[Haskell-cafe] WHNF versus HNF (was: Optimizing unamb by
determining the "state" of a thunk?)
bugfact at gmail.com
Wed Apr 22 17:57:08 EDT 2009
I'm having difficulty to understand the difference between WHNF and HNF.
Is this explanation<http://encyclopedia2.thefreedictionary.com/Weak+Head+Normal+Form>
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...
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