float rate;            /* e.g. 12.75 per hour */
       };

    Let's say we have a bunch of these structures in a disk file
and we want to read each one out and print out the first and last
name of each one so that we can have a list of the people in our
files.  The remaining information will not be printed out.  We
will want to do this printing with a function call and pass to
that function a pointer to the structure at hand.  For
demonstration purposes I will use only one structure for now. But
realize the goal is the writing of the function, not the reading
of the file which, presumably, we know how to do.

    For review, recall that we can access structure members with
the dot operator as in:

--------------- program 5.1 ------------------
/* Program 5.1 from PTRTUT02.TXT   12/14/95 */

#include <stdio.h>
#include <string.h>

struct tag{
       char lname[20];      /* last name */
       char fname[20];      /* first name */
       int age;             /* age */
       float rate;          /* e.g. 12.75 per hour */
       };

struct tag my_struct;       /* declare the structure my_struct */

int main(void)
{
  strcpy(my_struct.lname,"Jensen");
  strcpy(my_struct.fname,"Ted");
  printf("\n%s ",my_struct.fname);
  printf("%s\n",my_struct.lname);
  return 0;
}
-------------- end of program 5.1 --------------

    Now, this particular structure is rather small compared to
many used in C programs.  To the above we might want to add:

  date_of_hire;                  (data types not shown)
  date_of_last_raise;
  last_percent_increase;
  emergency_phone;
  medical_plan;
  Social_S_Nbr;
  etc.....

    If we have a large number of employees, what we want to do
manipulate the data in these structures by means of functions.
For example we might want a function print out the name of the
employee listed in any structure passed to it.  However, in the
original C (Kernighan & Ritchie, 1st Edition) it was not possible
to pass a structure, only a pointer to a structure could be
passed.  In ANSI C, it is now permissible to pass the complete
structure. But, since our goal here is to learn more about
pointers, we won't pursue that.

    Anyway, if we pass the whole structure it means that we must
copy the contents of the structure from the calling function to
the called function.  In systems using stacks, this is done by
pushing the contents of the structure on the stack.  With large
structures this could prove to be a problem.  However, passing a
pointer uses a minimum amount of stack space.

    In any case, since this is a discussion of pointers, we will
discuss how we go about passing a pointer to a structure and then
using it within the function.

    Consider the case described, i.e. we want a function that
will accept as a parameter a pointer to a structure and from
within that function we want to access members of the structure.
For example we want to print out the name of the employee in our
example structure.

    Okay, so we know that our pointer is going to point to a
structure declared using struct tag.  We declare such a pointer
with the declaration:

    struct tag *st_ptr;

and we point it to our example structure with:

    st_ptr = &my_struct;

    Now, we can access a given member by de-referencing the
pointer. But, how do we de-reference the pointer to a structure?
Well, consider the fact that we might want to use the pointer to
set the age of the employee.  We would write:

    (*st_ptr).age = 63;

    Look at this carefully.  It says, replace that within the
parenthesis with that which st_ptr points to, which is the
structure my_struct.  Thus, this breaks down to the same as
my_struct.age.

    However, this is a fairly often used expression and the
designers of C have created an alternate syntax with the same
meaning which is:

    st_ptr->age = 63;

    With that in mind, look at the following program:

------------ program 5.2 ---------------------
/* Program 5.2 from PTRTUT02.TXT   12/14/95 */

#include <stdio.h>
#include <string.h>

struct tag{                   /* the structure type */
       char lname[20];        /* last name */
       char fname[20];        /* first name */
       int age;               /* age */
       float rate;            /* e.g. 12.75 per hour */
       };

struct tag my_struct;         /* define the structure */

void show_name(struct tag *p);    /* function prototype */

int main(void)
{
  struct tag *st_ptr;         /* a pointer to a structure */
  st_ptr = &my_struct;        /* point the pointer to my_struct */
  strcpy(my_struct.lname,"Jensen");
  strcpy(my_struct.fname,"Ted");
  printf("\n%s ",my_struct.fname);
  printf("%s\n",my_struct.lname);
  my_struct.age = 63;
  show_name(st_ptr);          /* pass the pointer */
  return 0;
}


void show_name(struct tag *p)
{
  printf("\n%s ", p->fname);     /* p points to a structure */
  printf("%s ", p->lname);
  printf("%d\n", p->age);
}
-------------------- end of program 5.2 ----------------

    Again, this is a lot of information to absorb at one time.
The reader should compile and run the various code snippets and
using a debugger monitor things like my_struct and p while single
stepping through the main and following the code down into the
function to see what is happening.

->==================================================================
CHAPTER 6:  Some more on Strings, and Arrays of Strings

   Well, let's go back to strings for a bit.  In the following
all assignments are to be understood as being global, i.e. made
outside of any function, including main.

   We pointed out in an earlier chapter that we could write:

   char my_string[40] = "Ted";

which would allocate space for a 40 byte array and put the string
in the first 4 bytes (three for the characters in the quotes and
a 4th to handle the terminating '\0'.

    Actually, if all we wanted to do was store the name "Ted" we
could write:

      char my_name[] = "Ted";

and the compiler would count the characters, leave room for the
nul character and store the total of the four characters in memory
the location of which would be returned by the array name, in this
case my_string.

    In some code, instead of the above, you might see:

     char *my_name = "Ted";

which is an alternate approach.  Is there a difference between
these?  The answer is.. yes.  Using the array notation 4 bytes of
storage in the static memory block are taken up, one for each
character and one for the terminating nul character.  But, in the
pointer notation the same 4 bytes required, _plus_ N bytes to
store the pointer variable my_name (where N depends on the system
but is usually a minimum of 2 bytes and can be 4 or more).

    In the array notation, "my_name" is short for &myname[0]
which is the address of the first element of the array.  Since
the location of the array is fixed during run time, this is a
constant (not a variable).  In the pointer notation my_name is a
variable.  As to which is the _better_ method, that depends on
what you are going to do within the rest of the program.

    Let's now go one step further and consider what happens if
each of these declarations are done within a function as opposed
to globally outside the bounds of any function.

void my_function_A(char *ptr)
{
  char a[] = "ABCDE";
  .
  .
}

void my_function_B(char *ptr)
{
  char *cp = "ABCDE";
  .
  .
}

    Here we are dealing with automatic variables in both cases.
In my_function_A the automatic variable is the character array
a[]. In my_function_B it is the pointer cp.  While C is designed
in such a way that a stack is not required on those systems
which don't use them, my particular processor (80286) and
compiler (TC++) combination uses a stack.  I wrote a simple
program incorporating functions similar to those above and found
that in my_function_A the 5 characters in the string were all
stored on the stack.  On the other hand, in my_function_B, the 5
characters were stored in the data space and the pointer was
stored on the stack.

    By making a[] static I could force the compiler to place the
5 characters in the data space as opposed to the stack.  I did
this exercise to point out just one more difference between
dealing with arrays and dealing with pointers.  By the way, array
initialization of automatic variables as I have done in
my_function_A was illegal in the older K&R C and only "came of
age" in the newer ANSI C.  A fact that may be important when one
is considering portability and backwards compatibility.

    As long as we are discussing the relationship/differences
between pointers and arrays, let's move on to multi-dimensional
arrays.  Consider, for example the array:

    char multi[5][10];

    Just what does this mean?   Well, let's consider it in the
following light.

        char multi[5][10];
             ^^^^^^^^

    Let's take the underlined part to be the "name" of an array.
Then prepending the "char" and appending the [10] we have an
array of 10 characters.  But, the name "multi[5]" is itself an
array indicating that there are 5 elements each being an array of
10 characters.  Hence we have an array of 5 arrays of 10
characters each..

    Assume we have filled this two dimensional array with data of
some kind.  In memory, it might look as if it had been formed by
initializing 5 separate arrays using something like:

      multi[0] = {'0','1','2','3','4','5','6','7','8','9'}
      multi[1] = {'a','b','c','d','e','f','g','h','i','j'}
      multi[2] = {'A','B','C','D','E','F','G','H','I','J'}
      multi[3] = {'9','8','7','6','5','4','3','2','1','0'}
      multi[4] = {'J','I','H','G','F','E','D','C','B','A'}

At the same time, individual elements might be addressable using
syntax such as:

      multi[0][3] = '3'
      multi[1][7] = 'h'
      multi[4][0] = 'J'

    Since arrays are contiguous in memory, our actual memory
block for the above should look like:

    0123456789abcdefghijABCDEFGHIJ9876543210JIHGFEDCBA
    ^
    |_____ starting at the address &multi[0][0]

    Note that I did _not_ write  multi[0] = "0123456789". Had I
done so a terminating '\0' would have been implied since whenever
double quotes are used a '\0' character is appended to the
characters contained within those quotes.  Had that been the case
I would have had to set aside room for 11 characters per row
instead of 10.

    My goal in the above is to illustrate how memory is laid out
for 2 dimensional arrays. That is, this is a 2 dimensional array
of characters, NOT an array of "strings".

    Now, the compiler knows how many columns are present in the
array so it can interpret multi + 1 as the address of the 'a' in
the 2nd row above.  That is, it adds 10, the number of columns,
to get this location.  If we were dealing with integers and an
array with the same dimension the compiler would add
10*sizeof(int) which, on my machine, would be 20.  Thus, the
address of the "9" in the 4th row above would be &multi[3][0] or
*(multi + 3) in pointer notation.  To get to the content of the
2nd element in the 4th row we add 1 to this address and
dereference the result as in

    *(*(multi + 3) + 1)

    With a little thought we can see that:

    *(*(multi + row) + col)    and
    multi[row][col]            yield the same results.

    The following program illustrates this using integer arrays
instead of character arrays.

------------------- program 6.1 ----------------------
/* Program 6.1 from PTRTUT02.TXT   12/14/95 */

#include <stdio.h>

#define ROWS 5
#define COLS 10

int multi[ROWS][COLS];

int main(void)
{
  int row, col;
  for (row = 0; row < ROWS; row++)
    for(col = 0; col < COLS; col++)
      multi[row][col] = row*col;
  for (row = 0; row < ROWS; row++)
    for(col = 0; col < COLS; col++)
    {
      printf("\n%d  ",multi[row][col]);
      printf("%d ",*(*(multi + row) + col));
    }
  return 0;
}
----------------- end of program 6.1 ---------------------

    Because of the double de-referencing required in the pointer
version, the name of a 2 dimensional array is often said to be
equivalent to a pointer to a pointer.  With a three dimensional
array we would be dealing with an array of arrays of arrays and
some might say its name would be equivalent to a pointer to a
pointer to a pointer.  However, here we have initially set aside
the block of memory for the array by defining it using array
notation.  Hence, we are dealing with a constant, not a variable.
That is we are talking about a fixed address not a variable
pointer.  The dereferencing function used above permits us to
access any element in the array of arrays without the need of
changing the value of that address (the address of multi[0][0] as
given by the symbol "multi").

->================================================================
CHAPTER 7: More on Multi-Dimensional Arrays

    In the previous chapter we noted that given

  #define ROWS 5
  #define COLS 10

  int multi[ROWS][COLS];

we can access individual elements of the array "multi" using
either:

    multi[row][col]    or  *(*(multi + row) + col)

To understand more fully what is going on, let us replace

    *(multi + row)    with   X   as in:

    *(X + col)

    Now, from this we see that X is like a pointer since the
expression is de-referenced and we know that col is an integer.
Here the arithmetic being used is of a special kind called
"pointer arithmetic" is being used.  That means that, since we
are talking about an integer array, the address pointed to by
(i.e. value of)   X + col + 1   must be greater than the address
X + col   by and amount equal to sizeof(int).

    Since we know the memory layout for 2 dimensional arrays, we
can determine that in the expression     multi + row    as used
above,   multi + row + 1    must increase by value an amount
equal to that needed to "point to" the next row, which in this
case would be an amount equal to     COLS * sizeof(int).

    That says that if the expression     *(*(multi + row) + col)
is to be evaluated correctly at run time, the compiler must
generate code which takes into consideration the value of COLS,
i.e. the 2nd dimension.  Because of the equivalence of the two
forms of expression, this is true whether we are using the
pointer expression as here or the array expression
multi[row][col].

Thus, to evaluate either expression, a total of 5 values must be
known:

    1)  The address of the first element of the array, which is
        returned by the expression  "multi", i.e. the name of the
        array.

    2)  The size of the type of the elements of the array,  in
        this case sizeof(int).

    3)  The 2nd dimension of the array

    4)  The specific index value for the first dimension, "row"
        in this case.

    5)  The specific index value for the second dimension, "col"
        in this case.

    Given all of that, consider the problem of designing a
function to manipulate the element values of a previously
declared array. For example, one which would set all the elements
of the array "multi" to the value 1.

   void set_value(int m_array[][COLS])
   {
     int row, col;
     for(row = 0; row < ROWS; row++)
     {
       for(col = 0; col < COLS; col++)
       {
          m_array[row][col] = 1;
       }
     }
   }

And to call this function we would then use:

    set_value(multi);

    Now, within the function we have used the values #defined by
ROWS and COLS which set the limits on the for loops. But, these
#defines are just constants as far as the compiler is concerned,
i.e. there is nothing to connect them to the array size within
the function.  row and col are local variables, of course.  The
formal parameter definition informs the compiler that we are
talking about an integer array.  We really don't need the first
dimension and, as will be seen later, there are occasions where
we would prefer not to define it within the parameter definition
so, out of habit or consistency, I have not used it here.  But,
the second dimension _must_ be used as has been shown in the
expression for the parameter.  The reason is that it is needed in
the evaluation of   m_array[row][col]   as has been described.
The reason is that while the parameter defines the data type (int
in this case) and the automatic variables for row and column are
defined in the for loops, only one value can be passed using a
single parameter.  In this case, that is the value of "multi" as
noted in the call statement, i.e. the address of the first
element, often referred to as a pointer to the array. Thus, the
only way we have of informing the compiler of the 2nd dimension
is by explicitly including it in the parameter definition.

    In fact, in general all dimensions of higher order than one
are needed when dealing with multi-dimensional arrays.  That is
if we are talking about 3 dimensional arrays, the 2nd _and_ 3rd
dimension must be specified in the parameter definition.

->================================================================
CHAPTER 8: Pointers to Arrays

    Pointers, of course, can be "pointed at" any type of data
object, including arrays.  While that was evident when we
discussed program 3.1, it is important to expand on how we do
this when it comes to multi-dimensional arrays.

    To review, in Chapter 2 we stated that given an array of
integers we could point an integer pointer at that array using:

    int *ptr;

    ptr = &my_array[0];       /* point our pointer at the first
                                 integer in our array */

As we stated there, the type of the pointer variable must match
the type of the first element of the array.

    In addition, we can use a pointer as a formal parameter of a
function which is designed to manipulate an array.  e.g.

    Given:

      int array[3] = {'1', '5', '7'};

      void a_func(int *p);

we can pass the address of the array to the function by making
the call

        a_func(array);

This kind of code promotes the mis-conception that pointers and
arrays are the same thing.  Of course, if you have followed this
text carefully up to this point you know the difference between a
pointer and an array.  The function would be better written (in
terms of clarity) as       a_func(int p[]);     Note that here
we need not include the dimension since what we are passing is
the address of the array, not the array itself.

    We now turn to the problem of the 2 dimensional array.  As
stated in the last chapter, C interprets a 2 dimensional array as
an array of one dimensional arrays.  That being the case, the
first element of a 2 dimensional array of integers is a one
dimensional array of integers.  And a pointer to a two
dimensional array of integers must be a pointer to that data
type.  One way of accomplishing this is through the use of the
keyword "typedef".  typedef assigns a new name to a specified
data type.  For example:

   typedef unsigned char byte;

provides the name "byte" to mean type "unsigned char".  Hence

  byte b[10];     would be an array of unsigned characters.

Note that in the typedef declaration, the word "byte" has
replaced that which would normally be the name of our unsigned
char.  That is, the rule for using typedef is that the new name
for the data type is the name used in the definition of the data
type.  Thus in:

    typedef int Array[10];

Array becomes a data type for an array of 10 integers.  i.e.

    Array my_arr;

declares my_arr as an array of 10 integers and

    Array arr2d[5];

makes arr2d an array of 5 arrays of 10 integers each.

    Also note that      Array *p1d;    makes p1d a pointer to an
array of 10 integers.  Because *p1d points to the same type as
arr2d, assigning the address of the two dimensional array arr2d to
p1d, the pointer to a one dimensional array of 10 integers is
acceptable.  i.e.       p1d = &arr2d[0];     or   p1d = arr2d;
are both correct.

    Since the data type we use for our pointer is an array of 10
integers we would expect that incrementing p1d by 1 would change
its value by 10*sizeof(int), which it does.  That is sizeof(*p1d)
is 20.  You can prove this to yourself by writing and running a
simple short program.

    Now, while using typedef makes things clearer for the reader
and easier on the programmer, it is not really necessary.  What
we need is a way of declaring a pointer like p1d without the need
of the typedef keyword.  It turns out that this can be done and
that   int (*p1d)[10];  is the proper declaration, i.e. p1d here
is a pointer to an array of 10 integers just as it was under the
declaration using the Array type.  Note that this is different
than    int *p1d[10];     which would make p1d the name of an
array of 10 pointers to type int.

->================================================================
CHAPTER 9: Pointers and Dynamic Allocation of Memory

    There are times when it is convenient to allocate memory at
run time using malloc(), calloc(), or other allocation functions.
Using this approach permits postponing the decision on the size
of the memory block need to store an array, for example, until
run time.  Or it permits using a section of memory for the
storage of an array of integers at one point in time, and then
when that memory is no longer needed it can be freed up for other
uses, such as the storage of an array of structures.

    When memory is allocated, the allocating function (such as
malloc(), calloc(), etc.) returns a pointer.  The type of this
pointer depends on whether you are using an older K&R compiler or
the newer ANSI type compiler.  With the older compiler the type
of the returned pointer is char, with the ANSI compiler it is
void.

    If you are using an older compiler, and you want to allocate
memory for an array  of integers you will have to cast the char
pointer returned to an integer pointer.  For example, to allocate
space for 10 integers we might write:

    int *iptr;
    iptr = (int *)malloc(10 * sizeof(int));
    if(iptr == NULL)
    { .. ERROR ROUTINE GOES HERE .. }

    If you are using an ANSI compliant compiler, malloc() returns
a void pointer and since a void pointer can be assigned to a
pointer variable of any object type, the (int *) cast shown above
is not needed.  The array dimension can be determined at run time
and is not needed at compile time.  That is, the "10" above could
be a variable read in from a data file or keyboard, or calculated
based on some need, at run time.

    Because of the equivalence between array and pointer
notation, once iptr has been assigned as above, one can use the
array notation.  For example, one could write:

    int k;
    for(k = 0; k < 10; k++
       iptr[k] = 2;

to set the values of all elements to 2.

    Even with a reasonably good understanding of pointers and
arrays, one place the newcomer to C is likely to stumble at first
is in the dynamic allocation of multi-dimensional arrays.  In
general, we would like to be able to access elements of such
arrays using array notation, not pointer notation, wherever
possible. Depending on the application we may or may not know
both dimensions at compile time.  This leads to a variety of ways
to go about our task.

    As we have seen, when dynamically allocating a one
dimensional array the dimension can be determined at run time.
Now, when using dynamic allocation of higher order arrays, we
never need to know the first dimension at compile time.  Whether
we need to know the higher dimensions depends on how we go about
writing the code.  Here I will discuss various methods of
dynamically allocating room for 2 dimensional arrays of integers.

    First we will consider cases where the 2nd dimension is known
at compile time.

METHOD 1:

    One way of dealing with the problem is through the use of the
"typedef" keyword.  To allocate a 2 dimensional array of integers
recall that the following two notations result in the same object
code being generated:

multi[row][col] = 1;     *(*(multi + row) + col) = 1;

    It is also true that the following two notations generate the
same code:

    multi[row]            *(multi + row)

    Since the one on the right must evaluate to a pointer, the
array notation on the left must also evaluate to a pointer.  In
fact multi[0] will return a pointer to the first integer in the
first row, multi[1] a pointer to the first integer of the second
row, etc.  Actually, multi[n] evaluates to a pointer to that
array of integers which makes up the n-th row of our 2
dimensional array. That is, multi can be thought of as an array
of arrays and multi[n] as a pointer to the n-th array of this
array of arrays.  Here the word "pointer" is being used
to represent an address value.  While such usage is common in the
literature, when reading such statements one must be careful to
distinguish between the constant address of an array and a
variable pointer which is a data object in itself.

Consider now:
--------------- Program 9.1 --------------------------------
/* Program 9.1 from PTRTUT02.TXT   12/14/95 */

#include <stdio.h>
#include <alloc.h>

#define COLS 5

typedef int RowArray[COLS];
RowArray *rptr;

int main(void)
{
  int nrows = 10;
  int row, col;
  rptr = malloc(nrows * COLS * sizeof(int));
  for(row = 0; row < nrows; row++)
   for(col = 0; col < COLS; col++)
   {
     rptr[row][col] = 17;
   }
   return 0;
}
------------- End of Prog. 9.1 --------------------------------
    Here I have assumed an ANSI compiler so a cast on the void
pointer returned by malloc() is not required.  If you are using
an older K&R compiler you will have to cast using:

    rptr = (RowArray *)malloc(.... etc.

    Using this approach, "rptr" has all the characteristics of an
array name and array notation may be used throughout the rest of
the program.  That also means that if you intend to write a
function to modify the array contents, you must use COLS as a
part of the formal parameter in that function, just as we did
when discussing the passing of two dimensional arrays to a
function.

METHOD 2:

    In the METHOD 1 above, rptr turned out to be a pointer to
type "one dimensional array of COLS integers".  It turns out that
there is syntax which can be used for this type without the need
of typedef.  If we write:

int (char *xptr)[COLS];

the variable xptr will have all the same characteristics as the
variable rptr in METHOD 1 above, and we need not use the
"typedef" keyword.  Here xptr is a pointer to an array of
integers and the size of that array is given by the #defined
COLS.  The parenthesis placement makes the pointer notation
predominate, even though the array notation has higher
precedence.  i.e. had we written

int char *xptr[COLS];

we would have defined xptr as an array of pointers holding the
number of pointers equal to that #defined by COLS.  Which is not
the same thing at all.  However, arrays of pointers have their
use in the dynamic allocation of two dimensional arrays, as will
be seen in the next 2 methods.

METHOD 3:

   Consider the case where we do not know the number of elements
in each row at compile time, i.e. both the number of rows and
number of columns must be determined at run time.  One way of
doing this would be to create an array of pointers to type int
and then allocate space for each row and point these pointers at
each row.  Consider:
-------------- Program 9.2 ------------------------------------
/* Program 9.2 from PTRTUT02.TXT   12/14/95 */

#include <stdio.h>
#include <stdlib.h>

int main(void)
{
  int nrows = 5;     /* Both nrows and ncols could be evaluated */
  int ncols = 10;    /* or read in at run time */
  int row;
  int **rowptr;
  rowptr = malloc(nrows * sizeof(int *));
  if(rowptr == NULL)
  {
    puts("\nFailure to allocate room for row pointers.\n");
    exit(0);
  }
  printf("\n\n\nIndex   Pointer(hex)   Pointer(dec)   Diff.(dec)");

  for(row = 0; row < nrows; row++)
  {
    rowptr[row] = malloc(ncols * sizeof(int));
    if(rowptr[row] == NULL)
    {
      printf("\nFailure to allocate for row[%d]\n",row);
      exit(0);
    }
    printf("\n%d         %p         %d", row, rowptr[row], rowptr[row]);
    if(row > 0)
      printf("              %d",(int)(rowptr[row] - rowptr[row-1]));
  }
  return 0;
}
--------------- End 9.2 ------------------------------------

    In the above code rowptr is a pointer to pointer to type int.
In this case it points to the first element of an array of
pointers to type int.  Consider the number of calls to malloc():

  To get the array of pointers             1     call
  To get space for the rows                5     calls
                                        -----
                   Total                   6     calls

    If you choose to use this approach note that while you can
use the array notation to access individual elements of the
array, e.g.  rowptr[row][col] = 17;, it does not mean that the
data in the "two dimensional array" is contiguous in memory.

    But, you can use the array notation just as if it were a
continuous block of memory.  For example, you can write:

  rowptr[row][col] = 176;

just as if rowptr were the name of a two dimensional array
created at compile time.  Of course 'row' and 'col' must be
within the bounds of the array you have created, just as with an
array created at compile time.

    If it is desired to have a contiguous block of memory
dedicated to the storage of the elements in the array it can be
done as follows:

METHOD 4:

    In this method we allocate a block of memory to hold the
whole array first.  We then create an array of pointers to point
to each row.  Thus even though the array of pointers is being
used, the actual array in memory is contiguous.  The code looks
like this:

----------------- Program 9.3 -----------------------------------
/* Program 9.3 from PTRTUT02.TXT   12/14/95 */

#include <stdio.h>
#include <stdlib.h>
#include <conio.h>

int main(void)
{
  int **rptr;
  int *aptr;
  int *testptr;
  int k;
  int nrows = 5;     /* Both nrows and ncols could be evaluated */
  int ncols = 10;    /* or read in at run time */
  int row, col;
     /* we now allocate the memory for the array */
  aptr = malloc(nrows * ncols * sizeof(int *));
  if(aptr == NULL)
  {
    puts("\nFailure to allocate room for the array");
    exit(0);
  }
     /* next we allocate room for the pointers to the rows */
  rptr = malloc(nrows * sizeof(int *));
  if(rptr == NULL)
  {
    puts("\nFailure to allocate room for pointers");
    exit(0);
  }
     /* and now we 'point' the pointers */
  clrscr();
  for(k = 0; k < nrows; k++)
  {
     rptr[k] = aptr + (k * ncols);
  }
  printf("\n\n\nIndex   Pointer(hex)   Pointer(dec)   Diff.(dec)");

  for(row = 0; row < nrows; row++)
  {
    printf("\n%d         %p         %d", row, rptr[row], rptr[row]);
    if(row > 0)
      printf("              %d",(int)(rptr[row] - rptr[row-1]));
  }
  for(row = 0; row < nrows; row++)
  {
     for(col = 0; col < ncols; col++)
     {
       rptr[row][col] = row + col;
       printf("%d ", rptr[row][col]);
     }
     putchar('\n');
  }
  puts("\n\n\n");

     /* and here we illustrate that we are, in fact, dealing with
        a 2 dimensional array in a _contiguous_ block of memory. */

  testptr = aptr;
  for(row = 0; row < nrows; row++)
  {
    for(col = 0; col < ncols; col++)
    {
      printf("%d ", *(testptr++));
    }
    putchar('\n');
  }
  return 0;
}

------------- End Program 9.3 -----------------

Consider again, the number of calls to malloc()

    To get room for the array itself      1      call
    To get room for the array of ptrs     1      call
                                        ----
                         Total            2      calls

    Now, each call to malloc() creates additional space overhead
since malloc() is generally implemented by the operating system
forming a linked list which contains data concerning the size of
the block.  But, more importantly, with large arrays (several
hundred rows) keeping track of what needs to be freed when the
time comes can be more cumbersome.  This, combined with he
contiguousness of the data block which permits initialization to
all zeroes using memset() would seem to make the second
alternative the preferred one.


    As a final example on multidimensional arrays we will
illustrate the dynamic allocation of a three dimensional array.
This example will illustrate one more thing to watch when doing
this kind of allocation.  For reasons cited above we will use the
approach outlined in alternative two.  Consider the following
code:

------------------- Program 9.4 -------------------------------------
/* Program 9.4 from PTRTUT02.TXT   12/14/95 */

#include <stdio.h>
#include <stdlib.h>
#include <mem.h>
#include <conio.h>

int X_DIM=16;
int Y_DIM=8;
int Z_DIM=4;

int main(void)
{
  char ***space;
  char ***Arr3D;
  int y, z;
  ptrdiff_t diff;

  /* first we set aside space for the array itself */

  space = malloc(X_DIM * Y_DIM * Z_DIM * sizeof(char));

  /* next we allocate space of an array of pointers, each
     to eventually point to the first element of a
     2 dimensional array of pointers to pointers */

  Arr3D = malloc(Z_DIM * sizeof(char **));

  /* and for each of these we assign a pointer to a newly
     allocated array of pointers to a row */

  for(z = 0; z < Z_DIM; z++)
  {
     Arr3D[z] = malloc(Y_DIM * sizeof(char *));

     /* and for each space in this array we put a pointer to
        the first element of each row in the array space
        originally allocated */

     for(y = 0; y < Y_DIM; y++)
     {
       Arr3D[z][y] = ((char *)space + (z*(X_DIM * Y_DIM) + y*X_DIM));
     }
  }

  /* And, now we check each address in our 3D array to see if
     the indexing of the Arr3d pointer leads through in a
     continuous manner */

  for(z = 0; z < Z_DIM; z++)
  {
    printf("Location of array %d is %p\n", z, *Arr3D[z]);
    for( y = 0; y < Y_DIM; y++)
    {
       printf("  Array %d and Row %d starts at %p", z, y, Arr3D[z][y]);
       diff = Arr3D[z][y] - (char *)space;
       printf("    diff = %d  ",diff);
       printf(" z = %d  y = %d\n", z, y);
    }
    getch();
  }
  return 0;
}


------------------- End of Prog. 9.4 ----------------------------
    If you have followed this tutorial up to this point you
should have no problem deciphering the above on the basis of the
comments alone.  There is one line that deserves a bit of special
attention however.  It reads:

  Arr3D[z][y] = ((char *)space + (z*(X_DIM * Y_DIM) + y*X_DIM));

    Note that here "space" is cast to a character pointer, which
is the same type as Arr3D[z][y].  A thing to be careful of,
however, is where that cast is made.  If the cast were made
outside the overall parenthesis as in...

  Arr3D[z][y] = (char *)(space + (z*(X_DIM * Y_DIM) + y*X_DIM));

the code fails.  The reason is that the cast, in this case, is
not so much to make the types on each side of the assignment
operator match, as it is to make the pointer arithmetic work.
Recall that when dealing with pointer arithmetic in something
like:

int *ptr;
ptr = ptr + 1;

the second line increments the pointer by sizeof(int), which is 2
on MS-DOS machines.  Now looking at the mentioned line, it should
be obvious that

         (z*(X_DIM * Y_DIM) + y*X_DIM))

calculates the number of array elements This will turn out to be
an arithmetic constant after the calculation.  Now since we are
dealing with an array of characters the result of the pointer
arithmetic which adds this value to the pointer to the start of
the array should yield a value equal to the pointer value plus
this constant.  Were we using an int data type, i.e. casting our
"space" pointer to (int *), the actual value by which the pointer
would be incremented would be the calculated value times
sizeof(int).

->==================================================================
CHAPTER 10: Pointers to Functions

    Up to this point we have been discussing pointers to data
objects.  C also permits the declaration of pointers to
functions.  Pointers to functions have a variety of uses and some
of them will be discussed here.

    Consider the following real problem.  You want to write a
function that is capable of sorting virtually any collection of
data that can be stored in an array.  This might be an array of
strings, or integers, or floats, or even structures.  The sorting
algorithm can be the same for all.  For example, it could be a
simple bubble sort algorithm, or the more complex shell or quick
sort algorithm.  We'll use a simple bubble sort for demonstration
purposes.

    Sedgewick [1] has described the bubble sort using C code by
setting up a function which when passed a pointer to the array
would sort it.  If we call that function bubble(), a sort program
is described by bubble_1.c, which follows:

/*-------------------- bubble_1.c --------------------*/
/* Program bubble_1.c from PTRTUT02.TXT   12/14/95 */

#include <stdio.h>

int arr[10] = { 3,6,1,2,3,8,4,1,7,2};

void bubble(int a[], int N);

int main(void)
{
  int i;
  putchar('\n');
  for(i = 0; i < 10; i++)
  {
    printf("%d ", arr[i]);
  }
  bubble(arr,10);
  putchar('\n');
  for(i = 0; i < 10; i++)
  {
    printf("%d ", arr[i]);
  }
  return 0;
}

void bubble(int a[], int N)
{
 int i, j, t;
 for(i = N-1; i >= 0; i--)
   for(j = 1; j <= i; j++)
     if(a[j-1] > a[j])
       {
         t = a[j-1];
         a[j-1] = a[j];
         a[j] = t;
       }
}
/*---------------------- end bubble_1.c -----------------------*/

The bubble sort is one of the simpler sorts.  The algorithm scans
the array from the second to the last element comparing each
element with the one which precedes it. If the one that precedes
it is larger than the current element, the two are swapped so the
larger one is closer to the end of the array.  On the first pass,
this results in the largest element ending up at the end of the
array.  The array is now limited to all elements except the last
and the process repeated.  This puts the next largest element at
a point preceding the largest element.  The process is repeated
for a number of times equal to the number of elements minus 1.
The end result is a sorted array.

     Here our function is designed to sort an array of integers.
Thus in line 1 we are comparing integers and in lines 2 through 4
we are using temporary integer storage to store integers.  What
we want to do now is see if we can convert this code so we can
use any data type, i.e. not be restricted to integers.

    At the same time we don't want to have to analyze our
algorithm and the code associated with it each time we use it.
We start by removing the comparison from within the function
bubble() so as to make it relatively easy to modify the
comparison function without having to re-write portions related
to the actual algorithm.  This results in bubble_2.c:

/*---------------------- bubble_2.c -------------------------*/
/* Program bubble_2.c from PTRTUT02.TXT   12/14/95 */

   /* Separating the comparison function */

#include <stdio.h>

int arr[10] = { 3,6,1,2,3,8,4,1,7,2};

void bubble(int a[], int N);
int compare(int m, int n);

int main(void)
{
  int i;
  putchar('\n');
  for(i = 0; i < 10; i++)
  {
    printf("%d ", arr[i]);
  }
  bubble(arr,10);
  putchar('\n');
  for(i = 0; i < 10; i++)
  {
    printf("%d ", arr[i]);
  }
  return 0;
}

void bubble(int a[], int N)
{
 int i, j, t;
 for(i = N-1; i >= 0; i--)
   for(j = 1; j <= i; j++)
     if (compare(a[j-1], a[j]))
       {
         t = a[j-1];
         a[j-1] = a[j];
         a[j] = t;
       }
}

int compare(int m, int n)
{
  return (m > n);
}
/*--------------------- end of bubble_2.c -----------------------*/

If our goal is to make our sort routine data type independent,
one way of doing this is to use pointers to type void to point to
the data instead of using the integer data type.  As a start in
that direction let's modify a few things in the above so that
pointers can be used.  To begin with, we'll stick with pointers
to type integer.

/*----------------------- bubble_3.c -------------------------*/
/* Program bubble_3.c from PTRTUT02.TXT   12/14/95 */

#include <stdio.h>

int arr[10] = { 3,6,1,2,3,8,4,1,7,2};

void bubble(int *p, int N);
int compare(int *m, int *n);

int main(void)
{
  int i;
  putchar('\n');
  for(i = 0; i < 10; i++)
  {
    printf("%d ", arr[i]);
  }
  bubble(arr,10);
  putchar('\n');
  for(i = 0; i < 10; i++)
  {
    printf("%d ", arr[i]);
  }
  return 0;
}

void bubble(int *p, int N)
{
 int i, j, t;
 for(i = N-1; i >= 0; i--)
   for(j = 1; j <= i; j++)
     if (compare(&p[j-1], &p[j]))
       {
         t = p[j-1];
         p[j-1] = p[j];
         p[j] = t;
       }
}

int compare(int *m, int *n)
{
  return (*m > *n);
}
/*------------------ end of bubble3.c -------------------------*/

Note the changes. We are now passing a pointer to an integer (or
array of integers) to bubble().  And from within bubble we are
passing pointers to the elements of the array that we want to
compare to our comparison function.  And, of course we are
dereferencing these pointer in our compare() function in order to
make the actual comparison.  Our next step will be to convert the
pointers in bubble() to pointers to type void so that that
function will become more type insensitive.  This is shown in
bubble_4.

/*------------------ bubble_4.c ----------------------------*/
/* Program bubble_4.c from PTRTUT02.TXT   12/14/95 */

#include <stdio.h>

int arr[10] = { 3,6,1,2,3,8,4,1,7,2};

void bubble(int *p, int N);
int compare(void *m, void *n);

int main(void)
{
  int i;
  putchar('\n');
  for(i = 0; i < 10; i++)
  {
    printf("%d ", arr[i]);
  }
  bubble(arr,10);
  putchar('\n');
  for(i = 0; i < 10; i++)
  {
    printf("%d ", arr[i]);
  }
  return 0;
}

void bubble(int *p, int N)
{
 int i, j, t;
 for(i = N-1; i >= 0; i--)
   for(j = 1; j <= i; j++)
     if (compare((void *)&p[j-1], (void *)&p[j]))
       {
         t = p[j-1];
         p[j-1] = p[j];
         p[j] = t;
       }
}

int compare(void *m, void *n)
{
  int *m1, *n1;
  m1 = (int *)m;
  n1 = (int *)n;
  return (*m1 > *n1);
}
/*------------------ end of bubble_4.c ---------------------*/

Note that, in doing this, in compare() we had to introduce the
casting of the void pointer types passed to the actual type being
sorted.  But, as we'll see later that's okay.  And since what is
being passed to bubble() is still a pointer to an array of
integers, we had to cast these pointers to void pointers when we
passed them as parameters in our call to compare().

We now address the problem of what we pass to bubble().  We want
to make the first parameter of that function a void pointer also.
But, that means that within bubble() we need to do something
about the variable t, which is currently an integer.  Also, where
we use t = p[j-1];  the type of p[j-1] needs to be known in order
to know how many bytes to copy to the variable t (or whatever we
replace t with).

Currently, in bubble_4.c, knowledge within bubble() as to the
type of the data being sorted (and hence the size of each
individual element) is obtained from the fact that the first
parameter is a pointer to type integer.  If we are going to be
able to use bubble() to sort any type of data, we need to make
that pointer a pointer to type void.  But, in doing so we are
going to lose information concerning the size of individual
elements within the array.  So, in bubble_5.c we will add a
separate parameter to handle this size information.

These changes, from bubble4.c to bubble5.c are, perhaps, a bit
more extensive than those we have made in the past.  So, compare
the two modules carefully for differences.

/*---------------------- bubble5.c ---------------------------*/
/* Program bubble_5.c from PTRTUT02.TXT   12/14/95 */

#include <stdio.h>
#include <string.h>

long arr[10] = { 3,6,1,2,3,8,4,1,7,2};

void bubble(void *p, size_t width, int N);
int compare(void *m, void *n);

int main(void)
{
  int i;
  putchar('\n');
  for(i = 0; i < 10; i++)
  {
    printf("%d ", arr[i]);
  }
  bubble(arr, sizeof(long), 10);
  putchar('\n');
  for(i = 0; i < 10; i++)
  {
    printf("%d ", arr[i]);
  }
  return 0;
}

void bubble(void *p, size_t width, int N)
{
 int i, j;
 unsigned char buf[4];
 unsigned char *bp = p;
 for(i = N-1; i >= 0; i--)
   for(j = 1; j <= i; j++)
     if (compare((void *)(bp + width*(j-1)), (void *)(bp + j*width)))  /* 1 */
       {
/*         t = p[j-1];   */
        memcpy(buf, bp + width*(j-1), width);
/*         p[j-1] = p[j];   */
        memcpy(bp + width*(j-1), bp + j*width , width);
/*         p[j] = t;   */
        memcpy(bp + j*width, buf, width);
       }

}

int compare(void *m, void *n)
{
  long *m1, *n1;
  m1 = (long *)m;
  n1 = (long *)n;
  return (*m1 > *n1);
}
/*--------------------- end of bubble5.c ---------------------*/

Note that I have changed the data type of the array from int to
long to illustrate the changes needed in the compare() function.
Within bubble I've done away with the variable t (which we would
have had to change from type int to type long).  I have added a
buffer of size 4 unsigned characters, which is the size needed to
hold a long (this will change again in future modifications to
this code).  The unsigned character pointer *bp is used to point
to the base of the array to be sorted, i.e. to the first element
of that array.

We also had to modify what we passed to compare(), and how we do
the swapping of elements that the comparison indicates need
swapping.  Use of memcpy() and pointer notation instead of array
notation work towards this reduction in type sensitivity.

Again, making a careful comparison of bubble5.c with bubble4.c
can result in improved understanding of what is happening and
why.

We move now to bubble6.c where we use the same function bubble()
that we used in bubble5.c to sort strings instead of long
integers.  Of course we have to change the comparison function
since the means by which strings are compared is different from
that by which long integers are compared.  And,in bubble6.c we
have deleted the lines within bubble() that were commented out in
bubble5.c.

/*--------------------- bubble6.c ---------------------*/
/* Program bubble_6.c from PTRTUT02.TXT   12/14/95 */

#include <stdio.h>
#include <string.h>

#define MAX_BUF 256

long arr[10] = { 3,6,1,2,3,8,4,1,7,2};

char arr2[5][20] = {  "Mickey Mouse",
                      "Donald Duck",
                      "Minnie Mouse",
                      "Goofy",
                      "Ted Jensen" };

void bubble(void *p, int width, int N);
int compare(void *m, void *n);

int main(void)
{
  int i;
  putchar('\n');
  for(i = 0; i < 5; i++)
  {
    printf("%s\n", arr2[i]);
  }
  bubble(arr2, 20, 5);
  putchar('\n\n');
  for(i = 0; i < 5; i++)
  {
    printf("%s\n", arr2[i]);
  }
  return 0;
}

void bubble(void *p, int width, int N)
     {
      int i, j, k;
      unsigned char buf[MAX_BUF];
      unsigned char *bp = p;
      for(i = N-1; i >= 0; i--)
        for(j = 1; j <= i; j++)
        {
          k = compare((void *)(bp + width*(j-1)), (void *)(bp + j*width));
          if (k > 0)
            {
             memcpy(buf, bp + width*(j-1), width);
             memcpy(bp + width*(j-1), bp + j*width , width);
             memcpy(bp + j*width, buf, width);
            }
        }
    }

int compare(void *m, void *n)
{
   char *m1 = m;
   char *n1 = n;
   return (strcmp(m1,n1));
}
/*------------------- end of bubble6.c ---------------------*/

But, the fact that bubble() was unchanged from that used in
bubble5.c indicates that that function is capable of sorting a
wide variety of data types.  What is left to do is to pass to
bubble() the name of the comparison function we want to use so
that it can be truly universal.  Just as the name of an array is
the address of the first element of the array in the data
segment, the name of a function decays into the address of that
function in the code segment.  Thus we need to use a pointer to a
function.  In this case the comparison function.

Pointers to functions must match the functions pointed to in the
number and types of the parameters and the type of the return
value.  In our case, we declare our function pointer as:

   int (*fptr)(const void *p1, const void *p2);

Note that were we to write:

    int *fptr(const void *p1, const void *p2);

we would have a function prototype for a function which returned
a pointer to type int.  That is because in C the parenthesis ()
operator have a higher precedence than the pointer * operator.
By putting the parenthesis around the string (*fptr) we indicate
that we are declaring a function pointer.

We now modify our declaration of bubble() by adding, as its 4th
parameter, a function pointer of the proper type.  It's function
prototype becomes:

  void bubble(void *p, int width, int N,
             int(*fptr)(const void *, const void *));

When we call the bubble(), we insert the name of the comparison
function that we want to use.  bubble7.c illustrate how this
approach permits the use of the same bubble() function for
sorting different types of data.

/*------------------- bubble7.c ------------------*/
/* Program bubble_7.c from PTRTUT02.TXT   12/14/95 */

#include <stdio.h>
#include <string.h>

#define MAX_BUF 256

long arr[10] = { 3,6,1,2,3,8,4,1,7,2};

char arr2[5][20] = {  "Mickey Mouse",
                      "Donald Duck",
                      "Minnie Mouse",
                      "Goofy",
                      "Ted Jensen" };

void bubble(void *p, int width, int N,
             int(*fptr)(const void *, const void *));
int compare_string(const void *m, const void *n);
int compare_long(const void *m, const void *n);
int main(void)
{
  int i;
  puts("\nBefore Sorting:\n");
  for(i = 0; i < 10; i++)               /* show the long ints */
  {
    printf("%ld ",arr[i]);
  }
  puts("\n");
  for(i = 0; i < 5; i++)                  /* show the strings */
  {
    printf("%s\n", arr2[i]);
  }
  bubble(arr, 4, 10, compare_long);          /* sort the longs */
  bubble(arr2, 20, 5, compare_string);     /* sort the strings */
  puts("\n\nAfter Sorting:\n");
  for(i = 0; i < 10; i++)             /* show the sorted longs */
  {
    printf("%d ",arr[i]);
  }
  puts("\n");
  for(i = 0; i < 5; i++)            /* show the sorted strings */
  {
    printf("%s\n", arr2[i]);
  }
  return 0;
}

void bubble(void *p, int width, int N,
             int(*fptr)(const void *, const void *))
     {
      int i, j, k;
      unsigned char buf[MAX_BUF];
      unsigned char *bp = p;
      for(i = N-1; i >= 0; i--)
        for(j = 1; j <= i; j++)
        {
          k = fptr((void *)(bp + width*(j-1)), (void *)(bp + j*width));
          if (k > 0)
            {
             memcpy(buf, bp + width*(j-1), width);
             memcpy(bp + width*(j-1), bp + j*width , width);
             memcpy(bp + j*width, buf, width);
            }
        }
    }

int compare_string(const void *m, const void *n)
{
   char *m1 = (char *)m;
   char *n1 = (char *)n;
   return (strcmp(m1,n1));
}

int compare_long(const void *m, const void *n)
{
  long *m1, *n1;
  m1 = (long *)m;
  n1 = (long *)n;
  return (*m1 > *n1);
}
/*----------------- end of bubble7.c -----------------*/

References for Chapter 10

  [1]  "Algorithms in C"
       Robert Sedgewick
       Addison-Wesley
       ISBN 0-201-51425-7

->==============================================================
EPILOG

    I have written the preceding material to provide an
introduction to pointers for newcomers to C.  In C, the more one
understands about pointers the greater flexibility one has in the
writing of code.  The above expands on my first effort at this
which was entitled ptr_help.txt and found in an early version of
Bob Stout's collection of C code SNIPPETS.  The content in this
version is the same as that in PTRTUTOT.ZIP included in
SNIP9510.ZIP with some minor typo corrections.

I am always ready to accept constructive criticism on this
material, or review requests for the addition of other relevant
material.  Therefore, if you have questions, comments,
criticisms, etc. concerning that which has been presented, I
would greatly appreciate your contacting me using one of the mail
addresses cited in the Introduction.