c ++ 11：如何编写包装器函数以制作`std :: function`对象

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...

Answer 2

In general you cannot solve it without the severe restriction that whatever you pass to make_function is only callable with exactly one signature.

What are you going to do with something like:

struct Generic
{
    void operator()() { /* ... */ }
    void operator()() const { /* ... */ }

    template<typename T, typename... Ts>
    T operator()(T&& t, Ts&&...) { /* ... */ }

    template<typename T, typename... Ts>
    T operator()(T&& t, Ts&&...) const { /* ... */ }
};

C++14 generic lambdas will have the same issue.

The signature in std::function is based on how you plan to call it and not on how you construct/assign it.

You cannot solve it for std::bind either, as that has indefinite arity:

void foo() { std::cout << "foo()" << std::endl; }
//...

auto f = std::bind(foo);
f();                 // writes "foo()"
f(1);                // writes "foo()"
f(1, 2, 3, 4, 5, 6); // writes "foo()"

Answer 3

The big reason why you want to be able to convert lambdas to std::function is because you want two overloads, each taking different signatures.

A good way to solve this involves std::result_of .

Suppose you are making a loop control structure that takes a lambda or other functional. If that functional returns void, you want to loop uncontrolled. If it returns bool or the like, you want to loop while it returns true . If it returns enum ControlFlow , you want to pay attention to the ControlFlow return value ( continue or break , say). The function in question takes either the element iterating over, and optionally some extra data (the index in the iteration, maybe some "location" meta-information about the element, etc).

std::result_of would let you pretend to invoke the passed in type with a different number of arguments. A traits class could then figure out which of the above signatures is the "best match", and then route to the implementation that handles that signature (possibly by wrapping the "simpler" cases in a lambda and calling the more complex cases).

Naively, your make_function would could this problem, because you could then simply overload on the various std::function< blah(etc) > cases. But with auto parameters coming down the pipe, and std::bind already doing perfect forwarding, this only handles the easiest cases.

std::result_of traits classes (and possibly related concept matching and requires clauses) and tag dispatching (or SFINAE as a last resort).

The big downside is that you end up having to manage the override order yourself semi-manually. I could see some utility in helper classes where you provide a list of signatures to match, and it either produces a boost::variant or you also produce a canonical output and a conversion method to that canonical output.

The short answer? std::bind 's implementation is implementation specific details, but it may involve the equivalent of perfect forwarding of variadic parameter packs, and as such is not suitable for your "get the address of the one and only concrete operator() " technique you are using.

As another example:

template <typename A,typename B> 
vector<B> map(std::function<B (A)> f, vector<A> arr) {
   vector<B> res;
   for (int i=0;i<arr.size();i++) res.push_back(f(arr[i]));
   return res;
}

should be written as:

template<typename expression>
using result = typename std::result_of<expression>::type;
template<typename expression>
using decayed_result = typename std::decay<result<expression>>::type;

template <typename function,typename B> 
vector<decayed_result<function(B)>> map(function&& f, vector<A> const& arr) {
   vector<decayed_result<function(B)>> res;
   res.reserve( arr.size() );
   for (A const& a : arr) {
     res.push_back( f(a) );
   }
   return res;
}

again, result_of is the right solution, not converting things needlessly to std::function .

For fold_right we get:

template<bool b, typename T=void>
using EnableIf = typename std::enable_if<b,T>::type;

template<typename function, typename src, typename dest>
EnableIf<
  std::is_convertible< result<function(src, dest)>, dest >::value,
  std::vector<dest>
>
fold_right( function&& f, std::vector<src> const& v, dest initial )

which again skips any type erasure on f . And if you really want to do type erasure on f , you can do:

template<typename T> struct identity { typedef T type; };
template<typename T> using do_not_deduce = typename identity<T>::type;

template<typename src, typename dest>
std::vector<dest> fold_right( do_not_deduce< std::function<dest(src,dest)> > f, std::vector<src> const& v, dest init );

std::function is a type erasure object. You type erase because you want to use a type somewhere you do not want to carry the type over to. Deducing from a type what kind of resulting type erasure object you should create is almost always the wrong answer, because you have all of the dependency of non-type erased cases, and all of the inefficiency of the type erasure.

make_function 's result is dependent the full type of the source, which makes type erasure output almost completely useless.