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1992-07-03
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11KB
Date: Tue Jun 16 1992 11:05:08
From: Ted Jensen
To: Michael Halcrow
Subj: Pointers & things
Attr:
international C echo
-------------------------------
Michael-
In reading your messages to Mike Dalsanto, Robert Place, and
All I see you need some help on pointers, arrays, and strings.
Perhaps the following will help:
A string is a one dimensional array of characters. A one
dimensional array of a given type is contiguous in memory with
each element following in memory the one preceding it.
A pointer is a variable much like any other variable. For
example, let's consider a more easily understood variable, the
integer:
int k;
defines an integer and reserves x bytes of space in memory
for its storage. The number of bytes reserved depends on the
system and/or compiler. For DOS systems an integer requires 2
bytes of storage. Defining an integer, as above, does not give
it a value. If it is defined as a global variable, it will be
intialized it to zero. If it is defined as an auto variable
(i.e. within a function) its value is indeterminate (i.e. a
random value). Before use, the safest thing to do with any
variable is to intialize it to some value (though in this case
most (all?) compilers will intialize it to zero if it is global.
Now, let's look at a pointer to the type integer which can be
defined as:
int *int_ptr;
defines a pointer and reserves x bytes of space in memory for
its storage. The number of bytes reserved depends on the system
and/or compiler. But, fundamentally, since a pointer is designed
to hold an address in memory, it must reserve the number of bytes
necessary to contain such an address. Defining a pointer, as
above, does not give it a value. If it is defined as a global
variable it will be initialzed to a NULL, i.e. the address it
will contain (which is the same as saying "the address to which
it points to") will be set to a value of NULL which is #defined
in <stdio.h>. In DOS systems, NULL is defined as the address of
the first byte of the data segment. In Borland products, the
first few bytes of the data segment contain a copyright notice.
Before use, a pointer _must_ be "pointed at something", i.e.
assigned a value equal to the address of some data item. More on
this later.
The key to understanding pointers is to remember that they
are variables which contain addresses and that when we say a
pointer "points to something", we mean that it contains the
address of another variable or some other point in memory (such
as the video screen buffer).
Suppose now, that I want to "point" my integer pointer,
int_ptr, at my integer k. We do this as follows:
int_ptr = &k;
Why use the '&'? Well, let's say k = 7;, were we to write
"int_ptr = k;" we would be "pointing our pointer" at memory
location 7. But what we want is to "point our pointer" to the
memory location where the value of k is stored, which is an
entirely different thing. By preceding the 'k' with the '&' we
get the that address where the value of k is stored and point the
pointer where we intend.
Now, let's move on to arrays, and eventually arrays of
characters which, in some languages, are called strings. But
first, let's discuss an array of integers as in:
int int_array[5];
defines an array capable of holding 5 integers. We could then
intialize these with statement such as: int_array[0] = 4; ,
int_array[1] = 7 , etc. Or, we could initialize the array at the
same time as we define it with:
int int_array[5] = { 4, 7, 2, -2, 6 };
The 5 memory locations for these integers will be contiguous
in memory, i.e. if the 4 is stored at memory location 1000, it is
gauranteed that the 7 will be stored at memory location 1002
(recall an integer takes 2 bytes of storage) and the 2 will be
stored at 1004, etc.
Recall that we stated that &k returned the address where the
value of the integer k was stored. Similarly
&int_array[0] will return the memory location of the
first integer in the array. Thus, if we want to point our
pointer at the array (i.e. the first element of the array) we can
write:
int_ptr = &int_array[0];
However, with arrays, there is a second option. The name of our
array will also return the memory location of the first element.
This means that we can write:
int_ptr = int_array;
and get exactly the same result.
With that as background, let's now turn to "strings", (called
character arrays in C). Everything we said about integer arrays
can also be said about character arrays. Thus:
char c_array[20]; /* reserves space for 20 characters */
char *c_ptr; /* an (unitialized) pointer to a
character */
c_ptr = &c_array[0]; /* points the pointer at the first
char of the array ("string") */
c_ptr = c_array; /* does the same */
Note that &c_array is meaningless.
char c_array[] = { 'M', 'i', 'c', 'h', 'a', 'e', 'l', '\0' };
This last line intializes c_array to hold "Michael".
However, with character arrays, the above line can be written
using a sort of shorthand as:
char c_array[] = "Michael";
Note that in the later case, the space reserved will be 8
bytes and the 8th byte will automatically be set to '\0' so that
this syntax is, in fact, shorthand for the longer method
preceding it.
Now, let's to back to pointers and discuss them in context
with functions. Consider the following function:
void do_something(int k, int *p);
If you have studied functions you know that parameters are
passed by _value_. That is, if k = 7, what gets passed as a
parameter is the value 7 (as distinguished from the address of
k). Similarly, what gets passed for the second parameter is the
value of p. But, the value of p is an address. Thus, the
following three calls to do_something are equivalent:
int k = 7;
int int_arr[5] = { 1, 2, 3, 4, 5 };
int *pk;
int *pa;
pk = &k;
pa = int_arr;
do_something( 3, pk );
do_something( 3, &k );
Similarly, these three calls to our function
do_something( 3, pa);
do_something( 3, &int_arr[0]);
do_something( 3, int_arr);
are identical, i.e. the _value_ of the second parameter is the
same in all three cases. This frees the programmer to choose
that approach which he/she is more comfortable with.
Now, let's talk about "strings" some more. There is one more
kind of "string" that we have yet to discuss. It is called the
"literal string". A literal string is any string in your code
that appears in quotes. Thus, in all the following cases the
string "l_string" is a literal string.
char s_arr[] = "l_string";
fopen("my_file", "wb");
strcpy(s_arr, "l_string");
printf("This is a test");
During the compilation process, whenever and where ever the
compiler sees a literal string it a) reserves space in the data
segment to hold the string, b) puts the string within the quotes
in that space and terminates it with the '\0' character, and c)
replaces the literal string in the statement with a pointer
"pointing to" the memory location where it is stored.
Thus, for example, if the compiler sees:
strcpy(s_arr, "l_string");
it moves the characters "l_string\0" to a memory location in the
data segment, lets say it is at DS:0205. It then "modifies" the
statement to read:
strcpy(s_arr, DS:0205); And if s_arr is located at DS:0097,
it becomes:
strcpy(DS:0097, DS:0097);
Thus, if you look at the prototype for strcpy(), located in
<string.h>, you will see something like:
char *strcpy(char *p1, char *p2);
which indicates that this function should be passed "pointers".
But, now we know that it _really_ means that the function should
be passed the _values_ of these pointers, which are _addresses_.
And because we also recognize that the compiler replaces literal
strings with addresses, a statment such as:
strcpy(s_arr, "Michael");
begins to make some sense (at least I hope it does!).
Michael, I have just scratched the surface here. Other things
that need "going into" include how and why the compiler _always_
converts array notation to pointer notation as in converting:
a[n] to *(a + n);
Why are arrays always based at zero? i.e. why is the first
element of an array always referred to as array[0] and not
array[1]? Why does something like:
char *p = "This is a test";
initialize a string much the same as;
char arr[] = "This is a test"; but with different results.
There are dozens of other issues that you will learn as you gain
experience. In the long run, you will find that C is much more
powerful than BASIC. But, having the power makes it more
difficult to use. One can add, subtract, multiply or divide with
pencil and paper, and this is the way we all learn in the
beginning. A little later we learn there is a machine that we
can use to do this called a calculator, and this does things
faster. Further more, calculators come in various degrees of
complexity, including some that can be programmed. When using
the more complex calculators we must have a little more knowledge
and take a little more care to avoid errors.
As we move up this chain of experience we eventually come to
computers where, once again, we can still add, subtract,
multiply, and divide, but to do so (in a program) requires still
more knowledge and care. With each increase in power comes the
need for more knowledge and the attainment of more flexibility
(ever try and do wordprocessing on a four function calculator?).
If you want to use sharper knives to make your task easier, you
just have to be more careful not to cut yourself since the cut
will go a lot deeper!
At the point you are in your development of skills with C, I
_strongly_ recommend you purchase:
"The C Programming Language" (2nd edition)
Kernighan and Ritchie
Prentice Hall
This is the 2nd edition of the book that started it all. The
authors were the ones who developed the C language and the book,
often referred to as K&R or K&R2, is the _true_ bible for C
users. If you start at page 1, take it slow, and read it
carefully, it will really pay off and it won't be any time at all
before you are answering the questions here that you are now
asking!
Hope this helps! ... and good luck!