What’s difference between char s[] and char *s in C?

Consider below two statements in C. What is difference between two?

   char s[] = "geeksquiz";
   char *s  = "geeksquiz";

The statements ‘char s[] = “geeksquiz”‘ creates a character array which is like any other array and we can do all array operations. The only special thing about this array is, although we have initialized it with 9 elements, its size is 10 (Compiler automatically adds ‘\0’)

#include <stdio.h>
int main()
{
    char s[] = "geeksquiz";
    printf("%lu", sizeof(s));
    s[0] = 'j';
    printf("\n%s", s);
    return 0;
}

Output:

10
jeeksquiz

The statement ‘char *s = “geeksquiz”‘ creates a string literal. The string literal is stored in read only part of memory by most of the compilers. The C and C++ standards say that string literals have static storage duration, any attempt at modifying them gives undefined behavior.
s is just a pointer and like any other pointer stores address of string literal.

#include <stdio.h>
int main()
{
    char *s = "geeksquiz";
    printf("%lu", sizeof(s));

    // Uncommenting below line would cause undefined behaviour
    // (Caused segmentation fault on gcc)
    //  s[0] = 'j';  
    return 0;
}

Output:

8

Running above program may generates a warning also “warning: deprecated conversion from string constant to ‘char*’”. This warning occurs because s is not a const pointer, but stores address of read only location. The warning can be avoided by pointer to const.

#include <stdio.h>
int main()
{
    const char *s = "geeksquiz";
    printf("%lu", sizeof(s));
    return 0;
}

This article is contributed by Abhay Rathi. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

GATE CS Corner    Company Wise Coding Practice

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