On 3 May 2007 23:36:11 -0700, Michael <[EMAIL PROTECTED]> wrote: > On May 2, 6:08 am, Carsten Haese <[EMAIL PROTECTED]> wrote: > > On Tue, 2007-05-01 at 22:21 -0700, Michael wrote: > > > Is there a reason for using the closure here? Using function defaults > > > seems to give better performance:[...] > > > > It does? Not as far as I can measure it to any significant degree on my > > computer. > > I agree the performance gains are minimal. Using function defaults > rather than closures, however, seemed much cleaner an more explicit to > me. For example, I have been bitten by the following before: > > >>> def f(x): > ... def g(): > ... x = x + 1 > ... return x > ... return g > >>> g = f(3) > >>> g() > Traceback (most recent call last): > File "<stdin>", line 1, in <module> > File "<stdin>", line 3, in g > UnboundLocalError: local variable 'x' referenced before assignment > > If you use default arguments, this works as expected: > >>> def f(x): > ... def g(x=x): > ... x = x + 1 > ... return x > ... return g > >>> g = f(3) > >>> g() > 4 >
You aren't getting "bit" by any problem with closures - this is a syntax problem. Assigning to x within the scope of g() makes x a local variable. x = x+1 doesn't work, then, because "x+1" can't be evaluated. If you just used "return x+1" or if you named the local variable in g something other than "x", the closure approach would also work as expected. -- http://mail.python.org/mailman/listinfo/python-list
