Question

Why does the following code give:

#include<stdio.h>

int voo()
{
    printf ("Some Code");
    return 0;
}


int main() {
    printf ("%zu", sizeof voo);
    return 0;
}

The following output:

1

Answer 1

The C language does not define sizeof for functions. The expression sizeof voo violates a constraint, and requires a diagnostic from any conforming C compiler.

gcc implements pointer arithmetic on function pointers as an extension. To support this, gcc arbitrarily assumes that the size of a function is 1, so that adding, say, 42 to the address of a function will give you an address 42 bytes beyond the function’s address.

They did the same thing for void, so sizeof (void) yields 1, and pointer arithmetic on void* is permitted.

Both features are best avoided if you want to write portable code. Use -ansi -pedantic or -std=c99 -pedantic to get warnings for this kind of thing.