Thread: Don't understand this 'sizeof' expression.

Probably you should make a small program and print to screen what sizeof(data) gives and what sizeof(struct dataEntry) gives

Run this

Code:

#include <stdio.h>

struct foo{
        int data;
};

int main(void)
{
        struct foo array[] = {{0}, {1}, {2}, {3}, {4}, {5}};
        printf("%u\n", sizeof(struct foo));
        printf("%u\n", sizeof(array));
        printf("%u\n", sizeof(array)/sizeof(struct foo));

        return 0;
}

Notice that I am not giving you the output, because we should learn to try things by yourself

It's more conveniently (and usefully) written as
sizeof(array)/sizeof(array[0])

I usually use this macro:

Code:

#define NELEMS(obj) (sizeof(obj)/sizeof(*obj))

Then you can just write NELEMS(array)

Suppose you write this

Code:

char buf[1000];
printf("%d\n", NELEMS(buf));

sizeof buf is 1000 and sizeof *buf is 1, so the result is 1000 elements. The nice thing about the macro is that it works for any object like an array of structs or doubles, etc.

Originally Posted by pyroknife

Wiat why is sizeof(*buf)=1? Was that something defined?

sizeof(*buf) is essentially sizeof(char), which is guaranteed to be 1 in all cases.

Originally Posted by pyroknife

Wiat why is sizeof(*buf)=1? Was that something defined?

In C, aggregates are treated as pointers. So buf is a pointer to char, *buf is a char, and sizeof(char) is 1, of course. If buf is an array of something else, then the size will change accordingly.

Code:

double buf[100];
printf("%d\n", NELEMS(buf));

You already did.

Code:

char buf[100];

If you reference *buf within this scope, the compiler knows that *buf is a char, and it also knows that sizeof buf is 100.