Question

char str[] = " http://www.ibegroup.com/";

char *p = str ;

void Foo ( char str[100]){

}

void *p = malloc( 100 );

What’s the sizeof str,p,str,p in the above 4 case in turn?

I’ve tested it under my machine(which seems to be 64bit) with these results:

25 8 8 8

But don’t understand the reason yet.

Answer 1

sizeof(char[]) returns the number of bytes in the string, i.e. strlen()+1 for null-terminated C strings filling the entire array. Arrays don’t decay to pointers in sizeof. str is an array, and the string has 25 characters plus a null byte, so sizeof(str) should be 26. Did you add a space to the value?

The size of a pointer is of course always determined just by the machine architecture, so both instances of p are 8 bytes on 64-bit architectures and 4 bytes on 32-bit architectures.

In function arguments, arrays do decay to pointers, so you’re getting the same result that you get for a pointer. Therefore, the following definitions are equivalent:

void foo(char s[42]) {};
void foo(char s[100]) {};
void foo(char* s) {};