Re: [Haskell] factoring `if'

[Colin Runciman]

Serge,
How do you think, is the program (1) equivalent to (2) 
in the meaning of Haskell-98 ?

(1)   (\ x -> (if p x then  foo (g x)  else  foo (h x))
             where
             p ... g ... h ... foo ...
     )
(2)   (\ x -> foo  ((if p x then  g x  else  h x)
                   where
                   p ... g ... h ... foo ...
                  )
     )
 

Both examples are illegal -- there are no where-expressions in Haskell, 
only where-equations. 

Ignoring the local definition part, however, the answer is no.  Lifting 
foo out of the branches of the conditional is only valid if foo is strict.

Colin R

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