Re: [Haskell-cafe] Re: Laziness question

Nicolas Pouillard schrieb:

On Wed, 04 Aug 2010 17:47:12 +0200, Janis Voigtländer wrote:

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Nicolas Pouillard schrieb:

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On Wed, 04 Aug 2010 17:27:01 +0200, Janis Voigtländer wrote:

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Nicolas Pouillard schrieb:

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However the rule is still the same when using an unsafe function you are on your own.

Clearer? Almost. What I am missing is whether or not you would then consider your genericSeq (which is applicable to functions) one of those "unsafe functions" or not. I would consider it as a safe function. Well, then I fear you have come full-circle back to a non-solution. It is not safe:

I feared a bit... but no

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Consider the example foldl''' from our paper, and replace seq therein by your genericSeq. Then the function will have the same type as the original foldl, but the standard free theorem for foldl does not hold for foldl''' (as also shown in the paper).

So foldl''' now has some Typeable constraints.

No, I don't see how it has that. Or maybe you should make explicit under what conditions a type (a -> b) is in Typeable. What exactly will the type of foldl''' be, and why? Ciao, Janis. -- Jun.-Prof. Dr. Janis Voigtländer http://www.iai.uni-bonn.de/~jv/ mailto:jv@iai.uni-bonn.de