测试lambda是否是无状态的？ [英] Test if a lambda is stateless?

问题描述

如果lambda是无状态的，如果它捕获任何东西，我该如何测试？
我的猜测是使用重载解析与函数指针重载，还是模板特化？

How would I go about testing if a lambda is stateless, that is, if it captures anything or not? My guess would be using overload resolution with a function pointer overload, or template specialization?

int a;
auto l1 = [a](){ return 1; };
auto l2 = [](){ return 2; };
// test l1 and l2, get a bool for statelessness.

推荐答案

根据标准，捕获任何变量，然后可以隐式转换为函数指针。

As per the Standard, if a lambda doesn't capture any variable, then it is implicitly convertible to function pointer.

基于这一点，我想出了 is_stateless< $ c>元函数，它告诉您lambda是否是无状态的

Based on that, I came up with is_stateless<> meta-function which tells you whether a lambda is stateless or not.

#include <type_traits>

template <typename T, typename U>
struct helper : helper<T, decltype(&U::operator())>
{};

template <typename T, typename C, typename R, typename... A>
struct helper<T, R(C::*)(A...) const> 
{
    static const bool value = std::is_convertible<T, R(*)(A...)>::value;
};

template<typename T>
struct is_stateless
{
    static const bool value = helper<T,T>::value;
};

这里是测试代码：

int main() 
{
    int a;
    auto l1 = [a](){ return 1; };
    auto l2 = [](){ return 2; };
    auto l3 = [&a](){ return 2; };

    std::cout<<std::boolalpha<<is_stateless<decltype(l1)>::value<< "\n";
    std::cout<<std::boolalpha<<is_stateless<decltype(l2)>::value<< "\n";
    std::cout<<std::boolalpha<<is_stateless<decltype(l3)>::value<< "\n";
}

输出：

false
true
false

在线演示。

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