Why does C string length change when passed to a function?

Question

Possible Duplicate:
C -> sizeof string is always 8

If I do sizeof("HELLO"); I get 6, which is expected, however if I have:

void printLength(char *s) {
    sizeof(s);
}

When passing "Hello" to to this function sizeof returns 8.

Why is this?

Answer 1

What you are getting is size of char *. To get it's actual length, use strlen.

Answer 2

It doesn't change. Arrays aren't pointers. sizeof("HELLO") gives you the size of the char array {'H','E','L','L','O','\0'}, and the sizeof(s) gives you the size of a pointer.

Answer 3

Because string literals and arrays declared to be of a fixed size are treated specially. The compiler knows exactly how big they are.

In your function, all the compiler knows is that it's a pointer to something. You are taking the size of the character pointer (which at that point, is all the compiler knows about it). You're on a 64 bit system, so you get an 8 byte pointer, no matter what you feed into the function.

There is a library function to do this for you:

#include <string.h>

int getLength(char * str)
{
    return strlen(str);
}