Simple python oo issue
Have a look a this simple example. I don't quite understand why o1 prints "Hello Alex" twice. I would think that because of the default self.a is always reset to the empty list. Could someone explain to me what's the rationale here? Thank you so much.
开发者_运维知识库class A(object):
def __init__(self, a=[]):
self.a = a
o = A()
o.a.append('Hello')
o.a.append('Alex')
print ' '.join(o.a)
# >> prints Hello Alex
o1 = A()
o1.a.append('Hello')
o1.a.append('Alex')
print ' '.join(o1.a)
# >> prints Hello Alex Hello Alex
Read this Pitfall about mutable default function arguments: http://www.ferg.org/projects/python_gotchas.html
In short, when you define
def __init__(self,a=[])
The list referenced by self.a by default is defined only once, at definition-time, not run-time. So each time you call o.a.append
or o1.a.append
, you are modifying the same list.
The typical way to fix this is to say:
class A(object):
def __init__(self, a=None):
self.a = [] if a is None else a
By moving self.a=[]
into the body of the __init__
function, a new empty list is created at run-time (each time __init__
is called), not at definition-time.
Default arguments in Python, like:
def blah(a="default value")
are evaluated once and re-used in every call, so when you modify a you modify a globally. A possible solution is to do:
def blah(a=None):
if a is None
a = []
You can read more about this issue on: http://www.ferg.org/projects/python_gotchas.html#contents_item_6
Basically, never use mutable objects, like lists or dictionaries on a default value for an argument.
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