How to pass a template method as a template argument? - c++

Question

I would like to pass a template method as a template argument.
I don't understand why I am getting this error:
no known conversion for argument 1 from '<unresolved overloaded function type>' to 'void (B::*&&)(int&&, double&&)
Here is the code:
struct A {
template <class Fn, class... Args>
void f(Fn&& g, Args&&... args) {
g(std::forward<Args>(args)...);
}
};
struct B {
template <class... Args>
void doSomething(Args&&... args) {}
void run() {
A a;
a.f(&doSomething<int, double>, 1, 42.); // error here
}
};
int main() {
B b;
b.run();
return 0;
}
Any ideas?

Answer 1

The root cause for the error is that you need an object to call the member function. However, with current code, the error is not that straightforward.
Change calling site to
a.f(&B::doSomething<int, double>, 1, 42.)
And you will see much better error:
error: must use '.' or '->' to call pointer-to-member function in 'g
(...)', e.g. '(... ->* g) (...)'

Answer 2

doSomething is a member function, as such, it cannot be called without an object, which you are trying to do
g(std::forward<Args>(args)...);
^
where is the instance?
One solution to this is to wrap doSomething in a lambda:
a.f([](B& instance, int a, int b) { instance.doSomething(a, b); }, *this, 1, 42.);
If you can use C++17, you could also use std::invoke:
template <class Fn, class... Args>
void f(Fn&& g, Args&&... args) {
std::invoke(g, std::forward<Args>(args)...);
}
and then when calling f:
a.f(&B::doSomething<int, double>, this, 1, 42.);