可变全局变量不会隐藏在python函数中,对吧？

Question

请参阅以下代码:

def good():
  foo[0] = 9         # why this foo isn't local variable who hides the global one

def bad():
  foo    = [9, 2, 3] # foo is local, who hides the global one

for func in [good, bad]:
  foo = [1,2,3]

  print('Before "{}": {}'.format(func.__name__, foo))
  func()
  print('After  "{}": {}'.format(func.__name__, foo))

结果如下:

# python3 foo.py
Before "good": [1, 2, 3]
After  "good": [9, 2, 3]
Before "bad" : [1, 2, 3]
After  "bad" : [1, 2, 3]

Answer 1

因为你没有设置foo,所以你在 foo中得到了一些东西(确切地说是foo [0]).

在bad你创建一个新的变量foo.在good你做的事情foo.set(0, 9)(设置项目0到值9).哪个使用变量,而不是定义新名称.