Using a empty list as default parameter in __init__, why the list in all instances refer to the same list?

Question

class Entry():
    def __init__(self,l=[]):
        self.list = l


a = Entry()
b = Entry()
a.list.extend([1,2])
assert a.list!=b.list   #assert error

if using

a = Entry([])
b = Entry([])
a.list.extend([1,2])
assert a.list!=b.list   #right

What's the difference between above two examples?

Answer 1

Consult: "Least Astonishment" and the Mutable Default Argument

But to solve your problem do this:

def __init__(self,l=None):
    self.list = l if l else []

alternatively, as has been suggested by an edit, (which should have been a comment). you can use or:

def __init__(self, l=None):
    self.list = l or []

Answer 2

Don't use [] as default parameter.

Use this:

class Entry():
    def __init__(self,l=list()):
        ...

Here the problem is that the same list is assigned to each Entry instance.

So what append is something like this:

lst = []
a = Entry(lst)
b = Entry(lst)

a.list == b.list # lst == lst -> True

Answer 3

This is because in the first case, you pass a default argument [ ], which references the same list object.

    class Entry():
        def __init__(self,l=[]):
        self.list = l

    a = Entry()
    b = Entry()
    a.list.extend([1,2])
    print a.list, b.list # [1, 2] [1, 2]
    print id(a.list), id(b.list) # 36993344 36993344