[Haskell-cafe] Function hanging in infinite input

[Matthew Korson]

The problem is that tAppend is too strict; it evaluates both its arguments
before producing anything. This is because you are pattern matching on
those arguments. You could use lazy patterns or make Temporal a newtype to
avoid that. Or you could rewrite to something like

    tAppend as bs = Temporal $ toList as ++ toList bs


On Sun, Apr 26, 2015 at 11:22 AM, martin <martin.drautzburg at web.de> wrote:

> Am 04/26/2015 um 07:05 PM schrieb Jerzy Karczmarczuk:
> > Martin reacts to my non-answer:
>
> > Martin I sent you a private follow-up. I repeat it here.
> >
> > I uncommented those lines.
> > Your program goes until the end of the list, and returns the /*last*/
> element (modified).  The form
> >
> > *tBind tpr f
> >  ...
> >  (tTail tpr `tBind` f) *
> >
> > loops until ...
> >
> > Now, I know about laziness... It seems that it doesn't help. Most
> probably your hd is simply empty, and the tail gets
> > stuck in an idle loop.
>
> Thanks a lot for taking the time to look into my code.
>
> I had made a mistake when I stripped down my code. In tUntil the
> inequality is wrong. I updated the example and put it here:
>
>
> https://www.dropbox.com/s/836hykwhhsb0n55/Function_hanging_in_infinite_input.hs?dl=0
>
> The strange thing is: I can set an upper limit to "outer" and I get the
> result
>
>         Temporal [(DPast,1),(T 3,3)]
>
> When I push the upper limit to later times, the result doesn't change.
> This is as expected, because I am only taking
> everything until (T 5). It looks like Haskell doesn't know that and
> believes that later recursions might contribute to
> the result. But I don't see why. tUntil is basically an innocent takeWhile.
>
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-- 
Matthew Korson
mjkorson at gmail.com
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