PyQt6中button.clicked.connect绑定Lambda回调输出异常问题

问题场景 #

编写了如下PyQt6计算器代码，预期点击除“C”和“=”之外的按钮时，打印对应的按键内容，但实际点击任意数字或运算符按钮时，始终输出“=”。使用functools.partial绑定回调则功能正常，需要排查回调的问题。

import sys
from functools import partial
from PyQt6.QtCore import Qt
from PyQt6.QtWidgets import (
    QApplication,
    QWidget,
    QMainWindow,
    QGridLayout,
    QVBoxLayout,
    QLineEdit,
    QPushButton
)

ErrorMessage = "ERROR"


class CalculatorUi(QMainWindow):
    def __init__(self):
        super().__init__()
        self.setWindowTitle("Calculator")
        self.setFixedSize(300, 300)
        self.generatLayout = QVBoxLayout()
        central_widget = QWidget()
        central_widget.setLayout(self.generatLayout)
        self.setCentralWidget(central_widget)
        self._createDisplay()
        self._createButtons()

    def _createDisplay(self):
        self.display = QLineEdit()
        self.display.setReadOnly(True)
        self.display.setMaxLength(12)
        self.display.setAlignment(Qt.AlignmentFlag.AlignCenter)
        self.display.setFixedHeight(50)
        self.generatLayout.addWidget(self.display)

    def _createButtons(self):
        self.buttonMap = {}
        buttonLayout = QGridLayout()
        keyboard = [
            ["7", "8", "9", "/", "C"],
            ["4", "5", "6", "*", "("],
            ["1", "2", "3", "-", ")"],
            ["0", "00", ".", "+", "="]
        ]


        for row, keys in enumerate(keyboard):
            for col,key in enumerate(keys):
                self.buttonMap[key] = QPushButton(key)
                self.buttonMap[key].setFixedSize(50, 50)
                buttonLayout.addWidget(self.buttonMap[key], row, col)

        self.generatLayout.addLayout(buttonLayout)

    def _getDisplayText(self):
        return self.display.text()

    def _setText(self, text):
        self.display.setText(text)

    def _clearDisplay(self):
        self.display.setText('')


class Calculate:
    def __init__(self, model, view):
        self._evaluate = model
        self._view = view
        self._connectSignalsAndSLot()


    def _connectSignalsAndSLot(self):
        for key, button in self._view.buttonMap.items():
            if key not in ("C", "="):
                button.clicked.connect(
                    lambda: print(key)
                )



def evaluate_expression(expression):
    try:
        result = str(eval(expression))
    except Exception:
        result = ErrorMessage
    return result


def main():
    app = QApplication([])
    calculator = CalculatorUi()
    calculator.show()
    Calculate(evaluate_expression, calculator)
    sys.exit(app.exec())

if __name__ == "__main__":
    main()

问题根源 #

问题出在lambda的变量延迟绑定特性上：

• 在循环中定义lambda时，lambda并没有直接捕获当前循环迭代的key值，而是保存了对key变量的引用。
• 当按钮被点击时，lambda才会去读取key的值，而此时循环已经执行完毕，key变量最终指向的是循环最后一次迭代的元素——也就是“=”（尽管代码里过滤了C和=，但循环遍历了所有按钮，最后一个元素是“=”，lambda引用的是全局的key变量）。

解决方法 #

方法1：使用functools.partial绑定参数 #

这是你已经验证可行的方式，通过partial将当前循环的key值直接绑定到回调函数中，避免延迟引用：

def _connectSignalsAndSLot(self):
    for key, button in self._view.buttonMap.items():
        if key not in ("C", "="):
            button.clicked.connect(
                partial(print, key)
            )

方法2：给lambda添加默认参数捕获当前值 #

通过给lambda设置默认参数，让每次循环时都将当前的key值赋值给默认参数，形成闭包保存该值：

def _connectSignalsAndSLot(self):
    for key, button in self._view.buttonMap.items():
        if key not in ("C", "="):
            button.clicked.connect(
                lambda k=key: print(k)
            )

内容的提问来源于stack exchange，提问作者mind overflow