[Haskell-cafe] Fortran mixed mode arithmetic expressions ->
daniel.is.fischer at web.de
Sun Oct 25 23:41:28 EDT 2009
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
> 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?
> Prelude> let mm = 2
> Prelude> let k1 = 3*mm+2
> Prelude> let k2 = (/) 150 119
> Prelude> let k = k1 - k2
> 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
In this case,
does the trick. In general, you have to use fromInteg[er/ral] and other conversion
let k = fromIntegral k1 - k2
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