如何在推导上下文中将 std::bind（或 lambda）转换为 std::function？

Question

我正在尝试将指向成员函数的指针（使用 std::bind 或 lambda）转换为 std::function。我的尝试（根据的回答）看起来像这样：

#include <functional>

template<typename T>
struct AsFunction :
  AsFunction<decltype(&T::operator())>
{};

template<class ReturnType, class... Args>
struct AsFunction<ReturnType(Args...)> {
  using type = std::function<ReturnType(Args...)>;
};

template<class ReturnType, class... Args>
struct AsFunction<ReturnType(*)(Args...)> {
  using type = std::function<ReturnType(Args...)>;
};

template<class Class, class ReturnType, class... Args>
struct AsFunction<ReturnType(Class::*)(Args...) const> {
  using type = std::function<ReturnType(Args...)>;
};

template<class F>
auto toFunction( F f ) -> typename AsFunction<F>::type {
  return {f};
}

struct MyStruct {
  int x,y;
  void f(int){};
};

int main(){
  MyStruct m;

  {
    // this works
    auto f = std::bind(&MyStruct::f, &m, std::placeholders::_1);
    f(2);
  }

  {
    // this doesn't
    auto f = toFunction(std::bind(&MyStruct::f, &m, std::placeholders::_1));
    f(2);                                                                                                                                                                                                                                                                      
  }

  { 
    // .. neither does this
    auto f = toFunction([m](int x) mutable { m.f(x); });
    f(2);
  } 
}

但我从编译器收到以下错误消息：

// first not working 
main.cpp:24:6: note:   substitution of deduced template arguments resulted in errors seen above
main.cpp: In instantiation of ‘struct AsFunction<std::_Bind<std::_Mem_fn<void (MyStruct::*)(int)>(MyStruct*, std::_Placeholder<1>)> >’:
main.cpp:24:6:   required by substitution of ‘template<class F> typename AsFunction<F>::type toFunction(F) [with F = std::_Bind<std::_Mem_fn<void (MyStruct::*)(int)>(MyStruct*, std::_Placeholder<1>)>]’
main.cpp:44:75:   required from here
main.cpp:4:8: error: decltype cannot resolve address of overloaded function
 struct AsFunction :
        ^~~~~~~~~~
main.cpp: In function ‘int main()’:
main.cpp:44:75: error: no matching function for call to ‘toFunction(std::_Bind_helper<false, void (MyStruct::*)(int), MyStruct*, const std::_Placeholder<1>&>::type)’
     auto f = toFunction(std::bind(&MyStruct::f, &m, std::placeholders::_1));
                                                                           ^
main.cpp:24:6: note: candidate: template<class F> typename AsFunction<F>::type toFunction(F)
 auto toFunction( F f ) -> typename AsFunction<F>::type {
      ^~~~~~~~~~
main.cpp:24:6: note:   substitution of deduced template arguments resulted in errors seen above

// second non working braces with lambda
main.cpp: In instantiation of ‘struct AsFunction<void (main()::<lambda(int)>::*)(int)>’:
main.cpp:4:8:   required from ‘struct AsFunction<main()::<lambda(int)> >’
main.cpp:24:6:   required by substitution of ‘template<class F> typename AsFunction<F>::type toFunction(F) [with F = main()::<lambda(int)>]’
main.cpp:50:55:   required from here
main.cpp:5:23: error: ‘operator()’ is not a member of ‘void (main()::<lambda(int)>::*)(int)’
   AsFunction<decltype(&T::operator())>
                       ^~
main.cpp:50:55: error: no matching function for call to ‘toFunction(main()::<lambda(int)>)’
     auto f = toFunction([m](int x) mutable { m.f(x); });
                                                       ^
main.cpp:24:6: note: candidate: template<class F> typename AsFunction<F>::type toFunction(F)
 auto toFunction( F f ) -> typename AsFunction<F>::type {
      ^~~~~~~~~~
main.cpp:24:6: note:   substitution of deduced template arguments resulted in errors seen above

Answer 1

我认为你有点矫枉过正了。如果你真的想要一个明确的 std::function 类型作为你的 f，你可以直接用一些技巧来指定它：

int main(){
  MyStruct m;

  {
    std::function<decltype(std::declval<MyStruct>().f(int()))(int)> f(std::bind(&MyStruct::f, &m, std::placeholders::_1));
    f(2);                                                                                                                                                                                                                                                                      
  }
}

话虽如此，到目前为止最简单的事情就是简单地使用 auto f 而不指定类型。我假设您已经想到了这一点，但在您的情况下不会这样做，但我还是以防万一。

Answer 2

很难从 lambda 或 std::bind return 类型中推断出参数，因此您可能希望延迟 toFunction 内的绑定，例如有：

template<typename C, typename Ret, typename... Args, typename... BArgs>
auto toFunction(Ret (C::*f)(Args...), BArgs&&... bargs) {
  return std::function<Ret(Args...)>(std::bind(f, std::forward<BArgs>(bargs)...));
}

这样，您可以使用以下方法检索 std::function：

auto f = toFunction(&MyStruct::f, &m, std::placeholders::_1);
f(2);

如何在推导上下文中将 std::bind（或 lambda）转换为 std::function？

How to convert std::bind (or lambda) to std::function in deduced context?

c++

lambda

template-meta-programming

c++11

std-function