[Haskell-cafe] The Trivial Monad
Michael Vanier
mvanier at cs.caltech.edu
Fri May 4 03:55:28 EDT 2007
The -> in type signatures associates to the right, so the type signatures
> fmap :: (a -> b) -> (W a -> W b)
> bind :: (a -> W b) -> (W a -> W b)
are the same as:
> fmap :: (a -> b) -> W a -> W b
> bind :: (a -> W b) -> W a -> W b
Sometimes people put in the extra parentheses because they want to
emphasize a particular way to use the function.
I'm assuming you understand that a function that takes two arguments and
returns a (possibly non-function) value is equivalent to a function that
takes one argument that returns a function that takes the other argument
and returns a value.
HTH,
Mike
Adrian Neumann wrote:
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> I've read this blogpost about the "trivial monad"
> http://sigfpe.blogspot.com/2007/04/trivial-monad.html, because I still
> don't understand what this monad thingy is all about.
>
> The author defines three functions:
>
> data W a = W a deriving Show
>
> return :: a -> W a
> return x = W x
>
> fmap :: (a -> b) -> (W a -> W b)
> fmap f (W x) = W (f x)
>
> bind :: (a -> W b) -> (W a -> W b)
> bind f (W x) = f x
>
> and asks the reader to prove the tree monad laws for them. However I
> don't understand the type signatures for bind and fmap. I'd say (and
> ghci's type inference agrees) that bind and fmap have the type
>
> bind:: (a->W b) -> W a -> W b
> fmap:: (a->b) -> W a -> W b
>
> They take a function f and something and return what f does to that. I
> don't see why they should return a function.
>
> This of course makes it hard for me to prove the monad laws. The first
> however works nonetheless:
>
> 1) bind f (return a)= f a
>
> => bind f (return a)= bind f (W a) = f a
>
> Can someone explain bind and fmap (and possible law 2 and 3)?
>
> Thanks,
>
> Adrian
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