--- On Thu, 10/2/08, Andrew Coppin <[EMAIL PROTECTED]> wrote:
> I'm lost...
>
> (What does liftM have to do with fmap?)
They're (effectively) the same function.
i.e.
liftM :: (Monad m) => (a -> b) -> m a -> m b
fmap :: (Functor f) => (a -> b) -> f a -> f b
liftM turns a function from a to b into a function from m a to m b;
fmap turns a function from a to b into a function from f a to f b;
If your datatype with a Monad instance also has a Functor instance (which it
*can* have, you just need to declare the instance), then liftM is equivalent to
fmap.
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