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What is the equivalent F# cod e for this haskell code?
f#haskell

Here is the code i wanna convert to F# :

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integral fb =
    \ts@(t:ts') -> 0 : loop t 0 ts' (fb ts)
      where loop t0 acc (t1:ts) (a:as)
               = let acc' = acc + (t1-t0)*a
                  in acc' : loop t1 acc' evs ts as
.

Well it does not seem to typecheck unless I comment out "evs".

: becomes ::

\ becomes fun

name@pat becomes "pat as name"

as becomes az, since we just saw "as" is a keyword

"blah where blarg" becomes "blarg; blah"

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(*
integral fb = 
\ts@(t:ts') -> 0 : loop t 0 ts' (fb ts) 
where loop t0 acc (t1:ts) (a:as) 
= let acc' = acc + (t1-t0)*a 
in acc' : loop t1 acc' evs ts as 
*)
 
let integral fb = 
    let rec loop t0 acc (t1::ts) (a::az) =
        let acc' = acc + (t1-t0)*a 
        acc' :: loop t1 acc' (*evs*) ts az
    (fun ((t::ts') as ts) -> 0 :: loop t 0 ts' (fb ts))
By on 7/18/2010 5:49 PM ()

Although a bit obtuse to the untrained eye, I do prefer the Haskell syntax.

By on 6/23/2012 3:39 PM ()

@brianmcn

thnx

Saagar

By on 7/19/2010 7:51 PM ()

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