条件显式模板实例化 [英] Conditional explicit template instantiation

问题描述

除了预处理器之外，我如何有条件地启用/禁用显式模板实例化？

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

请考虑：

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

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

在某些编译条件下，Type3与Type1相同，Type4与Type2相同。当这种情况发生时，我得到一个错误。我想检测的类型是相同的，而不是实例化Type3和Type4如

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>;

我已经转移了自己尝试enable_if和SFINAE（我相信我知道他们为什么失败）只有预处理器工作了（ugh）。我正在考虑将类型放在元组或变量中，删除重复的，然后使用剩余的实例化。

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.

有条件启用/禁用显式基于模板参数类型的模板实例化？

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

推荐答案

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>;

这仍然会产生四个显式实例化，但是在 Type3 与 Type1 相同（除非 Type1 为 dummy< 3> ！）

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>!)

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