String Pointer in C - Sanfoundry

This Tutorial Explains Different Operations Performed on String Literals Using Pointers in C Programming

A string literal is a sequence of characters terminated by null byte ‘0’. For example,

/* strlit_op.c -- program shows operations on string literals */
#include <stdio.h>
#define NAME "What is your name?"
 
int main(void)
{
    char name[25] = "xyz";
    char * const p2name = "abc";
 
    "abc" += 1;
    "abc" + 1;
    "abc"[0];
    "abc"[1];
    "abc"[10];
    *("abc" + 2);
    *"abc";
 
    return 0;
}

If you compile the above program, compiler saves copy of #define sting some where in the memory and set a pointer to it. Compiler also terminates the string with null terminator. What about name character array? It’s a pointer constant. And p2name? It’s also a pointer constant. Let’s try to figure out how expressions in above program evaluate to,

    "abc" += 1;

It seems as if we add 1 to string literal “abc” and assigns it the result. What do you think, if you can figure it out. This statement is an error. Because value of string literal is a pointer constant. Like, for ex., exp.,

    5 += 5; /* 5 is constant */

can’t be evaluated. Therefore, statement

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    "abc" += 1;

is invalid. Consider the following statement,

    "abc" + 1;

Again we are trying to add 1 to the string literal “abc”. Sure, you can, this time, easily unravel this. Since, string literal “abc” is a pointer constant pointing to first character ‘a’. Type of “abc” is a pointer to char. Therefore, when you add 1 to “abc”, 1 is adjusted to type char * and added to pointer constant. This moves pointer to character ‘b’.

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Take another exp.,

    "abc"[0];

As we know that array name is a pointer constant pointing to first element of the array. Therefore, exp. “abc”[0] can be rewritten as

    *("abc" + 0);

or even as

    *"abc";

which results first character ‘a’ in the string “abc”. Let’s explore exp.

    "abc"[10];

Let’s take an analogy to evaluate the above exp.,

    int white[5]; /* white is an array of 5 integers */

What will happen if you attempt to access

    white[10];

Surely, subscript value 10 takes you off the end of the array white. Hence, an undefined result. Similarly, in exp.,

    "abc"[10];

you go out of bounds and therefore undefined result.

/* strlit_op.c -- program shows operations on string literals */
#include <stdio.h>
 
#define NAME "What is your name?"
 
int main(void)
{
    char name[25] = "xyz";
    char * const p2name = "abc";
 
    //"abc" += 1;
    printf("Result of \"abc\" + 1 is %c\n", "abc" + 1);
    printf("Result of *(\"abc\" + 1) is %c\n", *("abc" + 1));
    printf("Result of \"abc\"[0] is %c\n", "abc"[0]);
    printf("Result of \"abc\"[1] is %c\n", "abc"[1]);
    printf("Result of \"abc\"[10] is %c\n", "abc"[10]);
    printf("Result of *(\"abc\" + 2) is %c\n", *("abc" + 2));
    printf("Result of *\"abc\" is %c\n", *"abc");
 
    return 0;
}
 
Output of the above program is as follows,
 
<pre lang="C" cssfile="hk1_style">
Result of "abc" + 1 is
Result of *("abc" + 1) is b
Result of "abc"[0] is a
Result of "abc"[1] is b
Result of "abc"[10] is
Result of *("abc" + 2) is c
Result of *"abc" is a

Let’s see an application of string literal in a C program that converts an unsigned integer into equivalent hexadecimal representation,

/*
 * strlit_app.c -- program converts unsigned values into hexadecimal 
 * number
 */
#include <stdio.h>
 
#define MINUS1 4294967295 /* unsigned representation of -1 */
 
/* uint_to_hex() is a recursive function */
void uint_to_hex(unsigned);
 
int main(void)
{
    unsigned value;
 
    puts("\n**Program converts Unsigned int to Hexadecimal No.**\n");
    printf("User, enter an unsigned integer value or -1 to quit: ");
    while (scanf("%u", &value) == 1 && value != MINUS1) {
        /* call to uint_to_hex() */
        uint_to_hex(value);
        puts("");
        printf("more conversions, enter unsigned integer or -1 to quit:");
    }
    puts("Thank You!");
    return 0;
}
 
void uint_to_hex(unsigned value)
{
    unsigned quotient;
 
    quotient = value / 16;
    if (quotient != 0)
        uint_to_hex(quotient);
 
    putchar("0123456789ABCDEF"[value % 16]);
}

Output of the program as follows,

 
**Program converts Unsigned int to Hexadecimal No.**
 
User, enter an unsigned integer value or -1 to quit: 54
36
more conversions, enter unsigned integer or -1 to quit: 12
C
more conversions, enter unsigned integer or -1 to quit: 8667
21DB
more conversions, enter unsigned integer or -1 to quit: 12
C
more conversions, enter unsigned integer or -1 to quit: 3232
CA0
more conversions, enter unsigned integer or -1 to quit: -1
Thank You!