[Haskell-beginners] ap = liftM2 id

[Andres Löh]

There's no such thing as a fixed arity of a function in Haskell. The
type of id is

> id :: a -> a

If you instantiate "a" with "b -> c", then it becomes

> id :: (b -> c) -> b -> c

and it suddenly takes two arguments. In fact, it then behaves like
function application. Try

> id not True

or

> not `id` True

and you'll see that it works as if you had used function application
or $ instead. So ap lifts function application, and if it helps, you
can think of it as instead being defined as

> ap = liftM2 ($)

Cheers,
  Andres


On Thu, Mar 26, 2015 at 9:18 PM, martin <martin.drautzburg at web.de> wrote:
> Hello all,
>
> can someone explain
>
> ap  =  liftM2 id
>
> to me? liftM2 wants a binary funcation
>
> liftM2  :: (Monad m) => (a1 -> a2 -> r) -> m a1 -> m a2 -> m r
>
> but id is unary. How does this work?
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-- 
Andres Löh, Haskell Consultant
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