On Mon, Jul 9, 2018 at 7:27 AM, Giampaolo Rodola' <g.rodola at gmail.com> wrote: > 5) It has no keyword argument correspondence. If foo(x := 1) is > allowed then why this one is not? > >>> foo(x=(x := 1)) > (I don't think it should BTW: it's not pretty) Actually it is. Nothing wrong with that. It assigns to 'x' in the local scope, and also passes that as a keyword parameter named 'x'. ChrisA
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