How to read the template partial specialization?

Question

Suppose the following declaration:

template <typename T> struct MyTemplate;

The following definition of the partial specialization seems to use the same letter T to refer to different types.

template <typename T> struct MyTemplate<T*> {};

For example, let's take a concrete instantiation:

MyTemplate<int *> c;

Now, consider again the above definition of the partial specialization:

template <typename T> struct MyTemplate<T*> {};

In the first part of this line (i.e. template <typename T>), T is int *. In the second part of the line (i.e. MyTemplate<T*>), T is int!

So, how is the definition of the partial specialization read?

It is similar to normal argument: the `a` in `int foo(int a)`, `void bar(int a)` and `void foobar(std::string&a)` are not the same. — Jarod42, Aug 02 '15 at 10:07
I understand the effect of the definition. However, using the same letter for two different types in the same line of code does not make sense to me. It's like having `T(T)`, which gets resolved as `int *(int)`... I have edited the question to emphasize the point. — AlwaysLearning, Aug 02 '15 at 10:11
`In the first part of this line (i.e. template ), T is int *,`. It isn't. What makes you think so? If you assume that `int*` in the invication is somehow mspped to that T, then no, it is not. — n. m. could be an AI, Aug 02 '15 at 10:21

Kerrek SB · Accepted Answer · 2015-08-02T10:34:21.423

Read it like this:

The primary template says "MyTemplate is a class template with one type parameter":
```
template <typename> struct MyTemplate;
```
The partial specialization says, "whenever there exists a type T"...
```
template <typename T>
```
... such that a specialization of MyTemplate is requested for the type T *"...
```
struct MyTemplate<T *>
```
... then use this alternative definition of the template.
You could also define explicit specializations. For example, could say "whenever the specialization is requested for type X, use this alternative definition:
```
template <> struct MyTemplate<X> { /* ... */ };
```

Note that explicit specializations of class templates define types, wheras partial specializations define templates.

To see it another way: A partial class template specialization deduces, or pattern-matches, the structure of the class template arguments:

template <typename T> struct MyTemplate<T *>
//       ^^^^^^^^^^^^                  ^^^^^
//       This is a new template        Argument passed to the original class
//                                     template parameter

The parameter names of this new template are matched structurally against the argument of the original class template's parameters.

Examples:

MyTemplate<void>: The type parameter of the class template is void, and the primary template is used for this specialization.
MyTemplate<int *>: The type parameter is int *. There exists a type T, namely T = int, such that the requested type parameter is T *, and so the definition of the partial specialization of the template is used for this specialization.
MyTemplate<X>: The parameter type is X, and an explicit specialization has been defined for that parameter type, which is therefore used.

n. m. could be an AI · Answer 2 · 2015-08-02T10:53:18.260

The correct reading of the specialisation is as follows:

template <typename T> // a *type pattern* with a *free variable* `T` follows
struct MyTemplate<T*> // `T*` is the pattern

When the template is instantiated by MyTemplate<int*>, the argument is matched against the pattern, not the type variable list. Values of the type variables are then deduced from the match.

To see this more directly, consider a template with two arguments.

template <typename T1, typename T2>
struct A;

and its specialisation

template <typename T1, typename T2>
struct A<T1*, T2*>;

Now you can write the latter as

template <typename T2, typename T1>
struct A<T1*, T2*>;

(the variable list order is reversed) and this is equivalent to the previous one. Indeed, order in the list is irrelevant. When you invoke A<int*, double*> it is deduced that T1=int, T2=double, regardless of the order of T1 and T2 in the template head.

Further, you can do this

template <typename T>
struct A<T*, T*>;

and use it in A<int*, int*>. It is now plainly clear that the type variable list has no direct correspondence with the actual template parameter list.

Note: the terms "pattern", "type variable", "type pattern matching" are not standard C++ terms. They are pretty much standard almost everywhere else though.

There seems to be a typo: it says T2 twice in the reversed order. — AlwaysLearning, Aug 02 '15 at 10:52

score 0 · Answer 3 · answered Aug 02 '15 at 10:26

You have this line:

MyTemplate<int *> c;

Your confusion seems to come from assuming that the int * in < > somehow corresponds to the T in template <typename T>. It does not. In a partial specialisation (actually in every template declaration), the template parameter names are simply "free variable" (or perhaps "placeholder") names. The template arguments (int * in your case) don't directly correspond to these, they correspond to what is (or would be) in the < > following the template name.

This means that the <int *> part of the instantiation maps to the <T*> part of the partial specialisation. T is just a name introduced by the template <typename T> prefix. In the entire process, the T is int.

Spock77 · Answer 4 · 2015-08-02T10:35:50.243

There is no contradiction. T should be read as T, T* is T*.

template <typename T> struct MyTemplate<T*> {};

"In the first part of this line (i.e. template <typename T>), T is int *."

No - in template <typename T> T is int, in struct MyTemplate<T*> {}; T is also int.

"Note that when a partial specialization is used, a template parameter is deduced from the specialization pattern; the template parameter is not simply the actual template argument. In particular, for Vector<Shape*>, T is Shape and not Shape*." (Stroustrup C++, 4th ed, 25.3, p. 732.)

How to read the template partial specialization?

4 Answers4