On Wed, 22 Jun 2022 at 08:21, Carl Meyer via Python-ideas
<python-ideas@python.org> wrote:
>
>
>
> On Tue, Jun 21, 2022 at 4:10 PM David Mertz, Ph.D. <david.me...@gmail.com>
> wrote:
>>
>> On Tue, Jun 21, 2022 at 5:53 PM Brendan Barnwell <brenb...@brenbarn.net>
>> wrote:
>>>
>>> > In the example, we assume that the built-in function `type()` is special
>>> > in not counting as a reference to the binding for purpose of realizing a
>>> > computation. Alternately, some new special function like `isdeferred()`
>>> > might be used to
>>> > check for ``Deferred`` objects.
>>>
>>> I'll have to ponder my thoughts about the proposal as a whole, but this
>>> particular aspect seems dubious to me. As I understand it this would
>>> require some fairly deep changes to how evaluation works in Python.
>>> Right now in an expression like `type(blah)`, there isn't any way for
>>> the evaluation of `blah` to depend on the fact that it happens to occur
>>> as an argument to `type`.
>>
>>
>> I absolutely agree that this is a sore point in my first draft. I could
>> shift the magic from `type()` to `isdeferred()`, but that doesn't really
>> change anything for your examples. I suppose, that is, unless
>> `isdeferred()` becomes something other than a real function, but more like
>> some sort of macro. That doesn't make me happy either.
>>
>> However, I *would* like to be able to answer the question "Is this object a
>> DeferredObject?" somehow. For example, I'd like some way to write code
>> similar to:
>>
>> if isdeferred(expensive_result):
>> log.debug("The computationally expensive result is not worth calculating
>> here")
>> else:
>> log.debug(f"We already hit a path that needed the result, and it is
>> {expensive_result}")
>>
>> Any thoughts on what might be the least ugly way to get that?
>
>
> I think all it really requires is for isdeferred() to be a builtin
> implemented in C rather than Python. It will be much better IMO to have
> isdeferred() returning a bool and not have any deferred object/type visible
> to Python.
>
It would have to be non-assignable and probably a keyword. Otherwise,
the exact same issues will keep occurring.
What are the scoping rules for deferred objects? Do they access names
where they are evaluated, or where they are defined? Consider:
def f(x):
spam = 1
print(x)
def g():
spam = 2
f(later spam)
g()
Does this print 1 or 2?
This is a fundamental and crucial point, and must be settled early.
You cannot defer this. :)
ChrisA
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