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Why do closures do this?

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Somewhat apropos of the recent "function principle" thread, I was recently surprised by this:

funcs=[]
for n in range(3):
    def f():
        return n
    funcs.append(f)

[i() for i in funcs]

The last expression, IMO surprisingly, is [2,2,2], not [0,1,2]. Google tells me I'm not the only one surprised, but explains that it's because "n" in the function "f" refers to whatever "n" is currently bound to, not what it was bound to at definition time (if I've got that right), and that there are at least two ways around it: either make a factory function:

def mkfnc(n):
    def fnc():
        return n
    return fnc

funcs=[]
for n in range(3):
    funcs.append(mkfnc(n))

which seems roundabout, or take advantage of the "default values set at definition time" behaviour:

funcs=[]
for n in range(3):
    def f(n=n):
        return n
    funcs.append(f)

which seems obscure, and a side-effect.

My question is, is this an inescapable consequence of using closures, or is it by design, and if so, what are some examples of where this would be the preferred behaviour?

Regards,

John 

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Thread

Why do closures do this? John O'Hagan <research@johnohagan.com> - 2011-08-28 13:45 +1000
  Re: Why do closures do this? Carl Banks <pavlovevidence@gmail.com> - 2011-08-28 14:16 -0700
  Re: Why do closures do this? Carl Banks <pavlovevidence@gmail.com> - 2011-08-28 14:16 -0700

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