template<typename Thing, template< typename... > typename Primary>
struct libsemigroups::is_specialization_of< Thing, Primary >
This struct contains the unique member value which indicates whether or not the template parameter Thing is a specialization of the template template parameter Primary. The implementation is based on this.
- Template Parameters
-
| Thing | the type we are checking is a specialization of Primary. |
| Primary | the type we are checking Thing against. |
- Example
Struct for checking if a template parameter is a specialization of another.
Definition is_specialization_of.hpp:56
- Note
- The template parameters of
Primary must be types, so, for example, is_specialization_of<std::array<int, 2>, std::array> will not compile.
- See also
- is_specialization_of_v
