[Haskell-beginners] Re: Pointfree expeiments

[Ertugrul Soeylemez]

Ozgur Akgun <ozgurakgun at gmail.com> wrote:

> I don't know whether this is related or not, but I have the following
> somewhere in my code:
>
> -- the following is apparently eqivalent to (\ i -> foo (bar i) i).
> -- pointfree suggests so. i don't understand why.
> (foo =<< bar)
>
> Is this a similar situation?

Yes, it is.  In the reader monad a computation is a function of one
argument:

  foo :: a -> e -> b
  bar :: e -> a

Running such a computation means passing a specific argument 'e' to all
those functions and regarding monadic bindings like normal let bindings.
Let me write this in 'do' syntax, so it gets clearer:

  do x <- bar
     foo x

becomes:

  let x = bar e
  in foo x e

I found the reader monad most useful in applicative style:

  splits :: [a] -> [([a], [a])]
  splits = zip <$> inits <*> tails


Greets,
Ertugrul


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