10:43 a.m.
Hi,
I don't think this is a new bug, but who knows, and anyway:
<hansl>
function scalar rc(void)
loop i=1..2
return i
endloop
end function
rc()
</hansl>
I get an error saying the function doesn't return the proper type.
If I replace "return i" with the two lines:
out = i
return out
then it works.
This is with the brand new 2016d.
Thanks,
sven
12:20 p.m.
On Sun, 20 Nov 2016, Sven Schreiber wrote:
Hi,
I don't think this is a new bug, but who knows, and anyway:
<hansl>
function scalar rc(void)
loop i=1..2
return i
endloop
end function
rc()
</hansl>
I get an error saying the function doesn't return the proper type.
If I replace "return i" with the two lines:
out = i
return out
then it works.
This is with the brand new 2016d.
I don't think this is new. The point is that if the index "i" is not
declared outside of the loop it is considered an automatic variable
which exists only for the scope of the loop. So you could get the
effect you expected by sticking in
scalar i
before the loop. (Then the function will return 1).
Allin
1:08 p.m.
Am 20.11.2016 um 18:20 schrieb Allin Cottrell:
I don't think this is new. The point is that if the index "i" is not
declared outside of the loop it is considered an automatic variable
which exists only for the scope of the loop. So you could get the
effect you expected by sticking in
scalar i
before the loop. (Then the function will return 1).
So you don't think it's a bug? I mean the 'return' statement _is_ inside
the loop where the loop index is declared. I don't see the difference to
the general situation where 'return x' is written with 'x' existing only
locally in the function scope. Or to put it differently, shouldn't the
statement 'loop i= ...' be considered a declaration of the index variable i?
thanks,
sven
2:35 p.m.
On Sun, 20 Nov 2016, Sven Schreiber wrote:
Am 20.11.2016 um 18:20 schrieb Allin Cottrell:
>
> I don't think this is new. The point is that if the index "i" is not
> declared outside of the loop it is considered an automatic variable
> which exists only for the scope of the loop. So you could get the
> effect you expected by sticking in
>
> scalar i
>
> before the loop. (Then the function will return 1).
So you don't think it's a bug? I mean the 'return' statement _is_
inside the loop where the loop index is declared.
OK, I think that point should carry it. But perhaps, before I try to
fix that particular issue, we should consider whether the policy of
making a non-predeclared loop index local to the loop is sound and
consistent with other aspects of hansl.
Here's another example to consider. As things stand, the final
"print i" fails in the following fragment unless the initial
declaration is uncommented:
<hansl>
# scalar i
loop i=1..2
print i
endloop
print i
</hansl>
If we decide against default locality of the loop index, the
function return case will solve itself. Otherwise it will need
special treatment.
Allin
4:40 p.m.
Am 20.11.2016 um 20:35 schrieb Allin Cottrell:
On Sun, 20 Nov 2016, Sven Schreiber wrote:
> So you don't think it's a bug? I mean the
'return' statement _is_
> inside the loop where the loop index is declared.
OK, I think that point should carry it. But perhaps, before I try to fix
that particular issue, we should consider whether the policy of making a
non-predeclared loop index local to the loop is sound and consistent
with other aspects of hansl.
In all generality, I don't know. To me personally it has never occurred
to use the loop index outside the loop (well, except in the next loop,
again as the index).
Here's another example to consider. As things stand, the final
"print i"
fails in the following fragment unless the initial declaration is
uncommented:
...
endloop
print i
I think either way is defensible; you could say that i only exists
inside the loop, or you could say i lives on and keeps the value of the
final iteration. I guess I'm indifferent, as long as i is fully usable
inside the loop which is what I expected.
If we decide against default locality of the loop index, the
function
return case will solve itself. Otherwise it will need special treatment.
At least right now I don't see any argument against the non-locality --
also noting that already currently you may not use the loop index (say,
i) as a non-scalar type (say, matrix) _before_ the loop, because that
provokes a type mismatch error when getting to the "loop i=..." part. So
in that sense it's already kind of non-local.
thanks,
sven
6:25 p.m.
On Sun, 20 Nov 2016, Sven Schreiber wrote:
Am 20.11.2016 um 20:35 schrieb Allin Cottrell:
> On Sun, 20 Nov 2016, Sven Schreiber wrote:
>> So you don't think it's a bug? I mean the 'return' statement
_is_
>> inside the loop where the loop index is declared.
>
> OK, I think that point should carry it. But perhaps, before I try to fix
> that particular issue, we should consider whether the policy of making a
> non-predeclared loop index local to the loop is sound and consistent
> with other aspects of hansl.
In all generality, I don't know. To me personally it has never occurred to
use the loop index outside the loop (well, except in the next loop, again as
the index).
OK, for now I've left things as they are, except that an automatic
loop-index scalar is available as a function return value (for a
"return" statement within the loop). The case where the return value
is a function of such an index should also be handled. That's in
git; snapshots will follow tomorrow.
Allin
2874
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