[Python-Dev] Missing operator.call

[Guido van Rossum]

On Fri, Feb 6, 2009 at 3:20 AM, Greg Ewing <greg.ewing at canterbury.ac.nz> wrote:
> Stephen J. Turnbull wrote:
>>
>> Greg Ewing writes:
>>
>>  > The fact that yielding is going on is not of
>>  > interest in that situation
>>
>> But doesn't "yield" in the sense of "yield the right of way" mean
>> exactly that?
>
> I've no problem with using 'yield' when actually
> giving up control. But the code making the call doesn't
> think of itself as yielding. The called code may
> want to yield, but the caller doesn't care about
> that. It just wants to make the callee do its thing,
> whatever it is.
>
> Ideally the caller would be able to use a normal
> function call, but Python generators don't work
> that way. The next best thing is a slightly
> different form of call syntax.

The more I read your motivation the less comfortable I am with using
"call" for this purpose. It differs in so many details from a regular
call that you're doing your potential users a terrible disservice by
trying to pretend that it is something which it isn't, or that it
isn't something which it is. Those differences may not seem relevant
when sketching a solution, but they come back to haunt you when
debugging, for example.

It would be way too confusing to have "a different form of call" with
totally different semantics that nevertheless used the same
*terminology* as is used for regular calls. Imagine the confusion
happens when a newbie (maybe someone coming from Fortran :-) writes
"call foo(x)" instead of just "foo(x)" and their program totally
breaks. "But you told me to call foo" will be their rightful excuse.

You could go on, but so could I. A -1 it is.

-- 
--Guido van Rossum (home page: http://www.python.org/~guido/)