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From: "vincenzo.innocente at cern dot ch" <gcc-bugzilla@gcc.gnu.org>
To: gcc-bugs@gcc.gnu.org
Subject: [Bug c++/35688] template visibility not overridden by template arguments
Date: Tue, 08 Nov 2011 08:53:00 -0000
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http://gcc.gnu.org/bugzilla/show_bug.cgi?id=3D35688

--- Comment #12 from vincenzo Innocente <vincenzo.innocente at cern dot ch>=
 2011-11-08 08:49:22 UTC ---
much better!
for the test in comment 7 now I get
c++ -O0 -shared -fPIC -fvisibility=3Dhidden -o bha.so testICF.cpp=20
nm bha.so | c++filt | grep " T "
0000000000000e5c T go()
00000000000069ce T unsigned long const& std::max<unsigned long>(unsigned lo=
ng
const&, unsigned long const&)
00000000000010b2 T std::__lg(long)
00000000000010d8 T operator new(unsigned long, void*)

puzzled by the visibility of std::max above I modified the original test
further as

cat visTest.cc=20
namespace s __attribute__((visibility("default"))) {
  template <class T>
    class vector {
  public:
    vector() { }
  };
  template <class T>
    void foo(T t) {
  }
}

class A {
public:
  A() { }
  s::vector<int> v;
};

s::vector<A> v;

int main() {
  A a;
  s::foo(a);
  s::foo(v);
  s::foo(a.v);

  return 0;
}


and I get default visibility for s::vector<int>, as expected. not necessari=
ly
as wished=E2=80=A6
In my opinion this is consistent with spec, expect bug reports though!

 c++ -O0 -shared -fPIC -fvisibility=3Dhidden -o bha.so visTest.cc=20
 nm bha.so | c++filt=20
0000000000000d77 t __GLOBAL__sub_I_visTest.cc
0000000000000d4c t __static_initialization_and_destruction_0(int, int)
0000000000000d8c t A::A()
0000000000000dba t void s::foo<A>(A)
0000000000000dc0 t void s::foo<s::vector<A> >(s::vector<A>)
0000000000000dc6 T void s::foo<s::vector<int> >(s::vector<int>)
0000000000000db0 t s::vector<A>::vector()
0000000000000da6 T s::vector<int>::vector()
0000000000000d04 t __dyld_func_lookup
0000000000000d0a t _main
0000000000001038 d _v
                 U dyld_stub_binder
0000000000000cf0 t dyld_stub_binding_helper


p.s I verified that if "s" in NOT default visible  vector<int> will become
hidden