[Bug c++/44285] New: Need an option that will create symbols for all public c++ methods, not only for those which bodies are outside the class declaration

[yuri at tsoft dot com]

Consider the example below.
When compiled with flag -O3 methods meth_used and meth_unused aren't present in
the resulting assembly at all.

Now consider the situation when this class is a part of the package compiled
into the shared library and both meth_used and meth_unused are API of this
package.

Without the symbols for meth_used and meth_unused it's pretty much assumed that
the user will have c++ code that will recompile those function into the user
code. It's impossible to bind from other languages only knowing symbols without
c++ coding.

gcc should have an option, either compiler option, or the special keyword on
the method, that would guarantee the method presence in the object.

I looked into __attribute__ visibility. It only allows these values: "default",
"hidden", "protected" or "internal" and doesn't help in this situation.

Maybe "public" visibility should be added for this case?

-- example header: exp.h --
extern void zzz();

class Abc {
public:
  void meth_used() {
    zzz();
  }
  void meth_unused() {
    zzz();
  }
};

-- example header: exp.C --
#include "exp.h"

void my_exp() {
  (new Abc)->meth_used();
}


-- 
           Summary: Need an option that will create symbols for all public
                    c++ methods, not only for those which bodies are outside
                    the class declaration
           Product: gcc
           Version: 4.5.0
            Status: UNCONFIRMED
          Severity: enhancement
          Priority: P3
         Component: c++
        AssignedTo: unassigned at gcc dot gnu dot org
        ReportedBy: yuri at tsoft dot com


http://gcc.gnu.org/bugzilla/show_bug.cgi?id=44285