如何使用[]作为python中命名的函数参数的默认值？ [英] How do I use [] as a default value for a named function argument in python?

问题描述

可能重复：

Python中的最少惊讶：可变默认参数

我有这个代码

class Test(object):
  def __init__(self, var1=[]):
    self._var1 = var1

t1 = Test()
t2 = Test()

t1._var1.append([1])

print t2._var1

[[1]]。所以清楚t1._var1和t2._var1寻址相同的列表。如果我放

and I get "[[1]]" as the result. So clearly t1._var1 and t2._var1 are addressing the same list. If I put

t3 = Test()
print t3._var1

那么我得到[[1]]。所以var1 = []似乎永久绑定var1到一些列表。我尝试复制列表，

then I get "[[1]]" as well. So var1=[] seems to permanently bind var1 to the some list. I tried copying the list,

def __init__(self, var1=copy([])):

但结果相同，因此命名参数的表达式在init被调用之前被评估， var1一个空列表的副本，然后在实例之间共享。

but got the same result, so the expression for the named argument appears to be evaluated prior to init being called, and it just gave var1 a copy of the empty list which was then shared amongst the instances.

那么如何使用[]作为命名参数的默认值？

So how do I use [] as a default value for a named argument?

推荐答案

如果你想让每个对象都有一个对象，你不能直接使用 [] 空列表。我倾向于使用一个工作：

You can't use [] directly if you want each object to have an empty list. I tend to use a work around:

def __init__(self, var1=None):
    if var1 is None:
        var1 = []
    ....

t工作如果 var1 可以无，则需要使用不同的对象。

Naturally this won't work if var1 can be None, you would need to use a different object.

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