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 > 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. Jeremy
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