代码之家 › 专栏 › 技术社区 › Stefan Monov

将信号连接到插槽,但只调用插槽一次,然后自动断开连接

signals-slots observer-pattern qtquick2 qml qt

Stefan Monov · 技术社区 · 7 年前

function handleSig() {
    emitter.someSig.disconnect(handleSig);
    // do some work here
}

emitter.someSig.connect(handleSig);

理想情况下,我想要这样的东西:

emitter.someSig.connect(
    function() {
        // do some work here
    },
    Qt.SingleShotConnection
);

近似重复: Automatically disconnect after first signal emission -但这个问题是关于Python的,我的问题是关于QML和C++的。

3 回复 | 直到 7 年前

derM - not here for BOT dreams 7 年前

function connectOnce(sig, slot) {
    var f = function() {
        slot.apply(this, arguments)
        sig.disconnect(f)
    }
    sig.connect(f)
}

作为用法演示:

import QtQuick 2.7
import QtQuick.Controls 2.0

ApplicationWindow {
    id: myWindow
    visible: true
    width: 600
    height: 600
    color: 'white'

    signal action(string name)

    function slot(name) {
        console.log(name)
    }

    Button {
        text: 'connect'
        onClicked: {
            connectOnce(action, slot)
        }
    }

    Button {
        y: 80
        text: 'action'
        onClicked: {
            action('test')
        }
    }

    function connectOnce(sig, slot) {
        var f = function() {
            slot.apply(this, arguments)
            sig.disconnect(f)
        }
        sig.connect(f)
    }
}

slot 和 slot2 在单发模式下发送信号 action . 按钮行动 行动

connectOnce 把它放进图书馆,在任何需要的地方都可以使用。

该解决方案很容易扩展为更通用的形式,将连接要执行的函数 n

function connectN(sig, slot, n) {
    if (n <= 0) return
    var f = function() {
        slot.apply(this, arguments)
        n--
        if (n <= 0) sig.disconnect(f)
    }
    sig.connect(f)
}

Stefan Monov 7 年前

我在这里找到了C++的答案,虽然有点不雅观:

https://forum.qt.io/topic/67272/how-to-create-a-single-shot-one-time-connection-to-a-lambda/2

来自该链接的代码:

QMetaObject::Connection * const connection = new QMetaObject::Connection;
*connection = connect(_textFadeOutAnimation, &QPropertyAnimation::finished, [this, text, connection](){
    QObject::disconnect(*connection);
    delete connection;
});

我仍在等待更好的C++答案,以及QML答案。

Michael 3 年前

QObject::connect 呼叫

#ifndef  CONNECT_ONCE_H
# define CONNECT_ONCE_H

# include <QObject>
# include <memory>

template<typename EMITTER, typename SIGNAL, typename RECEIVER, typename... ARGS>
void connectOnce(EMITTER* emitter, SIGNAL signal, RECEIVER* receiver, void (RECEIVER::*slot)(ARGS...), Qt::ConnectionType connectionType = Qt::AutoConnection)
{
  auto connection = std::make_shared<QMetaObject::Connection>();
  auto onTriggered = [connection, receiver, slot](ARGS... arguments){
    (receiver->*slot)(arguments...);
    QObject::disconnect(*connection);
  };

  *connection = QObject::connect(emitter, signal, receiver, onTriggered, connectionType);
}

template<typename EMITTER, typename SIGNAL, typename RECEIVER, typename SLOT, typename... ARGS>
void connectOnce(EMITTER* emitter, SIGNAL signal, RECEIVER* receiver, SLOT slot, Qt::ConnectionType connectionType = Qt::AutoConnection)
{
  std::function<void (ARGS...)> callback = slot;
  auto connection = std::make_shared<QMetaObject::Connection>();
  auto onTriggered = [connection, callback](ARGS... arguments) {
    callback(arguments...);
    QObject::disconnect(*connection);
  };

  *connection = QObject::connect(emitter, signal, receiver, onTriggered, connectionType);
}

#endif

关于基于lambda的模板,我无法在直接声明时编译它 std::function 作为一个参数,因为编译器无法将lambda参数识别为有效的候选参数……这就是我声明的原因 SLOT 作为模板参数,然后将其强制转换为 std::函数 .虽然没有我想要的那么简洁,但它很管用。