[Bug c++/101844] New: Don't pretty print template parameter names that aren't needed

[Bug c++/101844] New: Don't pretty print template parameter names that aren't needed

[redi at gcc dot gnu.org]

https://gcc.gnu.org/bugzilla/show_bug.cgi?id=101844

            Bug ID: 101844
           Summary: Don't pretty print template parameter names that
                    aren't needed
           Product: gcc
           Version: 12.0
            Status: UNCONFIRMED
          Keywords: diagnostic
          Severity: normal
          Priority: P3
         Component: c++
          Assignee: unassigned at gcc dot gnu.org
          Reporter: redi at gcc dot gnu.org
  Target Milestone: ---

The diagnostic for this invalid code has ugly pretty printing:

struct A
{
  template<int> void f();
};

void A::f() { }


t.C:6:6: error: no declaration matches ‘void A::f()’
    6 | void A::f() { }
      |      ^
t.C:3:22: note: candidate is: ‘template<int <anonymous> > void A::f()’
    3 |   template<int> void f();
      |                      ^

There is no reason to print "<anonymous>" in the note here. Nothing else in the
declarator needs to refer to it (it can't refer to it, since it hasn't got a
name!) so printing it just makes the diagnostic harder to read.

Printing <anonymous> is strictly worse than printing nothing at all, but I
think we shouldn't print any template parameter names if they aren't used later
in the declarator. For example:

struct A
{
  template<int ReallyLongAndVerboseAndAlsoUglyName> void f();
};

void A::f() { }


This prints:

t.C:6:6: error: no declaration matches ‘void A::f()’
    6 | void A::f() { }
      |      ^
t.C:3:58: note: candidate is: ‘template<int
ReallyLongAndVerboseAndAlsoUglyName> void A::f()’
    3 |   template<int ReallyLongAndVerboseAndAlsoUglyName> void f();
      |                                                          ^

I don't see any benefit to printing the name in the note, it just adds noise.

Obviously there are some cases where we need to pretty print the template
parameter name, because it's used in another parameter:

  template<bool A, std::enable_if_t<A, bool> = true> void f();

Here we need to print A so that "enable_if<A, bool>" can use it:

t.C:5:59: note: candidate is: ‘template<bool A, typename std::enable_if<A,
bool>::type <anonymous> > void A::f()’

However, we don't need the <anonymous> for the second parameter.