Python递归时函数内变量无法引用的问题 - V2EX
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bearzx

Python递归时函数内变量无法引用的问题

  •  
  •   bearzx Oct 27, 2013 4496 views
    This topic created in 4580 days ago, the information mentioned may be changed or developed.
    hi各位好,
    我今天写code的时候遇到一个问题,考虑以下代码:

    def a():
    def b():
    if a != 0:
    print a
    a -= 1
    b()
    a = 100
    b()

    a()

    运行的话就会报出a变量没有创建的错误。但是这样是不是说这种情况下在递归的内部函数里就不能去引用外部函数的变量了?那如果有这种需求的话就只能定义全局变量了么。。。

    求大神们指导
    10 replies    1970-01-01 08:00:00 +08:00
    bearzx
        1
    bearzx  
    OP
       Oct 27, 2013
    我擦这缩进。。。
    bearzx
        2
    bearzx  
    OP
       Oct 27, 2013
    貌似可以用python3的nonlocal解决。。。
    mengzhuo
        3
    mengzhuo  
       Oct 27, 2013   1
    能用gist不……

    Python的变量规则好像是是Local Closure Global Buildin

    可以看看这个
    http://stackoverflow.com/questions/4851463/python-closure-write-to-variable-in-parent-scope
    bearzx
        4
    bearzx  
    OP
       Oct 28, 2013
    @bearzx v2ex可以用gist?
    bearzx
        5
    bearzx  
    OP
       Oct 28, 2013
    @mengzhuo 很有帮助,谢谢~
    est
        6
    est  
       Oct 28, 2013
    a 又是变量又是函数?
    bearzx
        7
    bearzx  
    OP
       Oct 28, 2013
    @est 没。。是我写的不清楚,可以把a函数里面的a是另一个东西
    VYSE
        8
    VYSE  
       Oct 28, 2013 via Android
    把变量放在tuple或list里传
    bearzx
        9
    bearzx  
    OP
       Oct 28, 2013
    @VYSE 嗯mengzhuo给的链接里用的是这么个方法,不过python3里面有一个nonlocal关键字也可以实现这个需求
    roricon
        10
    roricon  
       Oct 29, 2013
    为什么我先想到的是构造一个generator然后重构它的__sub__方法来实现……
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