Q:

"function returning a function" when inheriting constructors from std::function

Trying to derive a class from std::function, and for starters inherit the constructors.   This is what I guessed at:
#include <iostream>
#include <functional>
using namespace std;

template< class R, class... Args >
class MyFunction : public std::function<R(Args...)> {
public:
    using std::function<R(Args...)>::function;
};

int main() {
    std::function<void()> f_display_42 = []() { std::cout << 42; }; //works

    MyFunction<void()> mine = []() { std::cout << 42; }; //errors
}

Errors that come back are:
prog.cpp: In instantiation of ‘class MyFunction<void()>’:
prog.cpp:14:24:   required from here
prog.cpp:6:7: error: function returning a function
 class MyFunction : public std::function<R(Args...)> {
       ^
prog.cpp:8:38: error: function returning a function
     using std::function<R(Args...)>::function;
                                      ^
prog.cpp:8:38: error: using-declaration for non-member at class scope
prog.cpp: In function ‘int main()’:
prog.cpp:14:55: error: conversion from ‘main()::__lambda1’
     to non-scalar type ‘MyFunction<void()>’ requested
     MyFunction<void()> mine = []() { std::cout << 42; };

In GCC 4.8.2. 

A:

To add to chris' answer, if you want your current definition to work, use 
MyFunction<void> mine = []() { std::cout << 42; };

If you want the same nifty MyFunction<R(Args...)> syntax as std::function, then you must supply a primary class template that's unimplemented, and a specialization that derives from std::function (this is exactly what std::function does too). 
template< class R, class... Args >
class MyFunction;

template< class R, class... Args >
class MyFunction<R(Args...)> : public std::function<R(Args...)>
{ ... }

Live demo

