Question

这似乎是this question中lambda的情况所解决的。但那个是2011年的答案，我正在寻找一个普遍的案例：lambdas，常规函数和函子。并且，如果可能的话，通过最现代的c ++语言添加。（注意：g++ -std=c++1y test.cpp）。

所以，给定一个函数（或一个lambda），我试图找出它的返回类型是什么。例如，声明一个变量（简化的情况）。

using namespace std;

template<typename F>
void test (F h) {
  // any of the following should be equivalent to int a; int b; int c;
  decltype(h) a; // <<<<< wrong
  result_of(h) b; // <<<<<< wrong
  result_of<decltype(h)> c; // <<<<<< wrong
}

int a_function (int i)  { 
  return 2*i; 
}

int main ()  {
  test (a_function);
}

感谢。

Answer 1

std::result_of<h(int)>::type a;

（取自here。）

Answer 2

假设：

您只想要返回类型。
您不知道参数类型是什么/将是什么（因此decltype()和std::result_of<>都不是一个选项。
作为参数传递的仿函数对象不会有重载或通用operator()。

然后你可以使用下面的特性来推断任何仿函数对象的返回类型：

template <typename F>
struct return_type_impl;

template <typename R, typename... Args>
struct return_type_impl<R(Args...)> { using type = R; };

template <typename R, typename... Args>
struct return_type_impl<R(Args..., ...)> { using type = R; };

template <typename R, typename... Args>
struct return_type_impl<R(*)(Args...)> { using type = R; };

template <typename R, typename... Args>
struct return_type_impl<R(*)(Args..., ...)> { using type = R; };

template <typename R, typename... Args>
struct return_type_impl<R(&)(Args...)> { using type = R; };

template <typename R, typename... Args>
struct return_type_impl<R(&)(Args..., ...)> { using type = R; };

template <typename R, typename C, typename... Args>
struct return_type_impl<R(C::*)(Args...)> { using type = R; };

template <typename R, typename C, typename... Args>
struct return_type_impl<R(C::*)(Args..., ...)> { using type = R; };

template <typename R, typename C, typename... Args>
struct return_type_impl<R(C::*)(Args...) &> { using type = R; };

template <typename R, typename C, typename... Args>
struct return_type_impl<R(C::*)(Args..., ...) &> { using type = R; };

template <typename R, typename C, typename... Args>
struct return_type_impl<R(C::*)(Args...) &&> { using type = R; };

template <typename R, typename C, typename... Args>
struct return_type_impl<R(C::*)(Args..., ...) &&> { using type = R; };

template <typename R, typename C, typename... Args>
struct return_type_impl<R(C::*)(Args...) const> { using type = R; };

template <typename R, typename C, typename... Args>
struct return_type_impl<R(C::*)(Args..., ...) const> { using type = R; };

template <typename R, typename C, typename... Args>
struct return_type_impl<R(C::*)(Args...) const&> { using type = R; };

template <typename R, typename C, typename... Args>
struct return_type_impl<R(C::*)(Args..., ...) const&> { using type = R; };

template <typename R, typename C, typename... Args>
struct return_type_impl<R(C::*)(Args...) const&&> { using type = R; };

template <typename R, typename C, typename... Args>
struct return_type_impl<R(C::*)(Args..., ...) const&&> { using type = R; };

template <typename R, typename C, typename... Args>
struct return_type_impl<R(C::*)(Args...) volatile> { using type = R; };

template <typename R, typename C, typename... Args>
struct return_type_impl<R(C::*)(Args..., ...) volatile> { using type = R; };

template <typename R, typename C, typename... Args>
struct return_type_impl<R(C::*)(Args...) volatile&> { using type = R; };

template <typename R, typename C, typename... Args>
struct return_type_impl<R(C::*)(Args..., ...) volatile&> { using type = R; };

template <typename R, typename C, typename... Args>
struct return_type_impl<R(C::*)(Args...) volatile&&> { using type = R; };

template <typename R, typename C, typename... Args>
struct return_type_impl<R(C::*)(Args..., ...) volatile&&> { using type = R; };

template <typename R, typename C, typename... Args>
struct return_type_impl<R(C::*)(Args...) const volatile> { using type = R; };

template <typename R, typename C, typename... Args>
struct return_type_impl<R(C::*)(Args..., ...) const volatile> { using type = R; };

template <typename R, typename C, typename... Args>
struct return_type_impl<R(C::*)(Args...) const volatile&> { using type = R; };

template <typename R, typename C, typename... Args>
struct return_type_impl<R(C::*)(Args..., ...) const volatile&> { using type = R; };

template <typename R, typename C, typename... Args>
struct return_type_impl<R(C::*)(Args...) const volatile&&> { using type = R; };

template <typename R, typename C, typename... Args>
struct return_type_impl<R(C::*)(Args..., ...) const volatile&&> { using type = R; };

template <typename T, typename = void>
struct return_type
    : return_type_impl<T> {};

template <typename T>
struct return_type<T, decltype(void(&T::operator()))>
    : return_type_impl<decltype(&T::operator())> {};

template <typename T>
using return_type_t = typename return_type<T>::type;

试验：

#include <type_traits>

template <typename F>
void test(F h)
{
    static_assert(std::is_same<return_type_t<F>, int>{}, "!");

    return_type_t<F> i = 1;
}

int function(int i) { return 2*i; }

int c_varargs_function(...) { return 1; }

struct A
{
    int mem_function(double, float) { return 1; } 
};

int main()
{
    // Function
    test(function);

    // C-style variadic function
    test(c_varargs_function);

    // Non-generic lambda
    test([](int i) { return 2*i; });

    // Member function
    test(&A::mem_function);
}

DEMO

找出函数，lambda或函数的返回类型

2 个答案: