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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')
Output
10
jeeksquizThe 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.
Output
8Running 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.
Output
8