Qt-如何使用不兼容的信号和插槽之间的连接 [英] Qt- How to use connect between incompatible signal and slot

问题描述

我想做一个游戏。在这里我想调用一个函数，该函数将接收一个点。但是这个函数应该由定时器的超时信号调用。任何人都可以说如何实现这一点。以下是必需的/错误代码

I am trying to do a game. In this I want to call a function, that function will receive a point . But this function should be invoked by a timer's timedout signal. Can anybody say how to achieve this. Below is the required/error code

Point p(a,b);

connect(timer,SIGNAL(timedout()),this,startDestruction(p));

任何人都可以说如何实现这一点？

Can anybody say how to achieve this?

推荐答案

创建一个方法作为中间插槽使用。

Create a method to use as an intermediate slot, like this.

Point p(a,b);
connect(timer, SIGNAL(timedout()), this, q);

q() {
    startDestruction(this->p)
}

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