[Haskell-beginners] Parser composition

[David Virebayre]

2010/12/2 Angelos Bimpoudis <ampimpou at otenet.gr>

So I would like to ask about whether or not this:
>
> P1>>=λu1 ->
>
> P2>>=λu2 ->
>
> P3>>=λu3 ->
>
> P4>>=λu4 ->
>
> return (f u1 u2 u3 u4)
>
>
>
> can be rewritten to:
>
> P1>>=(λu1 -> P2>>=(λu2 -> P3>>=(λu3 -> P4>>=(λu4 -> return (f u1 u2 u3
> u4)))))
>

Yes, but the extra parenthesis you added are not necessary.

I am trying to understand what is the f in p>>=f in each one of above
> sequencing applications?
>

Amy probably explained it better that I will, and I'm not sure I understood
the question but :

In
P1>>=(λu1 -> P2>>=(λu2 -> P3>>=(λu3 -> P4>>=(λu4 -> return (f u1 u2 u3
u4)))))
--
-------------------------------------------------------------------------
p    f

In
P2>>=(λu2 -> P3>>=(λu3 -> P4>>=(λu4 -> return (f u1 u2 u3 u4)))))
--   ------------------------------------------------------------
p    f


Etc.



>
> So >>= gets you in this structure (is this the monad?), and return is the
> gate out of it?
>

(Someone please correct me if I'm not using the right words;)

>>= binds two monadic computation together, getting a value out of the first
one, and feeding that value to the second one.
return is more a gate into it, as it takes a value and 'returns' its monadic
equivalent, for example :

-- In the list monad
Prelude> return 1 :: [Int]
[1]

As you see, return didn't "gate out" anything of the list monad, on the
contrary it took a value into the list monad. Same with the Maybe monad :

Prelude> return 1 :: Maybe Int
Just 1

And it works that way for every monad, returns take a value, and return its
monadic version.

Best regards

David.
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