I've been experimenting with calling a templated friend function with no parameter, which is defined in class template. I found no solution for the exact that case and solved my problem the other way, but during my experiments I suddenly found some interesting piece of code. It works, but I don't know why. Here's this example:
#include <iostream>
template<class T> class A
{
public:
void test() const
{
std::cout << "A!" << std::endl;
}
template<class U> friend A fun()
{
return A();
}
};
int main(int argc, char* argv[])
{
A<double> aa; // 1
const auto& a = fun<int>();
a.test();
return 0;
}
As I said, it works for me on GCC 4.8.1. If I remove (1), it fails with the following:
main.cpp: In function 'int main(int, char**)':
main.cpp:20:18: error: 'fun' was not declared in this scope
const auto& a = fun<int>();
^
main.cpp:20:22: error: expected primary-expression before 'int'
const auto& a = fun<int>();
I suspect there's UB here, but it will be very interesting if anyone could clarify:
- Why it at all works while I didn't tell fun() which specialization of A it should use?
- If it's not UB, what type is T? I tried type_info and found just that it's not neither int nor double. typeid().name() didn't help as it returns 'FdvE' for me.
This seems to be a GCC-bug.
friend
function templates that are defined in a class can only be found via ADL: GCC apparently considersaa
during ADL for the call inmain
(for no reason) and callsA<double>
'sfun
, as confirmed through this static assertion:Demo.
Clang does not compile this code at all.