Note [Left and right sections]

Dealing with left sections (x *) and right sections (* x) is
surprisingly fiddly.  We expand like this
     (`op` e) ==> rightSection op e
     (e `op`) ==> leftSection  (op e)

Using an auxiliary function in this way avoids the awkwardness of
generating a lambda, esp if `e` is a redex, so we *don't* want
to generate `(\x -> op x e)`. See Historical
Note [Desugaring operator sections]

Here are their definitions:
   leftSection :: forall r1 r2 n (a::TYPE r1) (b::TYPE r2).
                  (a %n-> b) -> a %n-> b
   leftSection f x = f x

   rightSection :: forall r1 r2 r3 n1 n2 (a::TYPE r1) (b::TYPE r2) (c::TYPE r3).
                   (a %n1 -> b %n2-> c) -> b %n2-> a %n1-> c
   rightSection f y x = f x y

Note the wrinkles:

* We do /not/ use lookupSyntaxName, which would make left and right
  section fall under RebindableSyntax.  Reason: it would be a user-
  facing change, and there are some tricky design choices (#19354).
  Plus, infix operator applications would be trickier to make
  rebindable, so it'd be inconsistent to do so for sections.

  TL;DR: we still use the renamer-expansion mechanism for operator
  sections, but only to eliminate special-purpose code paths in the
  renamer and desugarer.

* leftSection and rightSection must be representation-polymorphic, to allow
  (+# 4#) and (4# +#) to work. See
  Note [Wired-in Ids for rebindable syntax] in GHC.Types.Id.Make.

* leftSection and rightSection must be multiplicity-polymorphic.
  (Test linear/should_compile/OldList showed this up.)

* Because they are representation-polymorphic, we have to define them
  as wired-in Ids, with compulsory inlining.  See
  GHC.Types.Id.Make.leftSectionId, rightSectionId.

* leftSection is just ($) really; but unlike ($) it is
  representation-polymorphic in the result type, so we can write
  `(x +#)`, say.

* The type of leftSection must have an arrow in its first argument,
  because (x `ord`) should be rejected, because ord does not take two
  arguments

* It's important that we define leftSection in an eta-expanded way,
  (i.e. not leftSection f = f), so that
      (True `undefined`) `seq` ()
      = (leftSection (undefined True) `seq` ())
  evaluates to () and not undefined

* If PostfixOperators is ON, then we expand a left section like this:
      (e `op`)  ==>   op e
  with no auxiliary function at all.  Simple!

* leftSection and rightSection switch on ImpredicativeTypes locally,
  during Quick Look; see GHC.Tc.Gen.App.wantQuickLook. Consider
  test DeepSubsumption08:
     type Setter st t a b = forall f. Identical f => blah
     (.~) :: Setter s t a b -> b -> s -> t
     clear :: Setter a a' b (Maybe b') -> a -> a'
     clear = (.~ Nothing)
   The expansion look like (rightSection (.~) Nothing).  So we must
   instantiate `rightSection` first type argument to a polytype!
   Hence the special magic in App.wantQuickLook.

Historical Note [Desugaring operator sections]
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
This Note explains some historical trickiness in desugaring left and
right sections.  That trickiness has completely disappeared now that
we desugar to calls to 'leftSection` and `rightSection`, but I'm
leaving it here to remind us how nice the new story is.

Desugaring left sections with -XPostfixOperators is straightforward: convert
(expr `op`) to (op expr).

Without -XPostfixOperators it's a bit more tricky. At first it looks as if we
can convert

    (expr `op`)

naively to

    \x -> op expr x

But no!  expr might be a redex, and we can lose laziness badly this
way.  Consider

    map (expr `op`) xs

for example. If expr were a redex then eta-expanding naively would
result in multiple evaluations where the user might only have expected one.

So we convert instead to

    let y = expr in \x -> op y x

Also, note that we must do this for both right and (perhaps surprisingly) left
sections. Why are left sections necessary? Consider the program (found in #18151),

    seq (True `undefined`) ()

according to the Haskell Report this should reduce to () (as it specifies
desugaring via eta expansion). However, if we fail to eta expand we will rather
bottom. Consequently, we must eta expand even in the case of a left section.

If `expr` is actually just a variable, say, then the simplifier
will inline `y`, eliminating the redundant `let`.

Note that this works even in the case that `expr` is unlifted. In this case
bindNonRec will automatically do the right thing, giving us:

    case expr of y -> (\x -> op y x)

See #18151.