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;
}
最佳答案
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);
关于C++ lambda/回调弹出窗口?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/47901447/