Some tidying up of type pretty-printing

Triggered by the changes in #13677, I ended up doing a bit of
refactoring in type pretty-printing.

* We were using TyOpPrec and FunPrec rather inconsitently, so
  I made it consisent.

* That exposed the fact that we were a bit undecided about whether
  to print
     a + b -> c + d   vs   (a+b) -> (c+d)
  and similarly
     a ~ [b] => blah  vs   (a ~ [b]) => blah

  I decided to make TyOpPrec and FunPrec compare equal
  (in BasicTypes), so (->) is treated as equal precedence with
  other type operators, so you get the unambiguous forms above,
  even though they have more parens.

  We could readily reverse this decision.
  See Note [Type operator precedence] in BasicTypes

* I fixed a bug in pretty-printing of HsType where some
  parens were omitted by mistake.

1 files changed, 3 insertions, 3 deletions

diff --git a/testsuite/tests/backpack/should_fail/bkpfail24.stderr b/testsuite/tests/backpack/should_fail/bkpfail24.stderr
index ef6a7d6c01..484ebf144b 100644
--- a/testsuite/tests/backpack/should_fail/bkpfail24.stderr
+++ b/testsuite/tests/backpack/should_fail/bkpfail24.stderr
@@ -7,15 +7,15 @@ bkpfail24.bkp:14:15: error:
     • Could not deduce: a ~ b
       from the context: {H1.T} ~ {H2.T}
         bound by the type signature for:
-                   f :: forall a b. {H1.T} ~ {H2.T} => a -> b
+                   f :: forall a b. ({H1.T} ~ {H2.T}) => a -> b
         at bkpfail24.bkp:13:9-34
       ‘a’ is a rigid type variable bound by
         the type signature for:
-          f :: forall a b. {H1.T} ~ {H2.T} => a -> b
+          f :: forall a b. ({H1.T} ~ {H2.T}) => a -> b
         at bkpfail24.bkp:13:9-34
       ‘b’ is a rigid type variable bound by
         the type signature for:
-          f :: forall a b. {H1.T} ~ {H2.T} => a -> b
+          f :: forall a b. ({H1.T} ~ {H2.T}) => a -> b
         at bkpfail24.bkp:13:9-34
     • In the expression: x
       In an equation for ‘f’: f x = x