c++ - c++11: How to write a wrapper function to make `std::function` objects

Question

I am trying to write a wrapper make_function, which like std::make_pair can create a std::function object out of suitable callable objects.

Just like make_pair, for a function pointer foo, auto f0 = make_function(foo); creates a std::function function object f0 of the right type signature. Just to clarify, I don't mind occasionally giving type parameters to make_function in case it is difficult (or impossible) to deduce the type entirely from the parameters.

What I came up with so far (code below) works fine for lambdas, some function pointers, and functors (I didn't consider volatiles). But I couldn't get it work for std::bind or std::bind<R> results. In the code below

auto f2 = make_function(std::bind(foo,_1,_2,_3)); //not OK

wouldn't compile/work, with gcc 4.8.1. I am guessing that I didn't capture the operator() for the bind result correctly, but I am not sure how to fix it.

Any help on how to fix this case or improvement in other corner cases is appreciated.

My question is, of course, how to fix the error in the example below.

For background, one of the cases I use this wrapper can be found at this question: How to make C++11 functions taking function<> parameters accept lambdas automatically. If you do not approve the use of std::function or my specific way of using it, please leave your comments in that post, and discuss technical issues here.

--- EDIT ---

From some of the comments, I learned that it's because of the ambiguity issue (ambiguity of the function call operator() of std::bind results). As pointed out by @Mooing Duck's answer, the solution is to give the parameter types explicitly. I have updated the code to combine the three functions in @Mooing Duck's answer (with slight change of type parameters), so that the make_function wrapper can now handle/type-deduce unambiguous cases as before, and allow specification of complete type signature when there is ambiguity.

(My original code for the unambiguous cases is at: https://stackoverflow.com/a/21665705/683218 and can be tested at: https://ideone.com/UhAk91):

#include <functional>
#include <utility>
#include <iostream>
#include <functional>
using namespace std;

// For generic types that are functors, delegate to its 'operator()'
template <typename T>
struct function_traits
  : public function_traits<decltype(&T::operator())>
{};

// for pointers to member function
template <typename ClassType, typename ReturnType, typename... Args>
struct function_traits<ReturnType(ClassType::*)(Args...) const> {
  enum { arity = sizeof...(Args) };
  typedef function<ReturnType (Args...)> f_type;
};

// for pointers to member function
template <typename ClassType, typename ReturnType, typename... Args>
struct function_traits<ReturnType(ClassType::*)(Args...) > {
  enum { arity = sizeof...(Args) };
  typedef function<ReturnType (Args...)> f_type;
};

// for function pointers
template <typename ReturnType, typename... Args>
struct function_traits<ReturnType (*)(Args...)>  {
  enum { arity = sizeof...(Args) };
  typedef function<ReturnType (Args...)> f_type;
};

template <typename L> 
static typename function_traits<L>::f_type make_function(L l){
  return (typename function_traits<L>::f_type)(l);
}

//handles bind & multiple function call operator()'s
template<typename ReturnType, typename... Args, class T>
auto make_function(T&& t) 
  -> std::function<decltype(ReturnType(t(std::declval<Args>()...)))(Args...)> 
{return {std::forward<T>(t)};}

//handles explicit overloads
template<typename ReturnType, typename... Args>
auto make_function(ReturnType(*p)(Args...))
    -> std::function<ReturnType(Args...)> {
  return {p};
}

//handles explicit overloads
template<typename ReturnType, typename... Args, typename ClassType>
auto make_function(ReturnType(ClassType::*p)(Args...)) 
    -> std::function<ReturnType(Args...)> { 
  return {p};
}

// testing
using namespace std::placeholders;

int foo(int x, int y, int z) { return x + y + z;}
int foo1(int x, int y, int z) { return x + y + z;}
float foo1(int x, int y, float z) { return x + y + z;}

int main () {
  //unambuiguous
  auto f0 = make_function(foo);
  auto f1 = make_function([](int x, int y, int z) { return x + y + z;});
  cout << make_function([](int x, int y, int z) { return x + y + z;})(1,2,3) << endl;

  int first = 4;
  auto lambda_state = [=](int y, int z) { return first + y + z;}; //lambda with states
  cout << make_function(lambda_state)(1,2) << endl;

  //ambuiguous cases
  auto f2 = make_function<int,int,int,int>(std::bind(foo,_1,_2,_3)); //bind results has multiple operator() overloads
  cout << f2(1,2,3) << endl;
  auto f3 = make_function<int,int,int,int>(foo1);     //overload1
  auto f4 = make_function<float,int,int,float>(foo1); //overload2

  return 0;
}

Ideone

Answer 1

The problem is your code doesn't handle lambdas, bind, or functionoids properly, your code assumes that all of these take no parameters. To handle these, you'll have to specify the parameter types:

//plain function pointers
template<typename... Args, typename ReturnType>
auto make_function(ReturnType(*p)(Args...))
    -> std::function<ReturnType(Args...)> 
{return {p};}

//nonconst member function pointers
template<typename... Args, typename ReturnType, typename ClassType>
auto make_function(ReturnType(ClassType::*p)(Args...)) 
    -> std::function<ReturnType(Args...)>
{return {p};}

//const member function pointers
template<typename... Args, typename ReturnType, typename ClassType>
auto make_function(ReturnType(ClassType::*p)(Args...) const) 
    -> std::function<ReturnType(Args...)>
{return {p};}

//qualified functionoids
template<typename FirstArg, typename... Args, class T>
auto make_function(T&& t) 
    -> std::function<decltype(t(std::declval<FirstArg>(), std::declval<Args>()...))(FirstArg, Args...)> 
{return {std::forward<T>(t)};}

//unqualified functionoids try to deduce the signature of `T::operator()` and use that.
template<class T>
auto make_function(T&& t) 
    -> decltype(make_function(&std::remove_reference<T>::type::operator())) 
{return {std::forward<T>(t)};}

Variables:

int func(int x, int y, int z) { return x + y + z;}
int overloaded(char x, int y, int z) { return x + y + z;}
int overloaded(int x, int y, int z) { return x + y + z;}
struct functionoid {
    int operator()(int x, int y, int z) { return x + y + z;}
};
struct functionoid_overload {
    int operator()(int x, int y, int z) { return x + y + z;}
    int operator()(char x, int y, int z) { return x + y + z;}
};
int first = 0;
auto lambda = [](int x, int y, int z) { return x + y + z;};
auto lambda_state = [=](int x, int y, int z) { return x + y + z + first;};
auto bound = std::bind(func,_1,_2,_3);

Tests:

std::function<int(int,int,int)> f0 = make_function(func); assert(f0(1,2,3)==6);
std::function<int(char,int,int)> f1 = make_function<char,int,int>(overloaded); assert(f1(1,2,3)==6);
std::function<int(int,int,int)> f2 = make_function<int,int,int>(overloaded); assert(f2(1,2,3)==6);
std::function<int(int,int,int)> f3 = make_function(lambda); assert(f3(1,2,3)==6);
std::function<int(int,int,int)> f4 = make_function(lambda_state); assert(f4(1,2,3)==6);
std::function<int(int,int,int)> f5 = make_function<int,int,int>(bound); assert(f5(1,2,3)==6);
std::function<int(int,int,int)> f6 = make_function(functionoid{}); assert(f6(1,2,3)==6);
std::function<int(int,int,int)> f7 = make_function<int,int,int>(functionoid_overload{}); assert(f7(1,2,3)==6);
std::function<int(char,int,int)> f8 = make_function<char,int,int>(functionoid_overload{}); assert(f8(1,2,3)==6);

http://coliru.stacked-crooked.com/a/a9e0ad2a2da0bf1f The only reason your lambda was succeeding is because it was implicitly convertible to a function pointer because your example doesn't capture any state. Note that my code requires the parameter types for overloaded functions, functionoids with overloaded operator() (including bind), but is now able to deduce all non-overloaded functionoids.

The decltype lines are complicated but they're used to deduce the return types. Notice that in NONE of my tests do I need to specify the return type. Let's break down make_function<short,int,int> down as if T is char(*)(short, int, int):

-> decltype(t(std::declval<FirstArg>(), std::declval<Args>()...))(FirstArg, Args...)
`std::declval<FirstArg>()` is `short{}` (roughly)
-> decltype(t(short{}, std::declval<Args>()...))(FirstArg, Args...)
`std::declval<Args>()...` are `int{}, int{}` (roughly)
-> decltype(t(short{}, int{}, int{})(FirstArg, Args...)
`t(short{}, int{}, int{})` is an `int{}` (roughly)
-> decltype(short{})(FirstArg, Args...)
`decltype(int{})` is `int`
-> int(FirstArg, Args...)
`FirstArg` is still `short`
-> int(short, Args...)
`Args...` are `int, int`
-> int(short, int, int)
So this complex expression merely figures out the function's signature
well, that should look familiar...