Re: [RFA 2/4] dwarf2_physname

[Tom Tromey <tromey@redhat.com>]

>>>>> "Daniel" == Daniel Jacobowitz <dan@codesourcery.com> writes:

>> +die_needs_namespace (struct die_info *die, struct dwarf2_cu *cu)
>> {
[...]
>> +    case DW_TAG_variable:
>> +      {
>> +	struct attribute *attr;
>> +	attr = dwarf2_attr (die, DW_AT_specification, cu);
>> +	if (attr)
>> +	  return 1;
>> +	attr = dwarf2_attr (die, DW_AT_external, cu);
>> +	if (attr == NULL && die->parent->tag != DW_TAG_namespace)
>> +	  return 0;
>> +	return 1;

Daniel> What's this about?  It needs an explanation, and I don't think
Daniel> DW_AT_specification has anything to do with die_needs_namespace.

I read through this sub-thread a little, and looked at the DWARF spec
too.

Here is what I believe to be the relevant text from the spec:

    If a type, variable, or function declared in a namespace is defined
    outside of the body of the namespace declaration, that type, variable,
    or function definition entry has a DW_AT_specification attribute whose
    value is a reference to the debugging information entry representing the
    declaration of the type, variable or function. Type, variable, or
    function entries with a DW_AT_specification attribute do not need to
    duplicate information provided by the declaration entry referenced by
    the specification attribute.

So, if a DIE has a DW_AT_specification, then we must use the namespace
indicated by the referenced DIE.  If it does not have DW_AT_specification,
then we use its own parentage.

It is not clear to me whether it is possible for a DIE to have a
DW_AT_specification referring to another DIE which is not in a
namespace.  Offhand this seems invalid, but if it does occur, it could
be supported in this function with just a little more effort.

I don't understand why die_needs_namespace only does this for
DW_TAG_variable and not other things.  I also don't understand why it
unconditionally returns 1 for enumerators, functions, and the like.
However, I didn't look to see how it is used, so perhaps it doesn't
matter.

After reading more of the thread, maybe I am missing the point.

Tom