范围函数里面的函数python3？

Question

，而我是阅读有关的名字空间和蟒蛇scopping，我读了

-the innermost scope, which is searched first, contains the local names 
-the scopes of any enclosing functions, which are searched starting with the nearest enclosing scope, contains non-local, but also non-global names 
-the next-to-last scope contains the current module’s global names 
-the outermost scope (searched last) is the namespace containing built-in names 

，但是当我尝试这样做：

def test() : 
    name = 'karim' 

    def tata() : 
     print(name) 
     name='zaka' 
     print(name) 
    tata() 

我收到此错误：

UnboundLocalError: local variable 'name' referenced before assignment 

时声明print（name） tata（）函数在当前范围中找不到名称，所以python会在范围内并在test（）函数范围内找到它没有？

Answer 1

在Python 3中，您可以使用nonlocal name来告诉python name未在本地作用域中定义，它应该在外部作用域中查找它。

这工作：

def tata(): 
    nonlocal name 
    print(name) 
    name='zaka' 
    print(name)