Re: [Haskell-beginners] ap = liftM2 id

Am 03/26/2015 um 09:33 PM schrieb Andres Löh: I'll have to think about this. But other than that I am laughing my ass off after reading your paper about "Konjunktiv III". I can hardly type.

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

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

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

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id not True

or

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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

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ap = liftM2 ($)

Cheers, Andres

On Thu, Mar 26, 2015 at 9:18 PM, martin wrote:

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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? _______________________________________________ Beginners mailing list Beginners@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners