traits用于测试func（args）是否格式良好并且需要返回类型 [英] traits for testing whether func(args) is well-formed and has required return type

问题描述

有许多类似的问题/答案，但我不能把这些答案在一起为我的目的服务。我想要一个traits

There are a number of similar questions/answers, but I couldn't quite put those answers together to serve my purposes. I want a traits

template<typename Func, typename ReturnType, typename... Args>
struct returns_a { static const bool value; };

以便

returns_a<F,T,Args>::value

c $ c> F（Args）形成良好，并返回 T 。经过一些更多的研究，我得到它的工作如下：

is true if F(Args) is well formed and returns a T. After some more research, I got it working as follows:

// value is true if Func(Args...) is well formed
template<typename Func, typename... Args>
class is_callable
{
  template <typename F>
  static decltype(std::declval<F>()(std::declval<Args>()...), void(), 0) test(int);
  template <typename>
  static void test(...);
public:
  static const bool value = !std::is_void<decltype(test<Func>(0))>::value;
};

// return_type<FunctionSignature>::type is the return type of the Function
template<typename>
struct return_type {};

template<typename ReturnType, typename... Args>
struct return_type<ReturnType(Args...)>
{ typedef ReturnType type; };

// helper class, required to use SFINAE together with variadic templates parameter
// generic case: Func(Args...) is not well-defined
template <typename Func, typename ReturnType, typename dummy, typename... Args>
struct returns_a_helper { static const bool value = false; };

// Func is a function signature
template <typename Func, typename ReturnType, typename... Args>
struct returns_a_helper<Func, ReturnType, typename
                        std::enable_if<std::is_function<Func>::value>::type, Args...>
{
  static const bool value =
    std::is_convertible<typename return_type<Func>::type,
                        ReturnType>::value;

};

// Func(Args...) not a function call, but well-defined 
template <typename Func, typename ReturnType, typename... Args>
struct returns_a_helper<Func,ReturnType,typename
                        std::enable_if<is_callable<Func>::value &&
                       !std::is_function<Func>::value
                                      >::type, Args...>
{
  static const bool value =
    std::is_convertible<typename std::result_of<Func(Args...)>::type,
                ReturnType>::value;
};

template <typename Func, typename ReturnType, typename... Args>
struct returns_a : returns_a_helper<Func, ReturnType, void, Args...> {};

现在可以正常工作和函数。这里是一个简单的测试：

which now works fine for functors and functions. Here is a simple test:

struct base { virtual bool member(int) const = 0; };
struct foo : base { bool member(int x) const { return x&2; } };
struct bar { foo operator()() { return foo(); } };
foo free_function() { return foo(); }

template<typename T, typename Func>
void test(Func const&func)
{
   std::cout << std::boolalpha << returns_a<Func,T>::value << std::endl;
}

int main()
{
   foo x;
   bar m;
   test<const base&>([&]() { return x; });
   test<const base&>(m);
   test<const base&>(free_function);
   return 0;
}

好吧，这个工作，但看起来有点麻烦。

Well, this works, but it seems a bit cumbersome. Anybody has better/more elegant/shorter solutions?

推荐答案

这是很尴尬的：我发现一个相当简单方法：

Well, this is embarrassing: I found a rather simple (hence elegant) way:

template <typename Func, typename ReturnType, typename... Args>
using returns_a = std::is_convertible<Func, std::function<ReturnType(Args...)>>;

< 。

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