default method parameter behavior

[jianbing.chen]

I ran into a similar situation like the following (ipython session).
Can anyone please explain why the behavior?
Thanks in advance.

In [11]: def foo(b=[]):
....: b.append(3)
....: return b
....:

In [12]: foo()
Out[12]: [3]

In [13]: foo()
Out[13]: [3, 3]

In [14]: foo([])
Out[14]: [3]

In [15]: foo([])
Out[15]: [3]

 

[Konstantin Veretennicov]

I ran into a similar situation like the following (ipython session).
Can anyone please explain why the behavior?

Of course.

From http://docs.python.org/ref/function.html:

Default parameter values are evaluated when the function definition is
executed. This means that the expression is evaluated once, when the
function is defined, and that that same ``pre-computed'' value is used
for each call. This is especially important to understand when a
default parameter is a mutable object, such as a list or a dictionary:
if the function modifies the object (e.g. by appending an item to a
list), the default value is in effect modified.

 

[Jerry Hill]

I ran into a similar situation like the following (ipython session).
Can anyone please explain why the behavior?

http://www.python.org/doc/faq/general/#why-are-default-values-shared-between-objects

Since you got bitten by this, you may also find it useful to take a
look at one of the pages that talks about common python problems, like
http://zephyrfalcon.org/labs/python_pitfalls.html or
http://www.ferg.org/projects/python_gotchas.html (both of which
mention the problems with mutable default arguments).

 

[Primoz Skale]

I ran into a similar situation like the following (ipython session).
Can anyone please explain why the behavior?
Thanks in advance.

In [11]: def foo(b=[]):
....: b.append(3)
....: return b
....:

In [12]: foo()
Out[12]: [3]

In [13]: foo()
Out[13]: [3, 3]

In [14]: foo([])
Out[14]: [3]

In [15]: foo([])
Out[15]: [3]

I think it has something to do with foo.func_defaults thing If parameter
that is assigned a default value is mutable then foo.func_defaults is
changed everytime there is no parameter passed to the function.

For example:

def f(b=[]):

b.append(3)
return b

f.func_defaults #default is [], because function was not yet called ([],)
f() #no args [3]
f.func_defaults #default is now b=[3], because b is mutable object ([3],)
f([]) #empty; default still == [3] [3]
f.func_defaults #as we can see here ([3],)
f() #again no args; default is changed to [3,3] [3, 3]
f.func_defaults ([3, 3],)
f([]) #no args [3]
f.func_defaults

([3, 3],)


As *I* understand it, it goes something like this. Because mutable objects
change globaly if not passed correctly, and because [] is mutable, python
creates a *def global* object, which is only seen by function that created
it, with a name b to which it assigns a default value of []. But when this
default [] is changed in function, it becomes [3], and so on.....

Correct me if I am wrong...

P.