python传递列表作为函数参数

folks,

乡亲，

The result of the following code is [] Why is it not ['0','1','2']? If I want to make psswd equal to number in the function foo, what should I do?

以下代码的结果是[]为什么不是['0'，'1'，'2']？如果我想让psswd等于函数foo中的数字，我该怎么办？

number = ['0','1','2']
def foo(psswd):
    psswd = number[:]

if __name__ == '__main__':
    psswd = []
    foo(psswd)
    print psswd

3 个解决方案

#1

Your code:

你的代码：

number = ['0','1','2']
def foo(psswd):
    psswd = number[:]

if __name__ == '__main__':
    psswd = []
    foo(psswd)
    print psswd

psswd = number[:] rebinds local variable psswd with a new list.

psswd = number [：]用新列表重新绑定局部变量psswd。

I.e., when you do foo(psswd), function foo is called, and local variable passwd inside it is created which points to the global list under the same name.

即，当你执行foo（psswd）时，会调用函数foo，并在其中创建局部变量passwd，该变量指向同名的全局列表。

When you do psswd = <something> inside foo function, this <something> is created/got and local name psswd is made to point to it. Global variable psswd still points to the old value.

当你在foo函数中执行psswd = 时，会创建/获取此，并使本地名称psswd指向它。全局变量psswd仍指向旧值。

If you want to change the object itself, not the local name, you must use this object's methods. psswd[:] = <smething> actually calls psswd.__setitem__ method, thus the object which is referenced by local name psswd is modified.

如果要更改对象本身而不是本地名称，则必须使用此对象的方法。 psswd [：] = 实际上调用了psswd .__ setitem__方法，因此修改了本地名称psswd引用的对象。

#2

You need to mutate instead of rebinding, with slice-assigning.

您需要使用切片分配进行变异而不是重新绑定。

psswd[:] = number[:]

#3

number = ['0','1','2']
def foo(psswd):
    psswd[:] = number  # swap the slice around here

if __name__ == '__main__':
    psswd = []
    foo(psswd)
    print psswd

#1