On Sun, Sep 4, 2016 at 2:22 AM, Chris Angelico <ros...@gmail.com> wrote: > On Sun, Sep 4, 2016 at 7:44 AM, Koos Zevenhoven <k7ho...@gmail.com> wrote: >>> I wonder if it would be different if you wrote that as a single expression: >>> >>> x = 1 if cond else 1.5 >>> >>> x = sum([1] + [0.5] * cond) >>> >>> What should type inference decide x is in these cases? Assume an >>> arbitrarily smart type checker that can implement your ideal; it's >>> equally plausible to pretend that the type checker can recognize an >>> if/else block (or even if/elif/else tree of arbitrary length) as a >>> single "assignment" operation. IMO both of these examples - and by >>> extension, the if/else of the original - should be assigning a Union >>> type. >> >> In the first case it would indeed again be Union[int, float] (the code >> is basically equivalent after all). >> >> In the second case, the checker would infer List[Union[int, float]]. >> No magic involved, assuming that list.__add__ is annotated precisely >> enough. > > I think you missed the sum() call there. Most likely, the sum of > List[Union[anything]] would be Union[anything], so these would all be > the same (which was my original point). >
Oh, you're right, I missed the sum. Suitable @overloads in the stubs for sum should lead to that being inferred as Union[int, float]. But if the annotations are not that precise, that might become just List (or List[Any]), and the programmer may want to annotate x or improve the stubs. -- Koos > ChrisA > _______________________________________________ > Python-ideas mailing list > Python-ideas@python.org > https://mail.python.org/mailman/listinfo/python-ideas > Code of Conduct: http://python.org/psf/codeofconduct/ -- + Koos Zevenhoven + http://twitter.com/k7hoven + _______________________________________________ Python-ideas mailing list Python-ideas@python.org https://mail.python.org/mailman/listinfo/python-ideas Code of Conduct: http://python.org/psf/codeofconduct/
