Specialization that is itself a template

Question

I have a template class that I have some specializations for.
But the next specialization is a template itself. How do you specify this:

template<typename T>
class Action
{
    public: void doStuff()  { std::cout << "Generic\n"; }
}

// A specialization for a person
template<>
class Action<Person>
{
    public: void doStuff()  { std::cout << "A Person\n";}
}


// I can easily specialize for vectors of a particular type.
// But how  do I change the following so that it works with all types of vector.
// Not just `int`
template<>
class Action<std::vector<int> >
{
    public: void doStuff()  { std::cout << "A Generic Vector\n";}
}

Now that I see the answer it is obvious, but I kept adding another template line in their. — Martin York, Jan 16 '12 at 18:47
Possible duplicate of [Templated class specialization where template argument is a template](https://stackoverflow.com/questions/4189945/templated-class-specialization-where-template-argument-is-a-template) — M.J. Rayburn, Sep 01 '17 at 19:37

score 19 · Accepted Answer · answered Jan 16 '12 at 07:30

19

Trivial partial specialization ?

template <typename T>
class Action<std::vector<T>> {
public:
  void doStuff() { std::cout << "A Generic Vector\n"; }
};

answered Jan 16 '12 at 07:30

Matthieu M.

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+1. Sometimes (often?) the obvious answer is the right answer :) – ereOn Jan 16 '12 at 08:02
@ereOn: the real question is --> why do I get so many votes for such a trivial answer :x ? – Matthieu M. Jan 16 '12 at 08:24
2

I guess people just love simplicity :) – ereOn Jan 16 '12 at 08:27

Specialization that is itself a template

1 Answers1