C++ lambda/回调弹出窗口？

Question

我有一个执行此操作的弹出系统 (GUI):

// creates popup with two possible answer-buttons
if (timeToEat())
  callPopup(ID_4, "What to eat?", "Cake", "Cookies!"); 

//elsewhere in code i check for key presses
if (popupAnswered(ID_4,0)) // clicked first button in popup
  eatCake();

if (popupAnswered(ID_4,1)) // clicked second button in popup
  eatCookiesDamnit();

我可以使用某种 lambda/回调来安排它，如下所示。这样该功能“保留”并且可以在按下按钮时激活(返回值)。

谢谢!

if (timeToEat())
   callPopup("What to eat?", "Cake", "Cookies!"){
       <return was 0 :>  eatCake(); break;
       <return was 1 :>  eatCookies(); break;
}

Answer 1

您可以向 callPopup 添加一个延续参数:

void callPopup(int id, std::function<void(int)> f)
{
    if (something)
        f(0);
    else
        f(1);
}

//...

callPopup(ID_4, [](int x) { if (x == 0) eatCake(); });

或者你可以添加另一个函数层并使用返回值:

std::function<void(std::function<void(int)>)> 
callPopup(int id)
{
        return [](std::function<void(int)> f) { f(something ? 0 : 1); }
}

// ...
callPopup(ID_4)([](int x) { if (x == 0) ... ;});
// or
void popupHandler(int);
auto popupResult = callPopup(ID_4);
// ...
popupResult(popupHandler);