CODEWITHSUNDEEP
c++templatesc++11
walrii
walrii

Reputation: 3522

Conditional explicit template instantiation

Other than the preprocessor, how can I conditionally enable/disable explicit template instantiations?

Consider:

template <typename T> struct TheTemplate{ /* blah */ };

template struct TheTemplate<Type1>;
template struct TheTemplate<Type2>;
template struct TheTemplate<Type3>;
template struct TheTemplate<Type4>;

Under some compilation conditions, Type3 is the same as Type1 and Type4 is the same as Type2. When this happens, I get an error. I'd like to detect that the types are the same and not instantiate on Type3 and Type4 as in

// this does not work
template struct TheTemplate<Type1>;
template struct TheTemplate<Type2>;
template struct TheTemplate<enable_if<!is_same<Type1, Type3>::value, Type3>::type>;
template struct TheTemplate<enable_if<!is_same<Type2, Type4>::value, Type4>::type>;

I've diverted myself trying enable_if and SFINAE (and I believe I know why they fail), but only the preprocessor has worked (ugh). I'm thinking about putting the types in a tuple or variadic, removing duplicates, and then use the remainder for instantiation.

Is there a way to conditionally enable/disable explicit template instantiation based on template argument types?

Upvotes: 7

Views: 2432

Answers (1)

Jonathan Wakely
Jonathan Wakely

Reputation: 171263

template <typename T> struct TheTemplate{ /* blah */ };

template<int> struct dummy { };

template struct TheTemplate<Type1>;
template struct TheTemplate<Type2>;
template struct TheTemplate<conditional<is_same<Type1, Type3>::value, dummy<3>, Type3>::type>;
template struct TheTemplate<conditional<is_same<Type2, Type4>::value, dummy<4>, Type4>::type>;

This still produces four explicit instantiations, but they won't be duplicates in the case where Type3 is the same as Type1 (unless Type1 is dummy<3>!)

Upvotes: 7

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