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comp.lang.haskell |
Frank Poettgen <Frank.Poett...@Post.RxWyTxH-AyAzCyHzEyN.de> schrieb: > expensive :: Int -> (Int,Int) > f :: Int -> Int > Will 'expensive x' be evaluated twice to compute 'f x'? f x = fst e + snd e where e = expensive or, using pattern matching, f x = l + r where (l, r) = expensive x or, more compact but less comprehensible, f = uncurry (+) . expensive
> Assume that the evaluation of a function 'expensive' takes a lot of That will depend on your compiler. Some compilers do common
> time. Hence I want to avoid unnecessary evaluations of 'expensive'.
> expensive x = (x,x)
> f x = fst (expensive x) + snd (expensive x)
subexpression elimination, others don't. The reason is that
this transformation can sometimes introduce space leaks. In
your case it wouldn't be dangerous, but you cannot rely on
your compiler recognizing this.
> If so, how can I avoid it? Rewrite the function by hand: