C lambda/回调弹出窗口

C++ lambda/callback popup?

本文关键字：窗口 回调 lambda      更新时间：2023-10-16

我有一个这样做的弹出系统（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/回调来安排像以下内容一样。因此，将函数"保留"，并在按下按钮时可以激活（返回值）。

谢谢！

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

您可以将持续参数添加到callPopup：

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(); });

，也可以添加另一个功能层并使用返回值：

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);

您可以将选择与操作相关联，然后执行与单击

关联的操作 
using PopupActions = std::map<std::string, std::function<void()>>;
void callPopup(int id, std::string prompt, PopupActions actions)
{
    for (auto & pair : actions)
        // create button with text from pair.first and on-click from pair.second
    // show buttons
}
if (timeToEat())
   callPopup(ID_4, "What to eat?", { 
       { "Cake!", [this]{ eatCake(); } }
       { "Cookies!", [this]{ eatCookies(); } }
   });
}
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