如何在PyQT中连接不同类的pyqtSignal

Question

如何在两个不同的对象（类）之间正确地连接 pyqtSignal？我想知道最佳实践是什么。

看看我为实现这个目标所做的：当 Pot 的温度升高时，Thermometer 类会收到通知：

from PyQt4 import QtCore

class Pot(QtCore.QObject):
    temperatureRaisedSignal = QtCore.pyqtSignal()

    def __init__(self, parent=None):
        super(Pot, self).__init__(parent)
        self.temperature = 1
    def Boil(self):
        self.temperature += 1
        self.temperatureRaisedSignal.emit()
    def RegisterSignal(self, obj):
        self.temperatureRaisedSignal.connect(obj)

class Thermometer():
    def __init__(self, pot):
        self.pot = pot
        self.pot.RegisterSignal(self.temperatureWarning)
    def StartMeasure(self):
        self.pot.Boil()
    def temperatureWarning(self):
        print("Too high temperature!")

if __name__ == '__main__':
    pot = Pot()
    th = Thermometer(pot)
    th.StartMeasure()

或者有没有更简单或更好的方法来做到这一点？

我还希望（如果可能的话）使用“新”风格的 PyQt 信号。

Answer 1

from PyQt4 import QtCore

class Pot(QtCore.QObject):

    temperatureRaisedSignal = QtCore.pyqtSignal()

    def __init__(self, parent=None):
        QtCore.QObject.__init__(self)
        self.temperature = 1

    def Boil(self):
        self.temperatureRaisedSignal.emit()
        self.temperature += 1

class Thermometer():
    def __init__(self, pot):
        self.pot = pot
        self.pot.temperatureRaisedSignal.connect(self.temperatureWarning)

    def StartMeasure(self):
        self.pot.Boil()

    def temperatureWarning(self):
        print("Too high temperature!")

if __name__ == '__main__':
    pot = Pot()
    th = Thermometer(pot)
    th.StartMeasure()

这是我根据文档的说明，应该这样做的方式：
http://www.riverbankcomputing.com/static/Docs/PyQt4/html/new_style_signals_slots.html