如何更改其他函数的locals（）结果？

Question

我有这样的代码：

def f():
    i = 5
    g(locals())
    print 'in f:', i, j

def g(env):
    env['j'] = env['i'] + 1
    print 'in g:', env['i'], env['j']

f()

我得到：

in g: 5 6
in f: 5---------------------------------------------------------------------------
NameError                                 Traceback (most recent call last)
NameError: global name 'j' is not defined

似乎g不能改变f中locals（）所得到的局部变量。还有其他函数中的局部变量可以改变吗？你知道吗

Answer 1

根据docs关于locals：

Note The contents of this dictionary should not be modified; changes may not affect the values of local and free variables used by the interpreter.

另一种解决方案是只返回j。你知道吗

def f():
    i = 5
    j = g(locals())
    print 'in f:', i, j

def g(env):
    j = env['i'] + 1
    print 'in g:', env['i'], j
    return j

f()