wt., 21 cze 2022 o 05:20 Steven D'Aprano <st...@pearwood.info> napisał(a):
> The point is, Rob thought (and possibly still does, for all I know) that > lazy evaluation is completely orthogonal to late-bound defaults. The PEP > makes that claim too, even though it is not correct. With a couple of > tweaks that we have to do anyway, and perhaps a change of syntax (and > maybe not even that!) we can get late-bound defaults *almost* for free > if we had lazy evaluation. That depends of lazy evaluation spec, if lazy expression would ever become a thing in python, it may be defined to have syntax like `lazy <expr>` which would be rough equivalent off `LazyObject(lambda: <expr>)` that would evaluate that lambda at most once, plus some interpreter tweaks to make LazyObject transparent to python code. so for this code (`??` replaced with different combination of early/late and not lazy/lazy) ``` x = [] def f(x, y, z ?? len(x)): x.append(y) print(z, end = ' ') x.append(0) f([1, 2, 3], 4) x.append(0) f([1, 2, 3, 4], 4) ``` I expect that for `??` = `=` I get `0 0 ` for `??` = `=>` I get `3 4 ` for `??` = `= lazy` I get `1 1` for '??' = `=> lazy` I get `4 5 ` That would be completely orthogonal. _______________________________________________ Python-ideas mailing list -- python-ideas@python.org To unsubscribe send an email to python-ideas-le...@python.org https://mail.python.org/mailman3/lists/python-ideas.python.org/ Message archived at https://mail.python.org/archives/list/python-ideas@python.org/message/GZ2OCWGI67FTZGTCNHHYAXB7Y54LBTPY/ Code of Conduct: http://python.org/psf/codeofconduct/
