Template specialization and instantiation

Question

These concepts is a bit unclear to me. Well, template instantiation's defined pretty well by N4296::14.7 [temp.spec]:

The act of instantiating a function, a class, a member of a class template or a member template is referred to as template instantiation.

That's if we have a function/variable/class template, instantiation of the template is just creating an object or function. For instance:

template<typename T> class U{ };
U<int> a; //instantiation

But N4296:14.7.1 [temp.inst] says (Emphasize mine):

Unless a class template specialization has been explicitly instantiated (14.7.2) or explicitly specialized (14.7.3), the class template specialization is implicitly instantiated when the specialization is referenced in a context that requires a completely-defined object type or when the completeness of the class type affects the semantics of the program.

What's the definition of the instantiation of the template specialization, not just the instantiation of a template?

For the first case, the compiler won't implicitly instantiate a template definition that has already been instantiated explicitly; for the second, the compiler won't instantiate a template definition at all if the given arguments match a more specialized one. — jthill

R Sahu R Sahu · Accepted Answer · 2015-02-12T05:38:10

Question:

What's the definition of the instantiation of the template specialization, not just the instantiation of a template?

My understanding:

There is no such thing as instantiation of a template. You always instantiate a template specialization.

If you have:

template <typename T> struct Foo {};

Foo<int> foo;

you have instantiated the template specialization Foo<int>, not the template Foo.

Update

Say you have the following class template:

template <typename T> struct Foo
{
   static int a;
};

int getNext()
{
   static int n = 0;
   return ++n;
}

template <class T> int Foo<T>::a = getNext();

Explicit template instantiation

You can create explicit instantiations of Foo<char> and Foo<int> by using:

template struct Foo<char>;
template struct Foo<int>;

Even if Foo<char> and Foo<int> are not used anywhere else in your code, the class template is instantiated for char and int.

Explicit template specialization

You can create explicit specializations of the class template by using:

template <> Foo<double> {};

Use of Foo

Now, let's see the use of Foo.

Foo<int> f1;    // An explicit instantiation has already been created.
                // No need for any further code creation.

Foo<double> f2; // An explicit specialization has already been created.
                // No need for any further code creation.

Foo<long> f3;   // There is no explicit instantiation or explicit specialization
                // Code needs to be created for Foo<long>

The third case, Foo<long> f3; triggers creation of the template specialization Foo<long>. I interpret the phrase "class template specialization is implicitly instantiated" to mean "creation of Foo<long> from the class template.".

Template specialization and instantiation

2 Answers

Update