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
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...
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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