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

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

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

Below are the key differences:

Aspect	char a[10] = "geek";	char *p = "geek";
Nature	`a` is an array	`p` is a pointer variable
Size	sizeof(a) = 10 bytes (32-bit Systems)	sizeof(p) = 4 bytes (32-bit Systems)
Address Comparison	`a` and `&a` are the same	`p` and `&p` are not the same
Memory Location	`"geek"` is stored in the stack section of memory	`p` is stored in stack, but `"geek"` is stored in code section
Reassignment	a = "hello"; → Invalid Reason: a is an address and cannot be reassigned to another address (string constant)	p = "india"; → Valid
Increment	a++ → Invalid	p++ → Valid
Modification	a[0] = 'b'; → Valid	p[0] = 'k'; → Invalid Reason: Code section is read-only

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 the 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

Running above program may generate 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 the read-only location. The warning can be avoided by the pointer to const.

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

Output

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

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