Am Montag 26 Oktober 2009 04:21:06 schrieb michael rice:
Translating Fortran mixed mode arithmetic expressions into Haskell is quite a challenge. Believe it or not
c=10.**(11-int(alog10(r)+10))
translates to
let c = (**) 10.0 $ fromIntegral $ subtract 11 $ truncate $ (+) (logBase 10.0 r) 10.0
No, subtract 11 x is x-11, not 11-x. let c = 10^^(11 - (truncate (logBase 10 r) + 10)) or, to make it a little simpler, let c = 10^^(1 - truncate (logBase 10 r)) Prelude> :t (^^) (^^) :: (Fractional a, Integral b) => a -> b -> a That is probably faster and more accurate than (**).
I finally broke the expression below into two parts (k1 & k2) to ease translation. I get it that Haskell is expecting to subtract two Integers and is instead being given an Integer and a Double. What must I do to make this work? Are there any guidelines for doing this kind of translation work?
Michael
================
Prelude> let mm = 2 Prelude> let k1 = 3*mm+2 Prelude> let k2 = (/) 150 119 Prelude> let k = k1 - k2
<interactive>:1:13: Couldn't match expected type `Integer' against inferred type `Double' In the second argument of `(-)', namely `k2' In the expression: k1 - k2 In the definition of `k': k = k1 - k2 Prelude> :t 150/119 150/119 :: (Fractional t) => t Prelude>
In this case, :set -XNoMonoMorphismRestriction does the trick. In general, you have to use fromInteg[er/ral] and other conversion functions explicitly. let k = fromIntegral k1 - k2