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";

Below are the key differences:

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



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