Why Does Using Lambda Expressions With PyQt Slots Lead to Unexpected Behavior?

Using Lambda Expressions to Connect Slots in PyQt

In PyQt, lambda expressions can be used to connect signals to slots. However, using lambda expressions in certain scenarios can lead to unexpected behavior.

Consider the following code:

class MainWindow(QtGui.QWidget):
    def __init__(self):
        ...
        self.buttons = []
        for idx in [3, 4]:
            button = QtGui.QPushButton('Button {} auto'.format(idx), self)
            button.clicked.connect(lambda x=idx: self.button_pushed(x))
            self.buttons.append(button)
            main_layout.addWidget(button)

When the buttons are clicked, instead of receiving the expected values of 3 and 4, the button_pushed method receives False.

The reason for this behavior is that the clicked signal of QPushButton emits an argument that indicates the button's state. When using a lambda expression in the slot, the assigned value (such as x=idx in this case) is overwritten by the emitted argument.

To resolve this issue, the lambda expression should ignore the button state and use an additional named parameter:

button.clicked.connect(lambda state, x=idx: self.button_pushed(x))

This modification ensures that the correct value is passed to the button_pushed method, regardless of the button's state.

Understanding the mechanism of lambda expressions is also crucial. In the context of slot connections, the lambda function is not evaluated when the signal fires. Instead, a pointer to the function (with the substituted parameters) is connected to the signal. When the signal is emitted, the connected function is called with the arguments that the signal passes.

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