如何为lambda分配重载operator =? [英] How can I overload operator= for lambda assignments?

问题描述

我正在尝试将一个函数插入到映射中，但我想先检查一下，所以我想重载std :: function的赋值操作，这可能吗?

我尝试重载赋值操作，因此，如果分配了预期以外的内容，则赋值运算符函数应将其包装在预期函数中并返回.

I am trying to insert a function into a map but I want to check it first so I would like to overload assignment operation for std::function, is this possible?

I try to overload the assignment operation, so if something other than expected is assigned, the assignment operator function should wrap it in the expected function and return it.

#include <iostream>
#include <map>
#include <functional>

class MyClass{
    public:
    std::map<int, std::map<int, std::function<void(int,int)>>> events;
    std::function<void(int,int)>& on(int type, int id){ return events[type][id]; };
    template<typename T> std::function<void(int,int)>& operator= (T&& fn){
        std::wcout << L"assigning correct function\n";
        return [&](int x, int y){
            if(typeid(fn)==typeid(std::function<void(int,std::wstring)>)) fn(x, L"two");
        };
    }
};

int main(int argc, char **argv)
{
    MyClass obj;
    obj.on(1,2) = [](int x, int y){ std::wcout << L"int " << x << L" " << y << std::endl; };  //this works but it's not calling the overload operator
    obj.on(1,2) = [](int x, std::wstring y){ std::wcout << L"string " << x << L" " << y << std::endl; }; //I need this to work too
    obj.events[1][2](2,3);
    return 0;
}

输出:

test.cpp:23:14: error: no match for 'operator=' (operand types are 'std::function<void(int, int)>' and 'main(int, char**)::<lambda(int, std::__cxx11::wstring)>')
obj.on(1,2) = [](int x, std::wstring y){ std::wcout << L"string " << x << L" " << y << std::endl; };
         ^

推荐答案

听起来像您需要的是代理类.问题是，当您从on()返回std::function<..>&时，最终会得到std::function.您不能覆盖该类的operator=，这是我认为您正在尝试做的.相反，您要覆盖MyClass::operator=-这是您从未真正调用过的函数.

Sounds like what you need is a proxy class. The problem is, when you return a std::function<..>& from on(), you end up with a std::function. You can't overwrite that class's operator=, which is what I think you're trying to do. Instead, you're overwriting MyClass::operator= - which is a function you're never actually calling.

相反，返回一个您可以控制的代理.像这样:

Instead, return a proxy whose assignment you can control. Something like this:

struct Proxy {
    std::function<void(int, int)>& f;
};

Proxy on(int type, int id){ return {events[type][id]}; };

然后我们可以为Proxy::operator=提供特殊的重载. 有效，正确的类型"情况:

And then we can provide special overloads for Proxy::operator=. The "valid, correct type" case:

template <typename F,
          std::enable_if_t<std::is_assignable<std::function<void(int, int)>&, F&&>::value>* = nullptr>
Proxy& operator=(F&& func) {
    f = std::forward<F>(func);
    return *this;
}

和wstring情况:

template <typename F,
          std::enable_if_t<std::is_assignable<std::function<void(int, std::wstring)>&, F&&>::value>* = nullptr>
Proxy& operator=(F&& func) {
    std::wcout << L"assigning correct function\n";
    f = [func = std::forward<F>(func)](int x, int ) {
        func(x, L"two");
    };
    return *this;
}

有了它，您原来的main()将会编译并执行您期望的操作.

With that, your original main() will compile and do what you expect.

这篇关于如何为lambda分配重载operator =?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！