Suppose I have the following class
class Example {
public:
using value_type = std::tuple<
uint8_t,
uint8_t,
uint16_t
>;
private:
value_type _value;
};
Now, I want to be able to create another class based upon this type that wraps each of the classes types in another type. Based upon Wrapping each type in a variadic template in a templated class, I know I can accomplish half my goal via:
template <typename T>
class Wrapper;
template <typename ... ARGS>
class ExampleWrapper {
private:
std::tuple<Wrapper<ARGS>...> _args;
};
However, what I can't figure out is how to get ARGS if all I know is T, where T is Example. I would like to be able to use ExampleWrapper as follows:
ExampleWrapper<Example> myWrapper;
CodePudding user response:
You can use template partial specialization to get ARGS:
template <typename T>
class Wrapper;
template <typename Tuple>
class ExampleWrapper;
template <typename ... ARGS>
class ExampleWrapper<std::tuple<ARGS...>> {
private:
std::tuple<Wrapper<ARGS>...> _args;
};
Then:
ExampleWrapper<Example::value_type> myWrapper;
CodePudding user response:
Make a static function to return the desired type and use decltype on it.
template <typename T>
class Wrapper{};
template<class T>
class ExampleWrapper {
public:
template<typename ...Args>
static std::tuple<Wrapper<Args>...> fnc(std::tuple<Args...>);
using value_type = decltype( fnc( std::declval<typename T::value_type>() ) );
private:
value_type _args;
};
template<typename T>
void test(T){ std::court << __PRETTY_FUNCTION__; }
int main(){
ExampleWrapper<Example>::value_type a;
test(a); /* void t(T) [with T = std::tuple<Wrapper<unsigned char>,
Wrapper<unsigned char>,
Wrapper<short unsigned int> >]
*/
// The `T` in the above example is the type of `ExampleWrapper<Example>::_args`
}
There's no need to make the static function declaration and value_type alias public. I made them public just for the sake of demonstration.
