runST and function composition

Why does this typecheck:

runST $ return $ True

While the following does not:

runST . return $ True

GHCI complains:

Couldn't match expected type `forall s. ST s c0'
            with actual type `m0 a0'
Expected type: a0 -> forall s. ST s c0
  Actual type: a0 -> m0 a0
In the second argument of `(.)', namely `return'
In the expression: runST . return

The short answer is that type inference doesn't always work with higher-rank types. In this case, it is unable to infer the type of (.), but it type checks if we add an explicit type annotation:

> :m + Control.Monad.ST
> :set -XRankNTypes
> :t (((.) :: ((forall s0. ST s0 a) -> a) -> (a -> forall s1. ST s1 a) -> a -> a) runST return) $ True
(((.) :: ((forall s0. ST s0 a) -> a) -> (a -> forall s1. ST s1 a) -> a -> a) runST return) $ True :: Bool

The same problem also happens with your first example, if we replace ($) with our own version:

> let app f x = f x
> :t runST `app` (return `app` True)
<interactive>:1:14:
    Couldn't match expected type `forall s. ST s t0'
                with actual type `m0 t10'
    Expected type: t10 -> forall s. ST s t0
      Actual type: t10 -> m0 t10
    In the first argument of `app', namely `return'
    In the second argument of `app', namely `(return `app` True)'

Again, this can be solved by adding type annotations:

> :t (app :: ((forall s0. ST s0 a) -> a) -> (forall s1. ST s1 a) -> a) runST (return `app` True)
(app :: ((forall s0. ST s0 a) -> a) -> (forall s1. ST s1 a) -> a) runST (return `app` True) :: Bool

What is happening here is that there is a special typing rule in GHC 7 which only applies to the standard ($) operator. Simon Peyton-Jones explains this behavior in a reply on the GHC users mailing list:

> This is a motivating example for type inference that can deal with > impredicative types. Consider the type of ($): > > ($) :: forall p q. (p -> q) -> p -> q > > In the example we need to instantiate p with (forall s. ST s a), and that's what > impredicative polymorphism means: instantiating a type variable with a > polymorphic type.
> > Sadly, I know of no system of reasonable complexity that can typecheck > [this] unaided. There are plenty of complicated systems, and I have > been a co-author on papers on at least two, but they are all Too > Jolly Complicated to live in GHC. We did have an implementation of > boxy types, but I took it out when implementing the new typechecker. > Nobody understood it. > > However, people so often write > > runST $ do ... > > that in GHC 7 I implemented a special typing rule, just for infix uses of ($). Just think of (f $ x) as a new > syntactic form, with the obvious typing rule, and away you go.

Your second example fails because there is no such rule for (.).

The runST $ do { ... } pattern is so common, and the fact that it normally wouldn't type-check is so annoying, that GHC included some ST-specific type-checking hacks to make it work. Those hacks are probably firing here for the ($) version, but not the (.) version.

The messages are a bit confusing the point (or so I feel). Let me rewrite your code:

runST (return True)   -- return True is ST s Bool
(runST . return) True  -- cannot work

Another way to put this is that the monomorphic m0 a0 (the result of return, if it would get an a0) cannot be unified with (forall s.ST s a).