exec 语句对 locals() 结果的影响有点不解, 请教一下!

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This topic created in 3285 days ago, the information mentioned may be changed or developed.

x = 30
def out():
    x = 50
    def inner():
        print x
        print locals()
    return inner
out()()    # 此处 print locals()  结果 {'x': 50}

============另一个 Python shell 中

x =30
def out():
    x = 50
    def inner():
        print x
        print locals()
        exec 'print x' in globals(), locals()
    return inner
out()()    # 此处 print locals()  结果 {}

locals

def

inner

2 replies • 2017-10-10 18:14:13 +08:00

babywhisper

Oct 10, 2017

exec 'print x' in locals(), globals()

这句话的意思是:
把 locals()返回的字典, 当做'print x'的全局域(global scope)
把 globals()返回的字典, 作为要执行的'print x'的 local scope

所以 exec 'print x' in globals(), locals() 就相当于 print globals()['x']
打印出的是 30

那么问题来了, 为什么`print locals()`是个空字典呢?
感觉是因为编译的时候?? local scope 被拿去占用了, 原来的被新生成的替代了.
执行完 exec 'print x' in globals(), locals()后, x 也消失了.

不知道楼主有没有什么新发现??

noobsheldon

Oct 10, 2017

@daya0576
x = 30
exec_local = {'x': 100}
def out():
x = 50
def inner():
print x
print locals()
exec 'print x' in globals(), exec_local
print exec_local
print x
return inner

out()() #===>

50
{}
100
{'x': 100}
50