如何以简单的方式声明可变参数模板类的类型

Question

I have a variadic templated struct in VS2013 that uses a templated function to allow automatic type deduction.

template<typename... T> struct BitVector {
   BitVector(T... args){...}
}
using BitVector_t = struct BitVector<T...>;

template<typename... T>
std::shared_ptr<BitVector_t<T...>>
CreateBitVector(T... args) {
   auto v = new BitVector_t<T...>(args...);
   return std::shared_ptr<BitVector_t<T...>(v);
}

With this, I can define the items and defaultvalues of my bitvector in one single line. Furthermore I can easily create a BitVector by calling

auto mybitvector = CreateBitVector("Item1", int(1), "Item2", std::string("defval")...);

where the types are deduced from the function parameters. This works great with auto , but if I need a certain BitVector as a class member, where no auto is allowed, all efforts to have a clean and simple API, where types and default values are just specified at a single location, seem to be gone, as a class member would require an additional declaration like

std::shared_ptr<
BitVector<char const*, int, 
char const*, std::basic_string<char, std::char_traits<char>, std::allocator<char>>>> 
mMyBitVector;

Can anyone imagine a simple workaround, to get the member declaration, without writing the parameter types explicitly a second time?

Answer 1

As suggested in comment, decltype may help:

#define AUTO_MEMBER(memberName, init) decltype(init) memberName = (init)

class C
{
public:    
    //...
private:
    AUTO_MEMBER(mMyBitVector, CreateBitVector("Item1", 1, "Item2", std::string("defval")));
};

Demo

but I would prefer be explicit on member type (and may be using some typedef ).

Answer 2

Create a type alias with all the template parameters.

using MyBitVector_t = std::shared_ptr<BitVector<char const*, int, std::string>>;

Use the type alias to declare a member.

MyBitVector_t mMyBitVector;

The type alias can be used to define the return types of functions and types of arguments of functions too.