Question

As a rule, code example first:

void f1(int)
{}

#define f2(a) f1(a)

template<class F>
void f3(F f)
{
    f(0);
}

int main()
{
    f3(f2); // error C2065: 'f2' : undeclared identifier
    return 0;
}

Compiled by VC++ 2012.

My question is:

Why does macro expansion follow template expansion? I think it is extremly counter-intuitive and error-prone.

Answer 1

Why does macro expansion follow template expansion?

Um, it doesn’t. Macro expansion is done by the preprocessor, template expansion is done by the parser/compiler stage (which runs only after preprocessing).

What you’re missing here is the style of the macro. f2() is a function-style macro. So if you write f2 (without the parentheses), the preprocessor won’t substitute it to f1. If you want such a substitution, simply define it as

#define f2 f1

Side note: as currently standing, this code doesn’t make much sense. Even if you used the parentheses and wrote f2(), you would get a compiler error, since the f2() macro takes exactly one argument. You should supply an argument to it if it’s a function-style macro that has an argument and is not variadic.