How to prevent change the status of pressed on a checkbutton?

Question

func _on_CheckButton_toggled(button_pressed):
    if button_pressed:
        $CheckButton.set_pressed(false)
    else:
        $CheckButton.set_pressed(true)

This function is connected with the toggled signal of the checkbutton. But when I click on the checkbutton, this error happens on the debugger:

Stack Overflow (Stack Size: 1024)

Answer 1

Solved by me!

func _process(delta):
    for i in self.get_children():
        if i is CheckButton:
            if $Label_Status_Valor.get_text() == "STATUS_PENDING_VALUE":
                i.set_disabled(false)
            else:
                i.set_disabled(true)

Answer 2

Stack overflow means that there is almost endless loop somewhere in code. Setting pressed to false in code triggers toggled signal again, which sets pressed to false and so on. Easiest way out is to use BUTTONPRESSED signal instead and setdeferred("pressed",false). Buttonpressed signal should be emitted once on being clicked, buttontoggled emits every time its state is changed from pressed to unpressed and vice versa.