Am I guaranteed to not be bitten by this ODR violation? - c++

Question

I have a header file that declares a template with a static variable and also defines it:
/* my_header.hpp */
#ifndef MY_HEADER_HPP_
#define MY_HEADER_HPP_
#include <cstdio>
template<int n>
struct foo {
static int bar;
static void dump() { printf("%d\n", bar); }
};
template<int n>
int foo<n>::bar;
#endif // MY_HEADER_HPP_
This header is included both by main.cpp and a shared library mylib. In particular, mylib_baz.hpp just includes this template and declares a function that modifies a specialization of the template.
/* mylib_baz.hpp */
#ifndef MYLIB_BAZ_HPP_
#define MYLIB_BAZ_HPP_
#include "my_header.hpp"
typedef foo<123> mylib_foo;
void init_mylib_foo();
#endif // MYLIB_BAZ_HPP_
and
/* mylib_baz.cpp */
#include "mylib_baz.hpp"
void init_mylib_foo() {
mylib_foo::bar = 123;
mylib_foo::dump();
};
When I make mylib.so (containing mylib_baz.o), the symbol for foo<123>::bar is present and declared weak. However, the symbol for foo<123>::bar is declared weak also in my main.o:
/* main.cpp */
#include "my_header.hpp"
#include "mylib_baz.hpp"
int main() {
foo<123>::bar = 456;
foo<123>::dump(); /* outputs 456 */
init_mylib_foo(); /* outputs 123 */
foo<123>::dump(); /* outputs 123 -- is this guaranteed? */
}
It appears that I am violating one definition rule (foo<123>::bar defined both in my_header.cpp and main.cpp). However, both with g++ and clang the symbols are declared weak (or unique), so I am not getting bitten by this -- all accesses to foo<123>::bar modify the same object.
Question 1: Is this a coincidence (maybe ODR works differently for static members of templates?) or am I in fact guaranteed this behavior by the standard?
Question 2: How could I have predicted the behavior I'm observing? That is, what exactly makes the compiler declare symbol weak?

Answer 1

There is no ODR violation. You have one definition of bar. It is here:
template<int n>
int foo<n>::bar; // <==
As bar is static, that indicates that there is one definition across all translation units. Even though bar will show up once in all of your object files (they need a symbol for it, after all), the linker will merge them together to be the one true definition of bar. You can see that:
$ g++ -std=c++11 -c mylib_baz.cpp -o mylib_baz.o
$ g++ -std=c++11 -c main.cpp -o main.o
$ g++ main.o mylib_baz.o -o a.out
Produces:
$ nm mylib_baz.o | c++filt | grep bar
0000000000000000 u foo<123>::bar
$ nm main.o | c++filt | grep bar
0000000000000000 u foo<123>::bar
$ nm a.out | c++filt | grep bar
0000000000601038 u foo<123>::bar
Where u indicates:
"u"
The symbol is a unique global symbol. This is a GNU extension to the standard set of ELF symbol bindings. For such a symbol the dynamic linker will make sure that in the entire process there is just one symbol with this name and type in use.