[Haskell-cafe] WHNF versus HNF (was: Optimizing unamb by
determining the "state" of a thunk?)
daniel.is.fischer at web.de
Wed Apr 22 18:22:13 EDT 2009
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
> 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` ()
> 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).
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