CHAPTER 3: Pointers 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-------------------------------------
#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. On the other hand, 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.

 

2006 onwards Programmershelp