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

 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，怎么样？ 
 
 基本上，我想要使用以下方式的lambda_traits ： 
 
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
 
 动机背后的是我想在函数模板中使用lambda_traits ，它接受一个lambda作为参数，我需要知道它是函数中的参数类型和返回类型： 
 
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!
   //...
}
 
 目前我们可以假设lambda只有一个参数。 
 
 最初，我试图使用std::function作为： 
 
template<typename T>
A<T> f(std::function<bool(T)> fun)
{
   return A<T>(fun);
}

f([](int){return true;}); //error
 
 但这显然会给出错误（ ideone ）。 所以我把它改成了TLambda版本的函数模板，并想在std::function内部构造std::function对象（如上图所示）。 

Given a lambda, is it possible to figure out it's parameter type and return type? If yes, how?
 
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
 
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!
   //...
}
 
For the time being, we can assume that the lambda takes exactly one argument.
 
Initially, I tried to work with std::function as:
 
template<typename T>
A<T> f(std::function<bool(T)> fun)
{
   return A<T>(fun);
}

f([](int){return true;}); //error
 
But it obviously would give error. 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).

原文：https://stackoverflow.com/questions/7943525