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Imagine we have three .h files: f.h : template <typename T> class Class {public: Class() {} T id(T x) { return x; }}; g.h : template <typename T> class Class {public: Class() {} T id(T x) { return x + 100; }}; h.h : template <typename T> class Class {public: Class(); T id(T x); }; Now, we also have three .cpp files: f.cpp : #include "f.h" int f(int x) { Class<int> t; return t.id(x); } g.cpp : #include "g.h" int g(int x) { Class<int> t; return t.id(x); } h.cpp : #include "h.h" int h(int x) { Class<int> t; return t.id(x); } Compiling them gives us f.o , g.o and h.o . Now let's throw in this main.cpp : #include <stdio> extern int f(int); extern int g(int); extern int h(int); int main() { std::cout << f(1) << std::endl; std::cout << g(2) << std::endl; std::cout << h(3) << std::endl; } A-a-and let's do g++ main.cpp f.o g.o h.o . Now comes my actual surprise: Since those three .o files contain three different definitions for int Class<int>::id(int) , I expect to get a linking error. However, what I get is a working a.out , which prints 1 2 3 . And if I reorder .o files in the command, it will print 101 102 103 . And now for the actual questions: How exactly does the linker perform linking in this case? How does it figures out what instantiation of Class<int> to keep and what to throw away? And why it does not complain about multiple definitions? nm</c

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