[Haskell-cafe] left to right pipeline of functions

[Theodore Lief Gannon]

Nitpick: (&) is flipped function application ($), not composition (.) --
though the signature you gave was correct.
On Jun 11, 2016 11:07 PM, "Csongor Kiss" <kiss.csongor.kiss at gmail.com>
wrote:

> Hi,
>
> (>>=) :: Monad m => m a -> (a -> m b) -> m b, and if I understand
> correctly, what you need is a ‘generic’ version, that works without the
> returns?
>
> 'return :: a -> m a' for some monad m. This m comes from the monad
> instance you use, but if we want a generic one, it’s only possible by
> fixing a specific m. The obvious choice is to use the Identity monad and
> have a specialised (>>=‘) :: Identity a -> (a -> Identity b) -> Identity b.
>
> For all `a`, `Identity a` is isomorphic to `a`, so with a bit of cheating,
> you could rewrite the above to (>>=‘’) :: a -> (a -> b) -> b,
> which starts to look really familiar, in fact, it’s just the function
> composition (.) with its arguments flipped, and is actually defined
> as (&) in Data.Function.
>
> g' = 2 & \n -> (n + 1) & \n -> (n + 3)
>
> This was obviously a really roundabout way of arriving at the result, but
> I think the analogy with function composition is really nice, as we can
> think of (a -> b) as a special case of (a -> m b), the so-called Kleisli
> arrow. What monads do is they compose these Kleisli arrows.
>
> --
> Csongor
>
> On 12 June 2016 at 06:47:13, Christopher Howard (ch.howard at zoho.com)
> wrote:
>
> Hi, I am learning about monads, and this question came to mind: is there
> a way to have a sequence of functions left to right like so:
>
> g = return 2 >>= \n -> return (n + 1) >>= \n -> return (n + 3)
>
> Either: some kind of generic monad that makes this legal, or some way to
> do this without monad (i.e., without the "returns").
>
> --
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