Question

当我在阅读关于命名空间和scthon的python时，我读到了

-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

但是当我尝试这个时：

def test() :
    name = 'karim'

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

我收到了这个错误：

UnboundLocalError: local variable 'name' referenced before assignment

当语句打印（名称）时，tata（）函数在当前范围内找不到名称，因此python会在范围内找到并在test（）函数范围内找到它吗？ / p>

Answer 1

在python 3中，您可以使用nonlocal name告诉python name未在本地范围内定义，它应该在外部范围内查找它。

这有效：

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

函数python3里面的范围函数？

1 个答案: