On 1/11/21 5:02 PM, Paul Bryan wrote:
Some more questions...
"Binding"," bound" and "unbound" code objects:
Is your use of "binding" terminology in the PEP identical to the
binding of a function to an object instance as a method during object
creation?
I'm not. In PEP 649 I think every reference of "binding" is talking
about binding a code object to a globals dict to produce a function
object. The process of binding a function to an object instance to
make a method is conceptually very similar, but distinct.
(and btw, functions aren't bound to their object to make methods during
object creation, it's done lazily at the time you ask for it--that's
what the "descriptor protocol" is all about!)
Function Annotations:
When binding an unbound annotation code object, a function will use
its own __globals__ as the new function's globals.
I'm having trouble parsing this. Does this mean the newly bound
__co_annotations__ function will inherit __globals__ from the function
it's annotating?
Yes. Though I wouldn't use "inherit", I'd just say it "uses" the
__globals__ from the function.
Exceptions:
It's quite possible for a __co_annotation__ function call to raise an
exception (e.g. NameError). When accessing __annotations__, if such an
exception is raised during the call to __co_annotations__, what is the
expected behavior?
If the function fails for any reason--throws an exception, or just
doesn't return an acceptable value--then the getter immediately exits,
and the internal state of the object is unchanged. If you wanted to,
you could catch the exception, fix the error, and get __annotations__
again, and it'd work.
s/__co_//?:
I'm probably naive, but is there a reason that one could not just
store a callable in __annotations__, and use the descriptor to resolve
it to a dictionary and store it when it is accessed? It would be one
less dunder in the Python data model.
That would work, but I think the API is a bit of a code smell.
__annotations__ would no longer be stable:
a.__annotations__ = o
assert a.__annotations__ == o
Would that assert fail? It depends on what type(o) is, which is surprising.
Cheers,
//arry/
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