How can I implement a class which can passed as a template argument?

Question

I want to implement a class, call it K, which will compile when passed as a template argument for template class C. Here is the code sample below:

template <typename T>
class C 
{
    typename T::counter_type counter;
public:
    C() : counter(0) {}
    void Do (T t) 
    {
        if (t>0) ++counter;
    }
};

class K
{

};

Can anyone help me to understand the implementation?

`struct K { typedef some_integral_type counter_type; };` or alternatively `struct K { using counter_type = some_integral_type; };` — user657267, Dec 06 '15 at 23:59
compile, add functionality as requested by the error messages. — Karoly Horvath, Dec 07 '15 at 00:02

score 0 · Answer 1 · edited May 23 '17 at 11:52

What you have is a template class. The template argument is basically a data type that defines the types of the internal data types.

Think about of an implementation of std::vector. It has an internal array, but of what type? Well, you can specify that type as the template argument.

In your class, you declare counter of the type typename T::counter_type (see more about typename here: When is the "typename" keyword necessary?), and then you have a constructor which initializes the counter to 0 and a method Do, which takes an argument of type T (specified at object declaration) and does something with it.

See templates as a shorthand. Instead of having:

class C_int
{
    int counter;
public:
    C() : counter(0) {}
    void Do (int t) 
    {
        if (t>0) ++counter;
    }
};

class C_long
{
    long counter;
public:
    C() : counter(0) {}
    void Do (long t) 
    {
        if (t>0) ++counter;
    }
};

you now have C<int> and C<long>, respectively.

How can I implement a class which can passed as a template argument?

1 Answers1