=On Mon, May 20, 2013 at 8:46 PM, Ned Batchelder <n...@nedbatchelder.com> wrote: > On 5/20/2013 6:09 AM, Chris Angelico wrote: >> >> Referencing a function's own name in a default has to have one of >> these interpretations: >> >> 1) It's a self-reference, which can be used to guarantee recursion >> even if the name is rebound >> 2) It references whatever previously held that name before this def >> statement. > > > The meaning must be #2. A def statement is nothing more than a fancy > assignment statement.
Sure, but the language could have been specced up somewhat differently, with the same syntax. I was fairly confident that this would be universally true (well, can't do it with 'print' per se in older Pythons, but for others); my statement about CPython 3.3 was just because I hadn't actually hunted down specification proof. > So your "apparently recursive" print function is no more > ambiguous "x = x + 1". The x on the right hand side is the old value of x, > the x on the left hand side will be the new value of x. > > # Each of these updates a name > x = x + 1 > > def print(*args,print=print,lock=Lock(),**kwargs): > with lock: > print(*args,**kwargs) Yeah. The decorator example makes that fairly clear. > Of course, if you're going to use that code, a comment might be in order to > help the next reader through the trickiness... Absolutely!! ChrisA -- http://mail.python.org/mailman/listinfo/python-list