标签：attr error: 上下 外部 证明 before erro lock ast
首先，函数里面是可以访问外部变量的
#scope.py def scope_test(): spam = ‘scope_test spam‘ def inner_scope_test(): spam = ‘inner_scope_test spam‘ def do_local(): spam = ‘local spam‘ def do_nonlocal(): nonlocal spam spam = ‘nonlocal spam‘ def do_global(): global spam spam = ‘global spam‘ do_local() print(‘after local assignment:‘, spam) do_nonlocal() print(‘after nonlocal assignment:‘, spam) do_global() print(‘after global assignment‘, spam) inner_scope_test() print(‘after inner_scope_test:‘, spam)
进行如下操作
>>> import scope
>>> scope.scope_test();... print(scope.spam) after local assignment: inner_scope_test spam after nonlocal assignment: nonlocal spam after global assignment nonlocal spam after inner_scope_test: scope_test spam global spam >>> print(spam) Traceback (most recent call last): File "<stdin>", line 1, in <module> NameError: name ‘spam‘ is not defined
- do_local() 对本地变量操作，并未影响外部变量
- do_nonlocal() 似乎是对外一层的 spam 进行了操作
- do_global() 则是对 scope.spam 进行了操作
- 同时可以发现 spam 未定义，这是因为 function 的 global 是相对于其定义所在的模块而言的，官方文档叙述如下
It is important to realize that scopes are determined textually: the global scope of a function defined in a module is that module’s namespace, no matter from where or by what alias the function is called.
若进行如下操作
>>> from scope import scope_test as f >>> f() after local assignment: inner_scope_test spam after nonlocal assignment: nonlocal spam after global assignment nonlocal spam after inner_scope_test: scope_test spam >>> print(spam) Traceback (most recent call last): File "<stdin>", line 1, in <module> NameError: name ‘spam‘ is not defined >>> import scope >>> scope.spam ‘global spam‘
可见即使只 from scope import scope_test as f ，执行 f() 后，当前命名空间仍然没有 spam ,甚至若在此时导入 scope 模块， 会发现存在 scope.spam ， 可见对 f() 而言，global 仍然是相对于 scope 而言的，即使 执行 f() 时 scope 模块并没有导入，为了证明 spam 确实是在执行 f() 时才被引入的，贴上下面的代码
Python 3.6.4 (v3.6.4:d48eceb, Dec 19 2017, 06:04:45) [MSC v.1900 32 bit (Intel)] on win32tel)] on win32 Type "help", "copyright", "credits" or "license" for more information. >>> import scope >>> scope.spam Traceback (most recent call last): File "<stdin>", line 1, in <module> AttributeError: module ‘scope‘ has no attribute ‘spam‘
还有一种情况需要说明，看下面代码
def scope_test(): spam = ‘scope_test spam‘ def inner_scope_test(): spam = ‘inner_scope_test spam‘ def do_local(): ############## print(spam) spam = ‘local spam‘ ############## def do_nonlocal(): nonlocal spam spam = ‘nonlocal spam‘ def do_global(): global spam spam = ‘global spam‘ do_local() print(‘after local assignment:‘, spam) do_nonlocal() print(‘after nonlocal assignment:‘, spam) do_global() print(‘after global assignment‘, spam) inner_scope_test() print(‘after inner_scope_test:‘, spam)
注意区别仅在于 do_local() 函数中，在 spam = ‘local spam‘ 前面加了一句 print(spam) ，本以为 spam 会先输出外部变量，再对本地变量赋值，但事实上被视为语法错误------先使用后定义错误
Python 3.6.4 (v3.6.4:d48eceb, Dec 19 2017, 06:04:45) [MSC v.1900 32 bit (Intel)] on win32Type "help", "copyright", "credits" or "license" for more information.
>>> import scope
>>> scope.scope_test()
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "E:\Coding\Python\scope.py", line 21, in scope_test
inner_scope_test()
File "E:\Coding\Python\scope.py", line 15, in inner_scope_test
do_local()
File "E:\Coding\Python\scope.py", line 7, in do_local
print(spam)
UnboundLocalError: local variable ‘spam‘ referenced before assignment
Python学习小记(3)---scope&namespace
标签：attr error: 上下 外部 证明 before erro lock ast
原文地址：https://www.cnblogs.com/liupy/p/9921980.html
