Nicola Paolucci durden at gmail.com
Mon Dec 11 14:57:07 EST 2006

```Hi All, Hi Cale,

Can you tell me if I understood things right ? Please see below ...

On 12/11/06, Cale Gibbard <cgibbard at gmail.com> wrote:
> The monad instance which is being used here is the instance for ((->)
> e) -- that is, functions from a fixed type e form a monad.
>
> So in this case:
> liftM2 :: (a1 -> a2 -> r) -> (e -> a1) -> (e -> a2) -> (e -> r)

> I bet you can guess what this does just by contemplating the type. (If
> it's not automatic, then it's good exercise) Now, why does it do that?

So the way I have to reason on the output I get from ghci is:

Prelude> :t liftM2
liftM2 :: (Monad m) => (a1 -> a2 -> r) -> m a1 -> m a2 -> m r

The m stands for ((->) e), that is like writing (e -> a1): a function
which will take an argument of type e and will return an argument of
type a1.

And so the above line has a signature that reads something like:
liftM2 will takes 3 arguments:
- a function (-) that takes two arguments and returns one result of type r.
- a function (fst) that takes one argument and returns one result.
- a function (snd) that takes one argument and returns one result.
- the result will be a certain function that will return the same type
r of the (-) function.
- Overall to this liftM2 I will actually pass two values of type a1
and a2 and will get a result of type r.

>From the type signature - correct me if I am wrong - I cannot actually
tell that liftM2 will apply (-) to the rest of the expression, I can
only make a guess. I mean I know it now that you showed me:

> liftM2 f x y = do
>    u <- x
>    v <- y
>    return (f u v)

If this is correct and it all makes sense, my next question is:
- How do I know - or how does the interpreter know - that the "m" of
this example is an instance of type ((->) e) ?
- Is it always like that for liftM2 ? Or is it like that only because
I used the function (-) ?

I am trying to understand this bit by bit I am sorry if this is either
very basic and easy stuff, or if all I wrote is completely wrong and I
did not understand anything. :D Feedback welcome.

Thanks again,
Regards,
Nick
```