Re: [Haskell-cafe] Why is it so different between 6.12.1 and 6.10.4_1 ?

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Why do you bother with the interior definition of f in there? Because i want to try a C code style not layout style without `do` syntax sugar .

Yusaku Hashimoto wrote:

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fac n = let {  f = foldr (*) 1 [1..n] } in f

Why do you bother with the interior definition of f in there?

fac = product . enumFromTo 1

let fac = do is_zero <- (==0); if is_zero then return 1 else liftM2 (*) id (fac . pred)

-nwn

On Sat, Mar 27, 2010 at 9:59 AM, Ivan Lazar Miljenovic wrote:

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zaxis writes:

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In 6.10.4_1 under freebsd

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let f x y z = x + y + z *Money> :t f f :: (Num a) => a -> a -> a -> a

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:t (>>=) . f (>>=) . f  :: (Monad ((->) a), Num a) => a -> ((a -> a) -> a -> b) -> a -> b ((>>=) . f) 1 (\f x -> f x) 2

<interactive>:1:1:     No instance for (Monad ((->) a))       arising from a use of `>>=' at <interactive>:1:1-5     Possible fix: add an instance declaration for (Monad ((->) a))     In the first argument of `(.)', namely `(>>=)'     In the expression: ((>>=) . f) 1 (\ f x -> f x) 2     In the definition of `it': it = ((>>=) . f) 1 (\ f x -> f x) 2

Some definitions and exports got changed, so in 6.12 the (-> a) Monad instance is exported whereas in 6.10 it isn't.

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fac n = let {  f = foldr (*) 1 [1..n] } in f

Why do you bother with the interior definition of f in there?

fac = product . enumFromTo 1

-- Ivan Lazar Miljenovic Ivan.Miljenovic@gmail.com IvanMiljenovic.wordpress.com _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe

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