[Haskell-cafe] Fortran mixed mode arithmetic expressions -> Haskell

[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

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> 




      
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