CODEWITHSUNDEEP
c++
user48956
user48956

Reputation: 15788

How to force a template code creation without creating object instance?

I have a template class that is only valid for couple of template parameters:

doIt.h:

// only int and float are valid T
template <typename T>
class doer
{
public:
   void doIt();
}

I want to hide the implementation inside the .cpp file (for faster compile and also because its proprietary):

doIt.cpp:

template <>
void doer<T>::doIt()  {  /* how to do it */ }

... and use it as follows: use.cpp:

int main( int, char** )
{
   doer<int>::doIt()
}

The above fails to link because the implementation of void doer::doIt(void) was never in scope at the place where it was called.

I can force the code to be generated into doItv2.obj, as follows:

doIt_v2.cpp:

template <>
void doer<T>::doIt()  {  /* how to do it */ }

doer<int> a;    
doer<real> b;

but this causes a variety of headaches (dynamic memory allocation before main is entered) and I actually don't want to make an instance -- I just want the object code for the template instantiations to be generated.

Any ideas?

Upvotes: 4

Views: 595

Answers (1)

Nemanja Trifunovic
Nemanja Trifunovic

Reputation: 24551

See the article How to Organize Template Source Code. I think you are after the second method described there: explicit template instantiation.

Upvotes: 6

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