Python闭包中的可变绑定时间

Question

从Python的Programming FAQ来看，这并不像许多人期望的那样有效：

>>> squares = []
>>> for x in range(5):
...     squares.append(lambda: x**2)
>>> squares[2]() # outputs 16, not 4

提供以下解释：

This happens because x is not local to the lambdas, but is defined in the outer scope, and it is accessed when the lambda is called --- not when it is defined.

这很好。然而，对我来说，这个解释听起来并不完全正确。考虑这一点：

def buildFunction():
    myNumber = 6 # Won't compile without this line
    def fun():
        return myNumber
    return fun

def main():
    myNumber = 3 # Ignored
    myFun = buildFunction()
    myNumber = 3 # Ignored
    print(myFun()) # prints 6, not 3

这是有道理的，但是有人能提供一个更准确/通用的定义来说明何时发生Python闭包的变量绑定吗

Answer 1

考虑如下：

def main():
    def buildFunction():
        def fun():
            return myNumber

        return fun

    myNumber = 3  # can use this...
    myFun = buildFunction()
    myNumber = 3  # ... or this
    print(myFun())  # prints 3 this time

这与lambda示例更具可比性，因为闭包函数嵌套在声明感兴趣变量的范围内。您的示例有两个不同的作用域，因此有两个完全不相关的myNumbers

如果您没有遇到nonlocal关键字，您能猜出这将打印什么吗

def main():
    def buildFunction():
        nonlocal myNumber
        myNumber = 6

        def fun():
            return myNumber

        return fun

    myNumber = 3
    myFun = buildFunction()
    # myNumber = 3
    print(myFun())