gdb: don't treat empty enums as flag enums

In C++ it is possible to use an empty enum as a strong typedef.  For
example, a user could write:

  enum class my_type : unsigned char {};

Now my_type can be used like 'unsigned char' except the compiler will
not allow implicit conversion too and from the native 'unsigned char'
type.

This is used in the standard library for things like std::byte.

Currently, when GDB prints a value of type my_type, it looks like
this:

  (gdb) print my_var
  $1 = (unknown: 0x4)

Which isn't great.  This gets worse when we consider something like:

  std::vector<my_type> vec;

When using a pretty-printer, this could look like this:

  std::vector of length 2, capacity 2 = {(unknown: 0x2), (unknown: 0x4)}

Clearly not great.  This is described in PR gdb/30148.

The problem here is in dwarf2/read.c, we assume all enums are flag
enums unless we find an enumerator with a non-flag like value.
Clearly an empty enum contains no non-flag values, so we assume the
enum is a flag enum.

I propose adding an extra check here; that is, an empty enum should
never be a flag enum.

With this the above cases look more like:

  (gdb) print my_var
  $1 = 4

and:

  std::vector of length 2, capacity 2 = {2, 4}

Which look much better.

Bug: https://sourceware.org/bugzilla/show_bug.cgi?id=30148

Reviewed-By: Tom Tromey <tom@tromey.com>

1 files changed, 2 insertions, 0 deletions

diff --git a/gdb/dwarf2/read.c b/gdb/dwarf2/read.c
index 69b2310be1f..038f31cb06b 100644
--- a/gdb/dwarf2/read.c
+++ b/gdb/dwarf2/read.c
@@ -13195,6 +13195,8 @@ update_enumeration_type_from_children (struct die_info *die,
       memcpy (type->fields (), fields.data (),
 	      sizeof (struct field) * fields.size ());
     }
+  else
+    flag_enum = 0;
 
   if (unsigned_enum)
     type->set_is_unsigned (true);