On Tue, Oct 26, 2021 at 8:29 PM Ethan Furman <[email protected]> wrote:
> Using the `bisect()` function as a stand-in for the 20+ years worth of > Python APIs in existence: > > def bisect_right(a, x, lo=0, hi=None, *, key=None): > if hi is None: > hi = len(a) > while lo < hi: > ... > > That function would be transformed to: > > def bisect_right(a, x, lo=0, @hi=len(a), *, key=None): > if hi is None: > hi = len(a) > while lo < hi: > > Notice that the `None` check is still in the body -- why? Backwards > compatibility: Of course. personally, I don't think there's any reason to go in and change the standard library at all. This is a feature for new code, new APIs. *maybe* a long time from now, some stdlib APIs could be updated, but no hurry. This seems like a lot of effort for a very marginal gain. > marginal gain to the stdlib, yes of course. Though now that you mention it -- there would be some marginal game even to functions like that, making the function signature a bit more clear. Interestingly, though, here's bisect's docstring: In [14]: bisect.bisect? Docstring: bisect_right(a, x[, lo[, hi]]) -> index Return the index where to insert item x in list a, assuming a is sorted. The return value i is such that all e in a[:i] have e <= x, and all e in a[i:] have e > x. So if x already appears in the list, i points just beyond the rightmost x already there Optional args lo (default 0) and hi (default len(a)) bound the slice of a to be searched. Type: builtin_function_or_method It's not actually documented that None indicates "use the default". Which, it turns out is because it doesn't :-) In [24]: bisect.bisect([1,3,4,6,8,9], 5, hi=None) --------------------------------------------------------------------------- TypeError Traceback (most recent call last) <ipython-input-24-65fd10e3a3b5> in <module> ----> 1 bisect.bisect([1,3,4,6,8,9], 5, hi=None) TypeError: 'NoneType' object cannot be interpreted as an integer I guess that's because in C there is a way to define optional other than using a sentinel? or it's using an undocumented sentinal? Note: that's python 3.8 -- I can't imagine anything;s changed, but ... -CHB -- Christopher Barker, PhD (Chris) Python Language Consulting - Teaching - Scientific Software Development - Desktop GUI and Web Development - wxPython, numpy, scipy, Cython
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