在Python中，为什么list []是自动全局的? [英] In Python, why is list[] automatically global?

问题描述

这是一个奇怪的行为.

尝试一下:

rep_i=0
print "rep_i is" , rep_i
def test():
  global rep_i #without Global this gives error but list , dict , and others dont
  if rep_i==0:
    print "Testing Integer %s" % rep_i
    rep_i=1
  return "Done"

rep_lst=[1,2,3]


def test2():
  if rep_lst[0]==1:
    print "Testing List %s" % rep_lst
  return "Done"


if __name__=="__main__":
  test()
  test2()

为什么列表不需要全局声明?它们是自动全局的吗?

Why list do not need to declare global? are they automatically global?

我发现它真的很奇怪，我大部分时间都使用列表，甚至根本不使用global作为全局.....

I find it really wierd , i use list most of the time and i dont even use global at all to us them as global .....

推荐答案

它不是自动全局的.

但是，rep_i=1和rep_lst[0]=1之间有区别-前者重新绑定了名称rep_i，因此需要global来防止创建相同名称的本地插槽.在后一种情况下，您只是在修改现有的全局对象，该对象可以通过常规名称查找找到(更改列表条目就像在列表上调用成员函数一样，而不是重新绑定名称).

However, there's a difference between rep_i=1 and rep_lst[0]=1 - the former rebinds the name rep_i, so global is needed to prevent creation of a local slot of the same name. In the latter case, you're just modifying an existing, global object, which is found by regular name lookup (changing a list entry is like calling a member function on the list, it's not a name rebinding).

要对其进行测试，请尝试在test2中分配rep_lst=[](即，将其设置为新列表).除非您声明rep_lst global，否则在test2外部将看不到效果，因为会创建相同名称的本地插槽并遮盖全局插槽.

To test it out, try assigning rep_lst=[] in test2 (i.e. set it to a fresh list). Unless you declare rep_lst global, the effects won't be visible outside test2 because a local slot of the same name is created and shadows the global slot.

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