getattr的最大递归深度误差 [英] Maximum recursion depth error with getattr

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问题描述

我有这个代码;

  class NumberDescriptor(object):
def __get __(self,instance,owner )
name =(hasattr(self,name)和self.name)
如果不是名字:
name = [attr for dir(owner)如果getattr(owner, attr)是自我] [0]
self.name = name
return getattr(instance,'_'+ name)
def __set __(self,instance,value):
name =(hasattr(self,name)和self.name)
如果不是名字:
owner = type(instance)
name = [attr for dir(owner)if getattr(owner,attr)is self] [$]
self.name = name
setattr(instance,'_'+ name,int(value))

class Insan (对象):
yas = NumberDescriptor()

a = Insan()
打印a.yas
a.yas =osman
打印我在线上获得最大的递归深度错误 name = [...] attr for attr in di r(owner)if getattr(owner,attr)is self] [0] 。我想要该行让我看到当前描述符实例使用的变量的名称。谁能看到我在这里做错什么?

解决方案

getattr() call调用你的 __ get __



解决这个问题的一种方法是明确地调用超类, 对象

  object .__ getattribute __(instance,name)

或者,更清楚:

  instance .__ dict __ [name] 


I have this code;

class NumberDescriptor(object):
    def __get__(self, instance, owner):
        name = (hasattr(self, "name") and self.name)
        if not name:
            name = [attr for attr in dir(owner) if getattr(owner,attr) is self][0]
            self.name = name
        return getattr(instance, '_' + name)
    def __set__(self,instance, value):
        name = (hasattr(self, "name") and self.name)
        if not name:
            owner = type(instance)
            name = [attr for attr in dir(owner) if getattr(owner,attr) is self][0]
            self.name = name
        setattr(instance, '_' + name, int(value))

class Insan(object):
    yas = NumberDescriptor()

a = Insan()
print a.yas
a.yas = "osman"
print a.yas

I am getting maximum recursion depth error in the line name = [attr for attr in dir(owner) if getattr(owner,attr) is self][0]. I want that line to get me the name of variable used for current descriptor instance. Can anyone see what am I doing wrong here?

解决方案

The getattr() call is calling your __get__.

One way to work around this is to explicitly call through the superclass, object:

object.__getattribute__(instance, name)

Or, clearer:

instance.__dict__[name]

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