Explicit Specialization of non-template member function is not called - c++

Question

...unless something else from the library is called. Here is a minimal example.
test1.cpp
#include <iostream>
void proofOfTwoLinked();
template <class T>
struct Foo
{
void bar(){ std::cout << "normal bar\n"; }
};
struct A{};
struct B{};
struct C{};
struct D{};
template <> void Foo<B>::bar(){ std::cout << "B bar\n"; }
int main()
{
Foo<A> a;
Foo<B> b;
Foo<C> c;
Foo<D> d;
a.bar();
b.bar();
c.bar();
d.bar();
//proofOfTwoLinked();
}
test2.cpp
#include <iostream>
struct C;
template <class T>
struct Foo
{
void bar(){ std::cout << "normal bar\n"; }
};
template <> void Foo<C>::bar(){ std::cout << "C bar\n"; }
void proofOfTwoLinked()
{
std::cout << "Yup, two is linked\n";
}
If I compile the two together, the program works as expected:
$ rm test; rm *.a; rm *.o; g++ -c test1.cpp; g++ -c test2.cpp; g++ -o test test1.o test2.o; ./test
normal bar
B bar
C bar
normal bar
If I compile test2, put it in an archive, and then link the program against that... the specialization for type C is not executed when c.bar() is called:
$ rm test; rm *.a; rm *.o; g++ -c test1.cpp; g++ -c test2.cpp; ar -r test2.a test2.o; g++ -o test test1.o test2.a; ./test
ar: creating test2.a
normal bar
B bar
normal bar
normal bar
But if I uncomment the last function call of test1 (proofOfTwoLinked) and then compile again, the specialization is executed.
$ rm test; rm *.a; rm *.o; g++ -c test1.cpp; g++ -c test2.cpp; ar -r test2.a test2.o; g++ -o test test1.o test2.a; ./test
ar: creating test2.a
normal bar
B bar
C bar
normal bar
Yup, two is linked
This strikes me as weird, and is certainly contrary to my expectation. Is this in fact normal behavior? Perhaps since there already exists some form of every function that is called in main() before the linker searches test2.a it skips the archive. Is there a way to force the linker to "look at the whole archive"?
I'm using gcc 4.6.1 and ar 2.21.53 (in ubuntu).

Answer 1

Using MSVC2010SP1 I get slightly different results:
compiling together as is I don't get "C bar". This is as expected because test1.cpp and test2.cpp are separate compilation units and with no forward declarations of the specializations that the other contains test1.cpp will instantiate its the default "normal bar" and test2.cpp will not instantiate the "C bar" because it can't see anything using it.
when I uncomment proofOfTwoLinked(); i get "Yup, two is linked" which is expected because "proofOfTwoLinked()" is forward declaired. I still don't get "C bar" which is as expected because it is not forward declaired in test1.cpp
When I compile again adding
template <> void Foo<C>::bar();
to test1.cpp I get a linker error because although the test1.cpp compilation unit now knows there is a
template <> void Foo<C>::bar()
out there somewhere, test2.cpp still does not know that anyone is using it.
When I compile again adding
template void Foo<C>::bar();
to test2.cpp everything works and I get "C bar". Note that
template void Foo<C>::bar();
must be BEFORE its definition.
As far as I can tell MSVC is acting correctly and gcc is acting weird in your case. I used http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2005/n1905.pdf section 14.7 as a reference, it might be helpful.