How to Correctly Connect Slots with Lambda Functions in PyQt?

Connecting Slots with Lambda Functions in PyQt

When attempting to connect slots with lambda functions, it's crucial to understand how arguments are handled. In the code sample provided, the issue arises when connecting multiple buttons in a loop.

In the first two buttons, the lambda expressions are correctly defined and connected:

button_1.clicked.connect(lambda x:self.button_pushed(1))
button_2.clicked.connect(lambda x:self.button_pushed(2))

For the buttons connected within the loop, the problem lies in the lambda expression:

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

In this case, the QPushButton.clicked signal emits an argument indicating the state of the button. However, when connecting to the lambda slot, the optional argument idx is overwritten by the button's state.

To resolve this, the connection should be defined as follows:

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

This ignores the button state and passes the correct value to the button_pushed method.

In summary, it's essential to be mindful of the arguments passed to lambda functions when connecting slots. By properly handling the emitted signal's arguments, you can ensure that the correct values are passed to your methods.

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