python局部赋值的规则

WBOY
Release: 2016-06-06 11:27:58
Original
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代码如下:


  myVar = 1

def myfunc():
    myVar += 1

myfunc()
  


   会提示错误:

UnboundlocalError: local variable 'myVar' referenced before assignment

Python提出如下假设:如果在函数体内的任何地方对变量赋值,则Python将名称添加到局部命名空间中。

语句myVar += 1对名称myVar赋值,则myVar是函数myfunc的局部命名空间的一部分,而它当前没有关联值,所以会产生错误。

解决方法:使用global语句

代码如下:


myVar = 1

def myfunc():
    global myVar
    myVar += 1

myfunc()

附作用域搜索规则:

L:本地的(Local)

E:封闭的(Eclosing)

G:全局的(Global)

B:内置的(Built-in) 

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