[Haskell-cafe] what is f=f (not) doing ?
bf3 at telenet.be
Sun Sep 23 05:13:06 EDT 2007
I'm not sure, but since it would require the detection of an evaluation
that does not terminate, it comes down to the halting problem, which is
not generally solvable. Maybe the experts can confirm my intuition?
Andrew Coppin wrote:
> Peter Verswyvelen wrote:
>> to confirm that?
> Would it be possible to somehow prevent this behavious? (E.g., by
> somehow annotating each black hole with *which* thread is evaluating
> it, so that if a thread reaches one of the black holes that it created
> itself, it can throw an error..?)
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