Deduce template parameter pack from function call - c++

Question

I have the following code, where I have a template class, and a type in it, which I would like to use in a separate template function.
template <typename... Types>
struct MyClass
{
enum SomeEnum { value0 = -1 };
};
template <typename... Types>
struct OtherClass
{
};
template <typename T, typename... Types>
T check(typename MyClass<Types...>::SomeEnum value)
{
OtherClass<Types...> obj;
T result;
// calculate result from obj;
return result;
}
int main() {
auto value = MyClass<int, bool>::value0;
// ...
int t = check<int>(value);
}
I tought that the compiler will be able to deduce the parameter pack from the function call, so I can use it in the function template also. Unfortunately the compiler can't deduce it:
$ g++ -std=c++11 op.cpp
op.cpp: In function ‘int main()’:
op.cpp:25:27: error: cannot convert ‘MyClass<int, bool>::SomeEnum’ to ‘MyClass<>::SomeEnum’ for argument ‘1’ to ‘T check(typename MyClass<Types ...>::SomeEnum) [with T = int; Types = {}; typename MyClass<Types ...>::SomeEnum = MyClass<>::SomeEnum]’
int t = check<int>(value);
Is there a solution to "transfer" the template parameter pack to the template function?

Answer 1

Template deduction is not possible, but maybe you can restructure your code in a way that MyClass defines the all necessary types and then you have a check function that takes MyClass as a template argument. That way, the checking function has access to all the necessary types.
template <typename... Types> struct OtherClass {};
template <typename... Types>
struct MyClass
{
typedef OtherClass<Types...> OtherClass_t;
typedef int result_t;
enum SomeEnum { value0 = -1 };
};
// version 1
template < typename C >
struct Checker {
typename C::result_t operator()(typename C::SomeEnum value)
{
typename C::OtherClass_t obj;
typename C::result_t result;
// calculate result from obj;
return result;
}
};
// version 2
template < typename C >
typename C::result_t check_fun(typename C::SomeEnum value)
{
typename C::OtherClass_t obj;
typename C::result_t result;
// calculate result from obj;
return result;
}
int main() {
typedef MyClass< int, bool > myclass_t;
auto value = myclass_t::value0;
// ...
Checker< myclass_t > check;
int t = check(value);
auto s = check_fun<myclass_t>(value);
}
The downside is of course, that you have to instantiate the checker class or call the function with the proper type of MyClass as template argument.

Answer 2

Template arguments cannot be deduced from nested types. This isn't new or changed with variadic templates.

Answer 3

The template parameter pack can be passed over using std::tuple. A wrapper class needed over SomeEnum type to store the parameter pack by creating a tuple type from them:
template <typename... Types>
struct MyClass
{
struct Value {
enum SomeEnum { value0 = -1 };
enum SomeEnum value;
typedef std::tuple<Types...> TypeTuple;
};
};
Than a helper class needed which feeds the template argument list of a template class from the tuple types:
template <
template <typename...> class Class,
typename Tuple, typename T, T... nums>
struct Helper_ : Class <
typename std::tuple_element<nums, Tuple>::type... >
{};
template <template <typename...> class Class, typename Tuple>
struct Helper : Helper_<
Class, Tuple,
make_integer_sequence<int, std::tuple_size<Tuple>::value > >
{};
The check function then uses this helper class to instanciate the other class:
template <typename T, typename V>
T check(V value)
{
Helper<OtherClass, typename V::TypeTuple> obj;
T result;
// calculate result from obj;
return result;
}
And the use of check function changes a bit becouse now it waits the wrapper type instead of the pure enum:
int main() {
MyClass<int, bool, double>::Value value;
value.value = MyClass<int, bool, double>::Value::value0;
int t = check<int>(value);
}