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[Python-ideas] Where-statement (Proposal for function expressions)

Jan Kaliszewski zuo at chopin.edu.pl
Fri Jul 17 10:09:20 CEST 2009 
17-07-2009, 04:25
Mike Meyer <mwm-keyword-python.b4bdba at mired.org> wrote:

> On Fri, 17 Jul 2009 14:02:34 +1200
> Greg Ewing <greg.ewing at canterbury.ac.nz> wrote:
>
>> Daniel Stutzbach wrote:
>>
>> > If the
>> > left-hand side were in the where-block's scope, then:
>> >
>> > x = 0
>> > x = i where:
>> >    i = 5
>> > print x
>> >
>> > Would print "0" not "5".
>>
>> My intuitive expectations are different for assignment
>> to a bare name vs. assignment to an item or attribute.
>> In
>>
>>    x = i where:
>>      i = 5
>>
>> I would expect x to be bound in the current scope,
>> not the where-block scope. But I would expect
>>
>>    x[i] = 5 where:
>>      x = y
>>      i = 7
>>
>> to be equivalent to
>>
>>    y[7] = 5
>>
>> i.e. both x and i are *evaluated* in the where-block
>> scope.
>>
>> I'm not sure where this leaves us with respect to
>> augmented assignment, though. If you follow it to
>> its logical conclusion, then
>>
>>    x += i where:
>>      i = 5
>>
>> should be equivalent to
>>
>>    x = x.__iadd__(i) where:
>>      i = 5
>>
>> and then the evaluation and rebinding of 'x' would
>> happen in different scopes!
[snip]
> I think the intuitive behavior doesn't involve position relative to
> the assignment, but binding vs. non-bindings. What you *expect* is
> that expressions left of the where will be evaluated in the inner
> scope (created by the where), but that any bindings that happen there
> will happen in the outer scope (before the where).
>
> That makes what you want to happen happen - *most* of the time:
>
> i = 3
> x += i where:
>    i = 5
>
> increments x by 5.
>
> i = 'a'
> x[i] = 'b' where:
>    i = 'c'
>
> sets x['c'] to 'b'.
>
> Things get really ugly when the bound variables start appearing in
> both scopes combined with augmented assignments:
>
> x = 3
> x += 4 where:
>     x = 5
> print x
>
> What should this print?

I think, a good rule could be that:

All *expressions* of the target line are evaluated within the
where-block scope; and -- moreover -- those names (variables) that
are *bound* in the target line, are "leaked" to the outer scope.

[target line == the line with where statement]

Then the former example code should print 9.

-- 
Jan Kaliszewski <zuo at chopin.edu.pl>
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