I don't see how nonlocal could fix this. Could you explain?

More specifically, given a better 'default arguments thing', how would you interpret:

    x = [2]
    def f(x=x*5):
        x.append(4)

With the earlier conversion it's:

    x = [2]
    def f(x=DefaultArg):
        if x is DefaultArg:
            x = x*5
        x.append(4)

This isn't going to work because the x inside of f() is different than the outside x, and you'll get the error message I mentioned.

If you add a nonlocal, as in:

    x = [2]
    def f(x=DefaultArg):
        nonlocal x
        if x is DefaultArg:
            x = x*5
        x.append(4)

then you'll get "SyntaxError: name 'x' is parameter and nonlocal".

What other solution are you thinking of?

Ah I see. Yeah nonlocal can't fix that.