Modifying the default argument of function

[Steve Jones]

On Tue, 21 Jan 2014 20:11:02 +0100
Mû <mu-- at melix.net> wrote:

> Hi everybody,
> 
> A friend of mine asked me a question about the following code:
> 
> [code]
> def f(x=[2,3]):
>      x.append(1)
>      return x
> 
> print(f())
> print(f())
> print(f())
> [/code]
> 
> The results are [2, 3, 1], [2, 3, 1, 1] and [2, 3, 1, 1, 1].
> 
> The function acts as if there were a global variable x, but the call of 
> x results in an error (undefined variable). I don't understand why the 
> successive calls of f() don't return the same value: indeed, I thought 
> that [2,3] was the default argument of the function f, thus I expected 
> the three calls of f() to be exactly equivalent.
> 
> I'm don't know much about python, does anybody have a simple explanation 
> please?

x is assigned to the list [2, 3] at the time the function is created not when the function is called, meaning that there's only ever 1 list created. When you call x.append this list is modified and the next time the function is called x still refers to this modified list.