friend, template, namespace - c++

Question

I want to have a templated friend function. However, I do not know how to make it works in the same way for no templated function.
Here is a sample code
#include <iostream>
namespace ns{
struct Obj {
friend void foo(Obj){std::cout << "no problem" << std::endl;}
template<typename T>
friend void bar(Obj){std::cout << "problem" << std::endl;}
};
}
int main() {
ns::Obj obj;
foo(obj); // Compile
bar<int>(obj); // Not compile
return 0;
}

Answer 1

Before C++20, you need to teach the compiler that bar is the name of a template so that it knows that < starts a template argument list and is not the less-than operator:
template<char> void bar() = delete;
int main() {
ns::Obj obj;
foo(obj); // Compile
bar<int>(obj); // Now compiles too
return 0;
}
Note that all the bar overload has to do is to be a function template. The signature doesn't matter as long as it's not so good as to interfere with the overload resolution; () is a good choice because by definition we are passing at least one argument, so a function template taking no parameters can never be viable.
Alternatively, you can redesign bar to deduce T from a tag argument:
template<class T>
struct type {};
namespace ns{
struct Obj {
// ...
template<typename T>
friend void bar(Obj, type<T>) { /* ... */ }
};
}
// ...
bar(obj, type<int>()); // OK
In C++20, the compiler will assume that bar names a template when it sees the < and name lookup finds nothing, so your code will just work.

Answer 2

The straightforward approach would be to add a forward declaration and let the function be located using qualified lookup:
namespace ns{
struct Obj;
void foo(Obj);
template<typename T>
void bar(Obj);
struct Obj {
friend void foo(Obj){std::cout << "no problem" << std::endl;}
template<typename T>
friend void bar(Obj){std::cout << "problem " << std::endl;}
};
} // namespace ns
int main() {
ns::Obj obj;
ns::foo(obj); // Ok
ns::bar<int>(obj); // Ok
return 0;
}