捕获的静态断言 lambda

Question

我目前正在学习并行编程和元编程，并且我很难对捕获变量的 lambda 进行编译时检查。

假设我们有这样的代码：

#include <type_traits>
#include <functional>
#include <iostream>

int main(){

    auto non_capture = [](){
        std::cout<<"this lambda is not capturing!"<<std::endl;
    };

    auto capture = [&](){
        std::cout<<"this lambda is capturing"<<std::endl;
    };

    //value is true for this one
    static_assert(std::is_convertible<decltype(non_capture), std::function<void()>>::value, "message1");

    //i need help for this
    static_assert(is_lambda_that_captures<decltype(capture)>::value, "this is some error message");
}

我需要帮助来定义is_lambda_that_captures类型特征，以便我可以传递第二个静态断言。我该怎么做？

拥有最新标准的解决方案是非常受欢迎的（当前最新的是 c++17）:)

Answer 1

这是我的解决方案。它依赖于两个原则：

1. 非捕获 lambda 可转换为函数指针。

2. 所有 lambda 表达式均可转换为 std::function。

 

#include <type_traits>
#include <functional>
#include <iostream>

template<class T, class Ret, class...Args>
struct is_non_capture_lambda : std::is_convertible<T, Ret (*)(Args...)>
{

};

template<class T, class Ret, class...Args>
struct is_capture_lambda
{
    static constexpr auto convertible = std::is_convertible<T, std::function<Ret(Args...)>>::value;
    using type = bool;
    static constexpr auto value = convertible && !is_non_capture_lambda<T, Ret, Args...>::value;


};

int main(){

    auto non_capture = [](){
        std::cout<<"this lambda is not capturing!"<<std::endl;
    };

    auto capture = [&](){
        std::cout<<"this lambda is capturing"<<std::endl;
    };

    //value is true for this one
    static_assert(is_non_capture_lambda<decltype(non_capture), void>::value, "message1");

    //i need help for this
    static_assert(is_capture_lambda<decltype(capture), void>::value, "this is some error message");
}