Force definition of symbol for a C++ template instance in a library -

i'd provide library provides template code. keep possible ownership of code (generated code) when can guess usage of different usual types of template. here example of trying do:

lib1.h

#include <iostream>  template<int n> void print_me() {     std::cout << "i function number " << n << std::endl; } 

lib1.cpp

#include "lib1.h"  /* force symbols defined here. */ template void print_me<0>(); template void print_me<1>(); 

i compile library using:

g++ -shared -fpic lib1.cpp -o lib1.so

and when use library:

main.cpp

#include <lib1.h>  int main() {     print_me<0>();     print_me<1>();     print_me<2>(); } 

compiled with:

g++ main.cpp -l1

here expect symbol print_me<0>() , print_me<1>() defined , used lib1.so , print_me<2>() defined , used executable (checked nm --defined-only). seems not case! symbols 0 , 1 defined in lib1.so weak symbols. , redefined in executable (0, 1 , 2) again, weak. implies code 0 , 1 executable taken main.cpp not want (i checked specification in main.cpp).

is there way (in lib1.h instance) @ compile time of main.cpp symbols defined somewhere , not need add these symbols?

c++11 solution: use extern templates. add these strings main.cpp file:

extern template void print_me<0>(); extern template void print_me<1>(); 

thus tell compiler not instantiate print_me function template in main.cpp (for template arguments 0 , 1). linker should search definition of void print_me<0>(); , void print_me<1>(); in other translation units.