python - PySide Slot Decorator 是必需的吗？

Question

我已经看到了一些使用 @QtCore.Slot 装饰器的 PySide 插槽示例代码，而有些则没有。自己测试了一下，好像没什么区别。我有理由应该或不应该使用它吗？例如，在以下代码中:

import sys
from PySide import QtCore

# the next line seems to make no difference
@QtCore.Slot()
def a_slot(s):
    print s

class SomeClass(QtCore.QObject):
    happened = QtCore.Signal(str)
    def __init__(self):
        QtCore.QObject.__init__(self)
    def do_signal(self):
        self.happened.emit("Hi.")

sc = SomeClass()
sc.happened.connect(a_slot)
sc.do_signal()

@QtCore.Slot 装饰器没有区别；我可以省略它，调用@QtCore.Slot(str)，甚至调用@QtCore.Slot(int)，它仍然很好地表示“嗨”。

PyQt 的 pyqtSlot 似乎也是如此。

Answer 1

This链接解释了关于 pyqtSlot 装饰器的以下内容:

Although PyQt4 allows any Python callable to be used as a slot when connecting signals, it is sometimes necessary to explicitly mark a Python method as being a Qt slot and to provide a C++ signature for it. PyQt4 provides the pyqtSlot() function decorator to do this.

和

Connecting a signal to a decorated Python method also has the advantage of reducing the amount of memory used and is slightly faster.

由于 pyqtSlot 装饰器可以接受额外的参数，例如 name，它允许不同的 Python 方法处理信号的不同签名。

如果您不使用插槽装饰器，则信号连接机制必须手动完成所有类型转换，以从底层 C++ 函数签名映射到 Python 函数。当使用 slot 装饰器时，类型映射可以是显式的。