Difference between template name and template id

Question

C++ Standard

Section 14/2 :

In a function template declaration, the declarator-id shall be a template-name (i.e.,

Answer 1

A declarator-id is the syntactical element that specifies the name in a simple-declaration ("type name;"). In the following "A" and "B::C" is the declarator-id

int A;
int B::C;
int A();
int *A;
int A[42];
template void A();

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A type-id syntactically is roughly a simple-declaration where the declarator-id is missing. A type-id is used as the syntactical element in a template type argument and in a cast.

int // type-id
int* // type-id
int[] // type-id
int() // type-id
int(*)() // type-id

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A template-name is the name of a template. Syntactically it appears before a template-argument list. The above quote misuses "template-name" and "declarator-id", because a template-name is a plain identifier and does not contain any qualifiers. C++0x has changed the text to

In a function template declaration, the last component of the declarator-id shall be a template-name or operator-function-id (i.e., not a template-id).

(The last part appears in cases such as operator+()). Even the C++0x text misses some cases - see this defect report.

The misuse of "declarator-id" happens in the note. The note was replaced by C++0x with

[ Note: in a class template declaration, if the class name is a ... — end note ]

In class template declarations, the name specified syntactically is a class-name instead of a declarator-id. The relation of class-name and declarator-id is as follows (very simplified...)

class class-name { ... } declarator-id;
class foo        { ... } bar;

In class template declarations, there may not be a declarator-id specified.

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A template-id is a template-name followed by a template-argument list.

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The quote means that in a function template declaration, the name must not be a template-id. In your example you declare a function instead of a template. There are still cases where an explicit specialization declares a template, though. But that can only happen for member function templates

template
struct A {
  template
  void f();
};

// this explicit specialization *contains* a template declaration and
// declares an identifier (a template-name) qualified by A:: 
template<> template 
void A::f() { }