Larry Hastings added the comment:
Thanks for noticing the "(/)", that's fixed.
Yes, the signature for type() was wrong. type() can accept either one
parameter or three parameters--in other words, it uses optional groups. And we
can't represent optional groups in an inspect.Signature signature in 3.4. And
I assert that it's better to not have a signature than to have a wrong
signature. So I've removed it.
I'm interested in your "other minor stuff". I'll try and leave the patch alone
this time :)
If you apply diff #2, you'll have a failure in test_inspect. This is because
there's a bug in inspect.Signature that it would take too long to fix, and I
have to leave right now-ish. I changed the tests so they expect the correct
results, but they get the wrong results at the moment.
Hopefully we can fix this really quickly.
--
Yuri: This change uncovered a lurking bug in inspect.Signature that I'm hoping
you can fix. I could probably figure it out given enough time but I won't have
time to look at it for about 24 hours.
Now that type() doesn't have a signature, I have discovered that some of the
logic in Signature is wrong.
class C(type): pass
print(str(inspect.signature(C)))
The signature of that *should* be the same as type(). But inspect.signature()
reports the signature as '()'.
The reason this happens: inspect.Signature gets to the "for base in mro" case
for handling classes. The first base it tries is type(), but type() doesn't
have a public signature so it keeps going. The next class in the MRO is
object(), which has a signature of "()", so it uses that.
It shouldn't keep going! I'm 99% certain that the first entry in the MRO will
always be callable. (Is it possible to have a type in Python that isn't
callable?)
I *think* what it should do is: simply try the first entry in the MRO. If that
has a signature, return it. If it doesn't have a signature, inspect.Signature
should raise ValueError.
I also *think* that all the code after this comment:
# No '__text_signature__' was found for the 'obj' class.
should be removed. If C doesn't define its own __new__ or __init__, and its
base class doesn't have __call__, then the signature of C *is* the signature of
its base class. If inspect.Signature can't read that signature, then it can't
return a signature for C.
----------
Added file:
http://bugs.python.org/file33953/larry.even.newerer.signature.syntax.2.diff
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