> On Mon, 2003-04-14 at 11:52, Phillip J. Eby wrote: > > If I understand correctly, it should also be breakable by deleting 'foo' > > from the outer function when you're done with it. E.g.: > > > > def bar(a): > > def foo(): > > return None > > x = a > > foo() > > > > del foo # clears the cell and breaks the cycle > From: Jeremy Hylton <jeremy@zope.com> > > You haven't tried this, have you? ;-) > > SyntaxError: can not delete variable 'foo' referenced in nested scope > > Since foo() could escape bar, i.e. become reachable outside of bar(), we > don't allow you to unbind foo. I don't see the reason for this semantic restriction. IMO it could just as well be a runtime error (e.g. raising UnboundLocalError). --Guido van Rossum (home page: http://www.python.org/~guido/)
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