跨编译单元的相同函数模板实例化的地址

I see similar SO questions stating that it does, but could someone explain it in more detail? Particularly, is this behavior protected by a standard?

i.h

#ifndef I_H_
#define I_H_

typedef void (*FuncPtr)();

template<typename T>
void FuncTemplate() {}

class C {};

#endif

a.cc

#include "i.h"

FuncPtr a() {
  return &FuncTemplate<C>;
}

b.cc

#include "i.h"

FuncPtr b() {
  return &FuncTemplate<C>;
}

m.cc

#include <iostream>

#include "i.h"

FuncPtr a();
FuncPtr b();

int main() {
  std::cout << (a() == b() ? "equal" : "not equal") << std::endl;

  return 0;
}

然后

$ g++ -c -o a.o a.cc
$ g++ -c -o b.o b.cc
$ g++ -c -o m.o m.cc
$ g++ a.o b.o m.o -o prog
$ ./prog
equal

将 -Wall -Wextra -Werror -ansi 扔到所有 g++ 调用上会产生相同的结果.

$ g++ -shared -o liba.so a.cc
$ g++ -shared -o libb.so b.cc
$ g++ -c -o m.o m.cc
$ g++ -L. -la -lb m.o -o prog
$ ./prog
equal