C++ passing a templated functor as a template typename

Question

I have a templated function:

template <typename my_func> int foo(int x, int y) {
 my_func bar;
 return bar(x, y);
}

which I can use with following functor:

struct _compare {
int operator()(int x, int y) { return x-y;  }
};

Usage: foo<_compare>(int x, int y)

In fact there is a tight loop in foo and thus i'm trying to pass pointer to functions that can be inlined at compile time.

Now I want to use a function from 3rd party lib, which is templated fuctor by itself. The new functor is like:

template<typename Scalar> struct _compare {
  Scalar operator()(const Scalar& a, const Scalar& b) const {
      return a-b;
  }
};

And I can't make it compile. I've tried foo<_compare<char>>(int x, int y) I've tried

const char t = 0;
foo<_compare<t>>(int x, int y);

I've tried to foo<_compare>(int x, int y) and change foo to

template <typename my_func> foo(int x, int y) {
 my_func<char> bar;
 bar(x, y);
}

As I need only _compare < char > for sure.
Any hints?

Answer 1

My preferred solution would be to do like the STL does: always pass the function object as parameter:

template<typename my_func>
void foo(int x, int y, my_func bar) {
    bar(x, y);
}

Then, it doesn't matter if my_func is a templated class, a lambda, a non templated class or even a function pointer. It just work.

Usage:

foo(int x, int y, _compare{});
foo(int x, int y, _compare<char>{});
foo(int x, int y, [](int, int){ return false; });

This Live example shows that C++ completely elide both callable, and the function call and its result.

Also, the GCC ABI has no storage for empty classes sent as parameter to a function, removing overhead if not inlined.