Oh wait, since the decorated version of the ufunc will be the one in the public numpy API it won't break. It would only break if the callable that was passed in *wasn't* the decorated version, so it kinda *has* to pass in the decorated function to preserve backward compatibility. Apologies for the noise.
On Tue, Jun 5, 2018 at 7:39 PM, Nathan Goldbaum <nathan12...@gmail.com> wrote: > Hmm, does this mean the callable that gets passed into __array_ufunc__ > will change? I'm pretty sure that will break the dispatch mechanism I'm > using in my __array_ufunc__ implementation, which directly checks whether > the callable is in one of several tuples of functions that have different > behavior. > > On Tue, Jun 5, 2018 at 7:32 PM, Marten van Kerkwijk < > m.h.vankerkw...@gmail.com> wrote: > >> Yes, the function should definitely be the same as what the user called - >> i.e., the decorated function. I'm only wondering if it would also be >> possible to have access to the undecorated one (via `coerce` or >> `ndarray.__array_function__` or otherwise). >> -- Marten >> >> >> _______________________________________________ >> NumPy-Discussion mailing list >> NumPy-Discussion@python.org >> https://mail.python.org/mailman/listinfo/numpy-discussion >> >> >
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