引用文章 如何在lambda中引入递归调用

// clang++ 3.5
// maybe gcc 4.9 support it, but I don't test it
#include<iostream>
int main()
{
    auto fac = [&](auto&& self, int x)->int{
        return x <  ?  : x * self(self, x - );
    };
    std::cout<<fac(fac, )<<std::endl; //
    return ;
}
 
作者：蓝色
链接：https://www.zhihu.com/question/27441424/answer/36643770
来源：知乎
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。

// clang++ 3.5
// maybe gcc 4.9 support it, but I don't test it
#include<iostream>
template<typename Functor>
struct wrapper_type
{
    Functor functor;
    template<typename... Args>
    decltype(auto) operator()(Args&&... args) const&
    {
        return functor(functor, std::forward<Args>(args)...);
    }
};
 
template<typename Functor>
wrapper_type<typename std::decay<Functor>::type> wrapper(Functor&& functor)
{
    return{ std::forward<Functor>(functor) };
}
 
int main()
{
    auto fac = wrapper([&](auto&& self, int x)->int{
        return x <  ?  : x * self(self, x - );
    });
    std::cout << fac() << std::endl; //
    return ;
}

作者：蓝色
链接：https://www.zhihu.com/question/27441424/answer/36643770
来源：知乎
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。

#include <iostream>
#include <typeinfo>
#include <functional>
using namespace std;
int main(void)
{
    std::function<int(int)> product;
    product = [&product](int n) -> int{ return n <= ?  : n * product(n - ); };
    cout << "The answer is: " << product() << endl;
} 
 
链接：https://blog.csdn.net/zenny_chen/article/details/6045596