On Tue, 21 Jan 2014 20:11:02 +0100 Mû <m...@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. -- https://mail.python.org/mailman/listinfo/python-list
