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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