C++11 variadic std::function parameter

Question

A function named test takes std::function<> as its parameter.

template<typename R, typename ...A>
void test(std::function<R(A...)> f)
{
    // ...
}

But, if I do the following:

void foo(int n) { /* ... */ }

// ...

test(foo);

Compiler(gcc 4.6.1) says no matching function for call to test(void (&)(int)).

To make the last line test(foo) compiles and works properly, how can I modify the test() function? In test() function, I need f with type of std::function<>.

I mean, is there any template tricks to let compiler determine the signature of function(foo in example), and convert it to std::function<void(int)> automatically?

EDIT

I want to make this work for lambdas(both stated and stateless) as well.

Answer 1

It looks like you want to use overloading

template<typename R, typename ...A>
void test(R f(A...))
{
    test(std::function<R(A...)>(f));
}

This simple implementation will accept most if not all the functions you will try to pass. Exotic functions will be rejected (like void(int...)). More work will give you more genericity.