If you don't, you get this warning....
warning: weak declaration of ‘foo’ after first use results in
unspecified behavior

I know moving it before it's first use stops the warning and works....
what I don't understand is why.

If mutually recursive functions are coded without being declared then at least one of the functions will be called before it has been defined.

A decent compiler should still be able to sort that out though.

1. Always declare your functions before you use them.
2. The compiler may generate different code if it knows that the called function is weak.

"Always declare your functions before you use them," regardless.

I think most of the responses in this thread are missing the question slightly... so let me elaborate.

====a.h====

void a(void);

===compiles without warning.c====
#include "a.h"
//Provide weak stub....
void __attribute__((weak)) a(void)
{
}

void b(void)
{
// Invoke a
a();
}

===compiles with warning.c====
#include "a.h"
void b(void)
{
// Invoke a
a();
}

// Get warning: weak declaration of ‘a’ after first use results in
unspecified behavior
void __attribute__((weak)) a(void)
{
}


I guess is that is the heart of the question... why would it?