[Haskell-cafe] WHNF versus HNF (was: Optimizing unamb by determining the "state" of a thunk?)

[Daniel Fischer]

Am Mittwoch 22 April 2009 23:57:08 schrieb Peter Verswyvelen:
> I'm having difficulty to understand the difference between WHNF and HNF.
>
> Is this
> explanation<http://encyclopedia2.thefreedictionary.com/Weak+Head+Normal+For
>m> the
> correct one? Or is WHNF and HNF equivalent in Haskell land?
>
> The GHC documentation of seq says:
> Evaluates its first argument to head normal form, and then returns its
> second argument as the result.
>
> Let's try in GHCi
>
> *Main> let f = trace "\\x" $ \x -> ((trace "\\y" $ \y -> trace "y" y +
> trace "x" x) $ trace "2" 2)
> *Main> f `seq` ()
> \x
> ()
> *Main>
>
> That did not evaluate anything inside the body of the first lambda, so
> according to the article, seq reduces to weak head normal form, not hnf...

Yes, error in the docs. seq evaluates to WHNF, not HNF (unless the two coincide).