Handling zero-argument variadic template in C++11

问题

Consider the following artificial example:

template <typename T, typename... Args>
struct A {
  typedef T Type;
};

Using A with 1 or more arguments works while using it with zero arguments fails as expected:

error: wrong number of template arguments (0, should be 1 or more)

Is it possible to make A handle the case of zero template arguments defining A::Type to int if there are no arguments and to the first template argument if there are?

回答1:

First define the primary template as the most general case — which also includes zero argument:

template <typename... Args>            //general : 0 or more 
struct A { using Type = int; }

Then partially specialize it for 1 or more parameters as:

template <typename T, typename... Args> //special : 1 or more
struct A<T,Args...>  { using Type = T; }

Once you have this specialization, the primary template would be used for zero-argument only!

Note that mathematically 1 or more is a special case of 0 or more — the latter is a more general case (not the other way round).

回答2:

You could just add a default to the first argument, no specializations necessary:

#include <type_traits>

template <typename T = int, typename... >
struct A {
    using type = T;
};

int main() {
    static_assert(std::is_same<A<>::type, int>::value, "");
    static_assert(std::is_same<A<double, char>::type, double>::value, "");
}

来源：https://stackoverflow.com/questions/29879564/handling-zero-argument-variadic-template-in-c11