Re: [Haskell-cafe] Expressing seq
Chad Scherrer wrote: There must be a subtlety I'm missing, right? What if the types are not instances of Eq? Jason Thanks, I figured it was something simple. Now I just to convince myself there's no way around that. Is there a proof around somewhere? Yes, there is a proof that seq :: a - b - b with the semantics as described in the Haskell report cannot be defined in Haskell minus seq. It goes as follows: If seq were so definable, then it would have to fulfill the free theorem derived from its type in Haskell minus seq (read: System F plus fix). But it doesn't. See Section 5 of http://wwwtcs.inf.tu-dresden.de/~voigt/seqFinal.pdf Ciao, Janis. -- Dr. Janis Voigtlaender http://wwwtcs.inf.tu-dresden.de/~voigt/ mailto:[EMAIL PROTECTED] ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
[Haskell-cafe] Expressing seq
I was reading on p. 29 of A History of Haskell (a great read, by the way) about the controversy of adding seq to the language. But other than for efficiency reasons, is there really any new primitive that needs to be added to support this? As long as the compiler doesn't optimize it away, why not just do something like this (in ghci)? Prelude let sq x y = if x == x then y else y Prelude 1 `sq` 2 2 Prelude (length [1..]) `sq` 2 Interrupted. There must be a subtlety I'm missing, right? -- Chad Scherrer Time flies like an arrow; fruit flies like a banana -- Groucho Marx ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Expressing seq
On 9/27/06, Chad Scherrer [EMAIL PROTECTED] wrote: I was reading on p. 29 of A History of Haskell (a great read, by the way) about the controversy of adding seq to the language. But other than for efficiency reasons, is there really any new primitive that needs to be added to support this? As long as the compiler doesn't optimize it away, why not just do something like this (in ghci)? Prelude let sq x y = if x == x then y else y Prelude 1 `sq` 2 2 Prelude (length [1..]) `sq` 2 Interrupted. There must be a subtlety I'm missing, right? What if the types are not instances of Eq? Jason ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Expressing seq
There must be a subtlety I'm missing, right? What if the types are not instances of Eq? Jason Thanks, I figured it was something simple. Now I just to convince myself there's no way around that. Is there a proof around somewhere? -- Chad Scherrer Time flies like an arrow; fruit flies like a banana -- Groucho Marx ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Expressing seq
Chad Scherrer wrote: I was reading on p. 29 of A History of Haskell (a great read, by the way) about the controversy of adding seq to the language. But other than for efficiency reasons, is there really any new primitive that needs to be added to support this? As long as the compiler doesn't optimize it away, why not just do something like this (in ghci)? Prelude let sq x y = if x == x then y else y Prelude 1 `sq` 2 2 Prelude (length [1..]) `sq` 2 Interrupted. There must be a subtlety I'm missing, right? The (sq x) function depends on x being an instance of typeclass Eq. Imagine a new typeclass Seq that is auto-defined for all types: class Seq a where seq :: a - (b - b) data Foo x = Bar x | Baz | Foo y x The instances always use a case to force just enough evaluation to compute the constructor, then return id: instance Seq Foo where seq a = case a of Bar _ - id Baz - id Foo _ _ - id foo1,foo2 :: Foo Int foo1 = undefined foo2 = Bar 1 Now (seq foo1) b will also be undefined which (seq foo2) b will be id b which is b. ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Expressing seq
Chad Scherrer wrote: Prelude let sq x y = if x == x then y else y Prelude 1 `sq` 2 2 Prelude (length [1..]) `sq` 2 Interrupted. There must be a subtlety I'm missing, right? Two, at least: First, your sq has a different type, as it requires an Eq instance: Prelude :t sq sq :: (Eq a) = a - t - t Prelude :t seq seq :: a - b - b Secondly, your sq is more akin to a deepSeq in that it forces all of its value instead of just evaluating to weak head normal form. Prelude [undefined] `seq` 1 1 Prelude [undefined] `sq` 1 *** Exception: Prelude.undefined You could implement seq explicitely for many types, for example, seqList []x = x seqList (_:_) x = x but not for function types. HTH, Bertram ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
