HPAVC/HPAVC CD-ROM.iso/CLESSONS.ZIP/LESSON2

Text File | 1985-12-28 | 15KB | 395 lines

.NT
 A NOTE ABOUT THE LESSONS in C
.b4-24
.R5C4
These were written while the author was ~Ilearning~N  the language and since
.R6C4
they  are  ~Ifree~N ( to  copy  and/or  distribute ) there  is  a money-back
.R7C4
guarantee on the accuracy of each and every statement in the lessons (!)
.R9C4
The  ~Idisplay~N  program was written ( in C ) in order to provide a vehicle
.R10C4
for displaying the lessons.
.R12C5
.B
P.J.Ponzo
.B
Dept. of Applied Math
.B
Univ. of Waterloo
.B
Ontario N2L 3G1
.K16,30
PonzoTUTOR
.WNT
   STRINGS 'N' THINGS

    When we declare ~b~Ichar x;~N we tell the C compiler that ~b~Ix~N is a
    ~b~Ichar~Nacter variable (so we may define ~b~Ix='Z';~N for example).

    But what if we wish to assign to ~b~Ix~N the ~Istring~N: 'hello' ?
    In order to identify a string (as opposed to a single character) we write:
    ~b~Ix="hello";~N  using the ~Idouble quote~N and we tell ~b~Iprintf()~N
    to print ~b~Ix~N using a ~b~Is~Ntring format, ~b~I%s~N:

~b~Imain() {~N
~b~Ichar x;~N                         declare ~b~Ichar x;~N as usual.
~b~Ix="hello";~N                      use ~Idouble quotes~N to define ~b~Ix~N.
~b~Iprintf("string x is %s",x);~N     use ~b~I%s~N to printf a ~b~Is~Ntring.
~b~I}~N
.W
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    The above program (when ~Icompiled~N and ~Irun~N) prints:

~r~Istring x is hello~N
.K19,60
  hello!
.WN


~b~Imain() {~N
~Vchar x;         ~N
~b~Ix="hello";~N                      use ~Idouble quotes~N to define ~b~Ix~N.
~b~Iprintf("string x is %s",x);~N     use ~b~I%s~N to printf a ~b~Is~Ntring.
~b~I}~N

.R9
  ~IThis declaration of a string ~b~Ix~N~I may NOT be acceptable to your C-compiler!~N
.b8-10



(It has the SAME format as the declaration of a single ~b~Ichar~Nacter variable!)

                            Lesson6 will explain .....
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patience!
.WNT
  Double Quotes and Strings : Single Quotes and Characters
    Consider the following:

~b~Imain() {~N
~b~Ichar x,y;~N                             ~b~Ix~N, ~b~Iy~N both ~b~Ichar~N.
~b~Ix="A";~N                                ~b~Ix~N is in ~Idouble~N quotes.
~b~Iy='A';~N                                ~b~Iy~N is in ~Isingle~N quotes.
~b~Iprintf("x is %s and y is %s",x,y);~N    BOTH printed as a ~b~I%s~Ntring!
~b~I}~N

    What do you think will be printed?
.W
.R15C1
~r~Ix is A and y is  ~N                     ~Ithe x-string is OK..but NOT the y!~N
.W
.R17C1
    If ~b~Ix~N is defined as a string (using the ~Idouble quotes~N) and
    is printed (via ~b~Iprintf()~N) using the ~b~I%s~Ntring format, then the
    printout is OK. BUT if ~b~Iy~N is defined as a ~Isingle character~N (using
    ~Isingle quotes~N) but is printed using the ~b~I%s~Ntring format then
    ~b~Iprintf()~N gets confused (..or maybe it's ~IWE~N who are confused!).
    ~w~BThis says something interesting about printf()!~N
.W
.N

    When we tell ~b~Iprintf()~N to print a ~b~I%s~Ntring (either ~b~Ix~N
    or ~b~Iy~N) ~b~Iprintf()~N expects the ~Iaddress~N of the first
    character of the string. When we ask to ~b~Iprintf()~N a ~b~I%c~Nharacter
    it expects the actual character itself.

    We normally don't worry about this...just define the variable ~b~Ix~N
    as a ~Istring~N (using ~Idouble quotes~N) and ask ~b~Iprintf()~N for the
    ~b~I%s~N format and - ~Iautomatically~N - the ~Iaddress~N of ~b~Ix~N is
    given to ~b~Iprintf()~N and printing begins with the first character
    found at that ~Iaddress~N and continues until the ~Ilast~N character
    (...and how does ~b~Iprintf()~N  know when it has reached the ~Ilast~N
    character ?? ... patience ...).

    If  we define ~b~Iy~N using ~Isingle quotes~N and use the ~b~I%c~N
    format in ~b~Iprintf()~N that's OK too. The C language looks after
    giving the ~Iactual character~N to ~b~Iprintf()~N (rather than the
    ~Iaddress~N).
.K19,60
let C doit
.WN

~b~Ix="Z";~N                            ~b~Ix~N is in ~Idouble quotes~N.
~b~Iprintf("x is %s",x);~N
~r~Ix is Z~N                            this is the printout..~IOK~N!

    If ~b~I%s~N is used in the ~b~Iprintf()~N statement, and if ~b~Ix~N is
    declared and defined as a ~Istring~N, then C will look after giving to
    ~b~Iprintf()~N the address where ~b~Ix~N is stored.
~b                                                                        ~N
    Now consider:

~b~Iy='Z';~N                            ~b~Iy~N is in ~Isingle quotes~N.
~b~Iprintf("y is %c",y);~N
~r~Iy is Z~N                            this is the printout..~IOK~N!

    Here ~b~Iy~N is in ~Isingle quotes~N (hence a ~Isingle~N character) and
    the ~b~I%c~N tells ~b~Iprintf()~N that the 'value' of ~b~Iy~N which it
    receives is to be interpreted as the actual character itself (in this
    example, the character ~b~IZ~N), so a ~r~IZ~N is (correctly) printed.
.WN


    Now consider:

~b~Iy='Z';~N                            ~b~Iy~N is in ~Isingle quotes~N.
~b~Iprintf("y is %s",y);~N
~r~Iy is ~N                             this is the printout ~INOT OK!~N.

    NOW ~b~Iy~N is a ~Isingle~N character (~b~IZ~N, in ~Isingle quotes~N)  but
    because we asked to have it printed as a ~b~I%s~Ntring, ~b~Iprintf()~N
    goes to the ~Imemory address~N given by the 'value' of ~b~Iy~N and prints
    characters on the screen (according to the numbers it finds in memory)!
    The 'value' given to ~b~Iprintf()~N was used as a ~w~Rpointer~N instead
    of the actual character to be printed...and ~b~Iprintf()~N went to some
    strange ~Iaddress~N in memory to find the ~b~I%s~Ntring to print!!

    Even if we do it right and define a string with ~Idouble quotes~N
    and use ~b~I%s~N in the ~b~Iprintf()~N format (so the ~Iaddress~N of the
    beginning of our string is passed to ~b~Iprintf()~N) ~ITHEN~N....
    ~w~Bhow does printf() know when it has come to the end of our string??~N
.WNT
 ..how does a string end?

    When we say ~b~Ix="hello";~N the C compiler will put the correct 'values'
    for the characters ~b~Ih~N, ~b~Ie~N, ~b~Il~N, ~b~Il~N, ~b~Io~N into memory
    and add, ~Iat the end~N (after the ~b~Io~N) the 'value' ~I0~N.
    It is this 'value' ~I0~N which signals the end of the string!

    NOTE: Every character such as ~b~Ia~N, ~b~Ib~N, ~b~IZ~N, ~b~I{~N, etc.
    has a certain 'value' (or 'number') associated with it. In ~IASCII~N
    (~IA~Nmerican ~IS~Ntandard ~IC~Node for ~II~Nnformation ~II~Nnterchange)
    the 'value' or 'number' associated with ~b~IA~N is ~I65~N (in decimal) and
    the 'value' or 'number' associated with ~b~I0~N is ~I48~N (in decimal).
    Notice that it is NOT the 'number' ~I0~N which is associated with the
    character ~b~I0~N but rather the 'number' ~I48~N! ~ISO~N...the ~I0~N which
    terminates strings cannot be confused with the character ~b~I 0~N.
    After all, the string may, in fact, be defined by: ~b~Ix="10"~N which
    will be stored in memory as the two 'numbers' associated with the two
    characters ~b~I1~N and ~b~I0~N (namely the two 'numbers' ~I49~N ~I48~N)
    followed by the terminating number ~I0~N
.WN





    It is possible to define a string by defining an ~Iarray of single~N
    ~Icharacters~N. Although we will talk (later) about ~Iarrays~N, now is
    an opportune time to talk briefly about such a definition of a string
    because, in this instance, we must ~Idefine the last single character~N
    ~Iin the array as the special terminating character, 0~N.
.K18,30
special 0
.WNT
  ...just one word about 'string arrays'

~b~Ichar x[10];~N                 defines an ~Iarray~N of ~b~I10~N elements.
~b~Ix[0]='A';~N                   the first element is the character ~b~IA~N.
~b~Ix[1]='b';~N                   the second element is the character ~b~Ib~N.
~b~Ix[2]='{';~N                   the third element is the character ~b~I{~N.
~b~Ix[3]='\0';~N                  the last element is the 'number' ~I0~N!!!!
~b~Iprintf("the string is %s",x);~N    print the string, up to the ~I0~N.

    ...and the printout would be: ~r~Ithe string is Ab{~N

   ~w~B                                                           ~N
   ~w~B but note the strange way we had to define the terminating ~N
   ~w~B element  so that C would  recognize it as the 'number' 0  ~N
   ~w~B and not the 'character' 0 ... we used  ~b~W\0~w~B  inside  single ~N
   ~w~B quotes in the statement: ~b~Wx[3]='\0';~w~B                       ~N
   ~w~B                                                           ~N
.WNT
  Other special characters like \0

    In order to define the special character ~I0~N which terminates a string
    we referred to it as ~b~I\0~N. The ~Ibackslash~N ~b~I\~N notifies C that
    the VERY NEXT CHARACTER is to be interpreted in a SPECIAL manner. (We've
    seen this kind of thing before: ~b~I%~N means the NEXT CHARACTER(s)
    is 'special' in a ~b~Iprintf()~N format....as in ~b~I%s~N).

    There are other ~b~I\character~N combinations in C. In each case they
    are used to define a character which cannot (normally) be typed into
    your text. For example you may have noticed that the statements:

~b~Ichar x,y;~N
~b~Ix="Z"; y='Z';~N
~b~Iprintf("x is %s y is %c",x,y);~N

    will print:

~r~Ix is Z y is Z~N
.K19,60
ZZZZZZZZZZ
.WN




    Now suppose we wanted to print:~r~Ix is Z~N and ~r~Iy is Z~N
    on two separate lines! Then we tell ~b~Iprintf()~N to print a 'special'
    character, ~b~I\n~N, meaning a ~b~In~New line:

~b~Ichar x,y;~N
~b~Ix="Z"; y='Z';~N
~b~Iprintf("x is %s ~N~V\n~b~I~W y is %c",x,y);~N   notice the ~V\n~N!

~r~Ix is Z ~N           this is the printout,
~r~I y is Z~N           notice the spaces.
.K19,60
\n=newline
.WNT
  SPECIAL \CHARACTERS

    We have the following 'special' characters:

~b~I\n~N      the ~In~New line.
~b~I\t~N      the ~It~Nab character.
~b~I\0~N      the ~I0~N terminator (NULL character), ASCII 'value' ~I0~N.
~b~I\b~N      the ~Ib~Nackspace.
~b~I\"~N      the double quote.
~b~I\\~N      the backslash character.

    What would the following print?

~b~Iint age;~N
~b~Iage=100;~N
~b~Iprintf("\"Sam\" is %d years old\t\ttoday",age);~N
.W
.R21C1
~r~I"Sam" is 100 years old          today~N          Notice the ~Iquotes~N
                                                     and the ~Itabs~N.
.R23C1
~b~Iprintf("~N~V\"~b~I~WSam~N~V\"~b~I~W is %d years old~N~V\t\t~b~I~Wtoday",age);~N
.WNT
  Other 'format' info you can give to printf()
    We've seen the ~b~I%d~N (for ~b~Id~Necimal number printouts) and we've
    seen ~b~I%f~N (for ~b~If~Nloating point numbers), and ~b~I%c~N and ~b~I%s~N
    for ~b~Ic~Nharacters and ~b~Is~Ntrings, but we have also ~b~I%o~N (for
    ~b~Io~Nctal numbers) and ~b~I%x~N (for he~b~Ix~Nadecimal numbers) and
    ~b~I%e~N (numbers ~b~Ie~Nxponential format, such as ~r~I-7.001100E+03~N).
    In each case you may qualify the above ~Iformat~N instruction with a
    ~Ifield width~N specification, such as:
.w
~b~Iint a=123;printf("%6d",a);~N         ~V   123~N

~b~Ifloat b=123;printf("%6.3f",b);~N     ~V123.000~N

~b~Ichar c='1';printf("%6c",c);~N        ~V     1~N

~b~Ichar d="123";printf("%6s",d);~N      ~V   123~N

~b~Iint e=123;printf("%6o",e);~N         ~V   173~N

~b~Ifloat f=123;printf("%6e",f);~N       ~V1.230000E+02~N

~b~Iint g=123;printf("%6x",g);~N         ~V    7B~N      in HEXADECIMAL
.K16,60
173=octal
.WNT
  good form .. bad form

.K5,60
bad form?!


    You may have noticed, in the compound statement:

~b~Ifloat b=123;printf("%6.3f",b);~N     ~V123.000~N

    that we declared ~b~Ib~N to be a ~b~Ifloat~N and, ~Iat the same time~N, we
    defined it to be ~b~I123~N (as in ~b~Ifloat b=123;~N). ~IThat's legal~N.

    BUT we also added ~b~Iprintf("%6.3f",b);~N ~Ion the same line~N. That's
    considered ~Ibad form~N (...but sure convenient if you want to see as
    much of your program as possible on a 25-line screen!)

    Note: for ~b~Ib=123~N, ~b~I6.~N~V3~b~I~Wf~N doesn't provide enough ~Ifield~N
    ~Iwidth~N (if we want ~V3~N decimal places) so ~b~Iprintf()~N expands
    the field width to ~b~I7~N (as required to accommodate ~V123.000~N).
.WN

    You will also have noticed that the statement:

~b~Ichar d="123";printf("%6s",d);~N      ~V   123~N

    ~Iright justified~N the ~r~I123~N in a ~Ifield width~N of 6. You may
    ask ~b~Iprintf()~N to ~Ileft-justify~N the ~r~I123~N by specifying:

~b~Ichar d="123";printf("%-6s",d);~N     ~V123   ~N

    (Notice the format specification ~b~I%-6s~N with ~b~I-~N meaning
     ~Ileft-justify~N).

    If ~b~Ib="1234567";~N then (as expected) ~b~Iprintf()~N will make the
    ~Ifield width~N ~b~I7~N (to accommodate ~r~I1234567~N)...even if we
    asked for ~b~I%6s~N! If you ~IREALLY MEAN 6~N (and want to cut off the
    string) then use the specification: ~b~I%.6s~N

~b~Ichar d="1234567";printf("%.6s",d);~N     ~V123456~N  note the ~b~I.6~N.
.WNT
    A note on format specifications



    If ~b~Is~N has been declared and defined as a string: ~b~Is="I'm a string"~N
    ( with 12 characters ) then we may select any number of characters and
    print then in a specified field width:

~b~Iprintf("%s",s);~N         gives   ~VI'm a string~N

~b~Iprintf("%20.12s",s);~N    gives   ~V        I'm a string~N

~b~Iprintf("%-20.12s",s);~N   gives   ~VI'm a string        ~N

~b~Iprintf("%20.10s",s);~N    gives   ~V          I'm a stri~N

~b~Iprintf("%-20.10s",s);~N   gives   ~VI'm a stri          ~N
                               field width = 20 
           print only 10 characters
.WNT
    More notes on format specifications

    If ~b~Ix~N has been declared and defined as an int: ~b~Ix=123~N
    then we have:

~b~Iprintf("%d",x);~N        gives   ~V123~N

~b~Iprintf("%10d",x);~N      gives   ~V       123~N

~b~Iprintf("%-10d",x);~N     gives   ~V123       ~N

~b~Iprintf("%010d",x);~N     gives   ~V0000000123~N

~b~Iprintf("%010x",x);~N     gives   ~V000000007B~N   in HEXADECIMAL
                              width=10
       this ~I0~N pads with 0's
.WNT
    ...and some final notes on format

    If ~b~Ix~N has been declared and defined as an float: ~b~Ix=123.456~N
    then we have:

~b~Iprintf("%f",x);~N        gives   ~V123.456000~N

~b~Iprintf("%10.4f",x);~N    gives   ~V  123.4560~N

~b~Iprintf("%-10.2f",x);~N   gives   ~V123.46    ~N
.WN




.T
    That's all folks!

.K16,32
au revoir!

.q