是否可以找出lambda的参数类型和返回类型？ [英] Is it possible to figure out the parameter type and return type of a lambda?

问题描述

给定一个lambda，是否可以找出它的参数类型和返回类型？如果是，如何？

Given a lambda, is it possible to figure out it's parameter type and return type? If yes, how?

基本上，我想要 lambda_traits ，可以通过以下方式使用：

Basically, I want lambda_traits which can be used in following ways:

auto lambda = [](int i) { return long(i*10); };

lambda_traits<decltype(lambda)>::param_type  i; //i should be int
lambda_traits<decltype(lambda)>::return_type l; //l should be long

后面的动机是我想使用 lambda_traits 在接受lambda作为参数的函数模板中，我需要知道它的参数类型和函数中的返回类型：

The motivation behind is that I want to use lambda_traits in a function template which accepts a lambda as argument, and I need to know it's parameter type and return type inside the function:

template<typename TLambda>
void f(TLambda lambda)
{
   typedef typename lambda_traits<TLambda>::param_type  P;
   typedef typename lambda_traits<TLambda>::return_type R;

   std::function<R(P)> fun = lambda; //I want to do this!
   //...
}

暂时，假设lambda只需要一个参数。

For the time being, we can assume that the lambda takes exactly one argument.

最初，我尝试使用 std :: function

template<typename T>
A<T> f(std::function<bool(T)> fun)
{
   return A<T>(fun);
}

f([](int){return true;}); //error

但它显然会给出错误（ideone ）。所以我把它改成了 TLambda 版本的函数模板，并希望在函数中构造 std :: function （如上所示）。

But it obviously would give error (ideone). So I changed it to TLambda version of the function template and want to construct the std::function object inside the function (as shown above).

推荐答案

有趣的是，我刚写了一个 function_traits com / questions / 2562320 / specializing-a-template-on-a-lambda-in-c0x>在C ++ 0x 中专门用于lambda的模板，它可以提供参数类型。这个问题的答案是使用lambda的 operator（）的 decltype 。

Funny, I've just written a function_traits implementation based on Specializing a template on a lambda in C++0x which can give the parameter types. The trick, as described in the answer in that question, is to use the decltype of the lambda's operator().

template <typename T>
struct function_traits
    : public function_traits<decltype(&T::operator())>
{};
// For generic types, directly use the result of the signature of its 'operator()'

template <typename ClassType, typename ReturnType, typename... Args>
struct function_traits<ReturnType(ClassType::*)(Args...) const>
// we specialize for pointers to member function
{
    enum { arity = sizeof...(Args) };
    // arity is the number of arguments.

    typedef ReturnType result_type;

    template <size_t i>
    struct arg
    {
        typedef typename std::tuple_element<i, std::tuple<Args...>>::type type;
        // the i-th argument is equivalent to the i-th tuple element of a tuple
        // composed of those arguments.
    };
};

// test code below:
int main()
{
    auto lambda = [](int i) { return long(i*10); };

    typedef function_traits<decltype(lambda)> traits;

    static_assert(std::is_same<long, traits::result_type>::value, "err");
    static_assert(std::is_same<int, traits::arg<0>::type>::value, "err");

    return 0;
}

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