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another monadic function question about translating from haskell to F#
f#haskell
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sequence       :: Monad m => [m a] -> m [a] 
sequence ms = foldr k (return []) ms
            where
              k m m' = do { x <- m; xs <- m'; return (x:xs) }

This function is pretty simple which is :

given a list of monadic values(I believe computation in F# terminology), run them in sequence(left to right) and collect the result into a list then return monadic list.

Is this doable in F# ?

.

In short, no.

You can't abstract over type constructors in F#, i.e. you can't express the fact that the m in your signature can be any monadic type. You would have to give a concrete type (Option, Seq,...)

You can of course write a sequence function for a specific computation expression, but you're going to have to copy and modify it for every different kind of computation expression.

Computation expressions in F# are more like syntactic sugar, they're not first class.

Kurt

By on 12/15/2008 4:06 AM ()

I was trying to use the ^a construct and member as well as new constraint but no luck there.

By on 12/15/2008 6:39 PM ()

Found this blog and seems that Cale Gibbard's comment sums it up nicely. And I believe that applies to C#'s LINQ too which is very similar to F#'s computation expression.

Nice syntax sugar but lacks the power of what monad in Haskell can do.<!--</cite>-->

[link:blog.matthewdoig.com]

Well after reading those blogs by Matthew, I would say that I incline to his feeling that F# is still tilt more towards the C# side than the functional side be it ocaml or haskell.

By on 12/17/2008 12:54 AM ()

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