Passing lambda as template parameter to templated by function-pointer function

Question

Looks like I cannot pass a no-capture lambda as a template parameter to a templated by function-pointer function. Am I doing it the wrong way, or is it impossible?

#include <iostream>

// Function templated by function pointer
template< void(*F)(int) >
void fun( int i )
{
    F(i);
}

void f1( int i )
{
    std::cout << i << std::endl;
}

int main()
{
    void(*f2)( int ) = []( int i ) { std::cout << i << std::endl; };

    fun<f1>( 42 ); // THIS WORKS
    f2( 42 );      // THIS WORKS
    fun<f2>( 42 ); // THIS DOES NOT WORK (COMPILE-TIME ERROR) !!!

    return 0;
}

Answer 1

It's mostly a problem in the language's definition, the following makes it more obvious:

using F2 = void(*)( int );

// this works:
constexpr F2 f2 = f1;

// this does not:
constexpr F2 f2 = []( int i ) { std::cout << i << std::endl; };

Live example

This basically means that your hope/expectation is quite reasonable, but the language is currently not defined that way - a lambda does not yield a function pointer which is suitable as a constexpr.

There is, however, a proposal to fix this issue: N4487.

Answer 2

This is not viable because f2 is not constexpr (i.e., is a runtime variable). As such it cannot be used as a template parameter. You could alter your code and make it more generic in the following manner:

#include <iostream>

template<typename F, typename ...Args>
void fun(F f, Args... args) {
  f(args...);
}

void f1( int i ) {
    std::cout << i << std::endl;
}

int main() {
    auto f2 = []( int i ) { std::cout << i << std::endl; };
    fun(f1, 42);
    f2( 42 );
    fun(f2, 42 );
    return 0;
}