When Is the "Typename" Keyword Necessary

When is the typename keyword necessary?

Short answer: Whenever referring to a nested name that is a dependent name, i.e. nested inside a template instance with unknown parameter.

Long answer: There are three tiers of entities in C++: values, types, and templates. All of those can have names, and the name alone doesn't tell you which tier of entity it is. Rather, the information about the nature of a name's entity must be inferred from the context.

Whenever this inference is impossible, you have to specify it:

template <typename> struct Magic; // defined somewhere else

template <typename T> struct A
{
  static const int value = Magic<T>::gnarl; // assumed "value"

  typedef typename Magic<T>::brugh my_type; // decreed "type"
  //      ^^^^^^^^

  void foo() {
    Magic<T>::template kwpq<T>(1, 'a', .5); // decreed "template"
    //        ^^^^^^^^
  }
};

Here the names Magic<T>::gnarl, Magic<T>::brugh and Magic<T>::kwpq had to be expliciated, because it is impossible to tell: Since Magic is a template, the very nature of the type Magic<T> depends on T -- there may be specializations which are entirely different from the primary template, for example.

What makes Magic<T>::gnarl a dependent name is the fact that we're inside a template definition, where T is unknown. Had we used Magic<int>, this would be different, since the compiler knows (you promise!) the full definition of Magic<int>.

(If you want to test this yourself, here's a sample definition of Magic that you can use. Pardon the use of constexpr in the specializaation for brevity; if you have an old compiler, feel free to change the static member constant declaration to the old-style pre-C++11 form.)

template <typename T> struct Magic
{
  static const T                    gnarl;
  typedef T &                       brugh;
  template <typename S> static void kwpq(int, char, double) { T x; }
};
template <> struct Magic<signed char>
{
  // note that `gnarl` is absent
  static constexpr long double brugh = 0.25;  // `brugh` is now a value
  template <typename S> static int kwpq(int a, int b) { return a + b; }
};

Usage:

int main()
{
  A<int> a;
  a.foo();

  return Magic<signed char>::kwpq<float>(2, 3);  // no disambiguation here!
}

Why do we need typename here?

In general, C++ needs typename because of the unfortunate syntax [*] it inherits from C, that makes it impossible without non-local information to say -- for example -- in A * B; whether A names a type (in which case this is a declaration of B as a pointer to it) or not (in which case this is a multiplication expression -- quite possible since A, for all you can tell without non-local information, could be an instance of a class that overloads operator* to do something weird;-).

In most cases the compiler does have the non-local information needed to disambiguate (though the unfortunate syntax still means the low-level parser needs feedback from the higher-level layer that keeps the symbol table info)... but with templates it doesn't (not in general, though in this specific case it might be technically illegal to specialize a std::list<T> so that its ::iterator is NOT a type name;-).

[*] not just my opinion, but also the opinion of Ken Thompson and Rob Pikes, currently my colleagues, who are busy designing and implementing a new programming language for internal use: that new programming language, while its syntax is mostly C-like, does NOT repeat C's syntax design errors -- it the new language (like e.g. in good old Pascal), syntax is sufficient to distinguish identifiers that must name a type from ones that must not;-).

Where and why do I have to put the template and typename keywords?

(See here also for my C++11 answer)

In order to parse a C++ program, the compiler needs to know whether certain names are types or not. The following example demonstrates that:

t * f;

How should this be parsed? For many languages a compiler doesn't need to know the meaning of a name in order to parse and basically know what action a line of code does. In C++, the above however can yield vastly different interpretations depending on what t means. If it's a type, then it will be a declaration of a pointer f. However if it's not a type, it will be a multiplication. So the C++ Standard says at paragraph (3/7):

Some names denote types or templates. In general, whenever a name is encountered it is necessary to determine whether that name denotes one of these entities before continuing to parse the program that contains it. The process that determines this is called name lookup.

How will the compiler find out what a name t::x refers to, if t refers to a template type parameter? x could be a static int data member that could be multiplied or could equally well be a nested class or typedef that could yield to a declaration. If a name has this property - that it can't be looked up until the actual template arguments are known - then it's called a dependent name (it "depends" on the template parameters).

You might recommend to just wait till the user instantiates the template:

Let's wait until the user instantiates the template, and then later find out the real meaning of t::x * f;.

This will work and actually is allowed by the Standard as a possible implementation approach. These compilers basically copy the template's text into an internal buffer, and only when an instantiation is needed, they parse the template and possibly detect errors in the definition. But instead of bothering the template's users (poor colleagues!) with errors made by a template's author, other implementations choose to check templates early on and give errors in the definition as soon as possible, before an instantiation even takes place.

So there has to be a way to tell the compiler that certain names are types and that certain names aren't.

The "typename" keyword

The answer is: We decide how the compiler should parse this. If t::x is a dependent name, then we need to prefix it by typename to tell the compiler to parse it in a certain way. The Standard says at (14.6/2):

A name used in a template declaration or definition and that is dependent on a template-parameter is
assumed not to name a type unless the applicable name lookup finds a type name or the name is qualified
by the keyword typename.

There are many names for which typename is not necessary, because the compiler can, with the applicable name lookup in the template definition, figure out how to parse a construct itself - for example with T *f;, when T is a type template parameter. But for t::x * f; to be a declaration, it must be written as typename t::x *f;. If you omit the keyword and the name is taken to be a non-type, but when instantiation finds it denotes a type, the usual error messages are emitted by the compiler. Sometimes, the error consequently is given at definition time:

// t::x is taken as non-type, but as an expression the following misses an
// operator between the two names or a semicolon separating them.
t::x f;

The syntax allows typename only before qualified names - it is therefor taken as granted that unqualified names are always known to refer to types if they do so.

A similar gotcha exists for names that denote templates, as hinted at by the introductory text.

The "template" keyword

Remember the initial quote above and how the Standard requires special handling for templates as well? Let's take the following innocent-looking example:

boost::function< int() > f;

It might look obvious to a human reader. Not so for the compiler. Imagine the following arbitrary definition of boost::function and f:

namespace boost { int function = 0; }
int main() { 
  int f = 0;
  boost::function< int() > f; 
}

That's actually a valid expression! It uses the less-than operator to compare boost::function against zero (int()), and then uses the greater-than operator to compare the resulting bool against f. However as you might well know, boost::function in real life is a template, so the compiler knows (14.2/3):

After name lookup (3.4) finds that a name is a template-name, if this name is followed by a <, the < is
always taken as the beginning of a template-argument-list and never as a name followed by the less-than
operator.

Now we are back to the same problem as with typename. What if we can't know yet whether the name is a template when parsing the code? We will need to insert template immediately before the template name, as specified by 14.2/4. This looks like:

t::template f<int>(); // call a function template

Template names can not only occur after a :: but also after a -> or . in a class member access. You need to insert the keyword there too:

this->template f<int>(); // call a function template

Dependencies

For the people that have thick Standardese books on their shelf and that want to know what exactly I was talking about, I'll talk a bit about how this is specified in the Standard.

In template declarations some constructs have different meanings depending on what template arguments you use to instantiate the template: Expressions may have different types or values, variables may have different types or function calls might end up calling different functions. Such constructs are generally said to depend on template parameters.

The Standard defines precisely the rules by whether a construct is dependent or not. It separates them into logically different groups: One catches types, another catches expressions. Expressions may depend by their value and/or their type. So we have, with typical examples appended:

Dependent types (e.g: a type template parameter T)
Value-dependent expressions (e.g: a non-type template parameter N)
Type-dependent expressions (e.g: a cast to a type template parameter (T)0)

Most of the rules are intuitive and are built up recursively: For example, a type constructed as T[N] is a dependent type if N is a value-dependent expression or T is a dependent type. The details of this can be read in section (14.6.2/1) for dependent types, (14.6.2.2) for type-dependent expressions and (14.6.2.3) for value-dependent expressions.

Dependent names

The Standard is a bit unclear about what exactly is a dependent name. On a simple read (you know, the principle of least surprise), all it defines as a dependent name is the special case for function names below. But since clearly T::x also needs to be looked up in the instantiation context, it also needs to be a dependent name (fortunately, as of mid C++14 the committee has started to look into how to fix this confusing definition).

To avoid this problem, I have resorted to a simple interpretation of the Standard text. Of all the constructs that denote dependent types or expressions, a subset of them represent names. Those names are therefore "dependent names". A name can take different forms - the Standard says:

A name is a use of an identifier (2.11), operator-function-id (13.5), conversion-function-id (12.3.2), or template-id (14.2) that denotes an entity or label (6.6.4, 6.1)

An identifier is just a plain sequence of characters / digits, while the next two are the operator + and operator type form. The last form is template-name <argument list>. All these are names, and by conventional use in the Standard, a name can also include qualifiers that say what namespace or class a name should be looked up in.

A value dependent expression 1 + N is not a name, but N is. The subset of all dependent constructs that are names is called dependent name. Function names, however, may have different meaning in different instantiations of a template, but unfortunately are not caught by this general rule.

Dependent function names

Not primarily a concern of this article, but still worth mentioning: Function names are an exception that are handled separately. An identifier function name is dependent not by itself, but by the type dependent argument expressions used in a call. In the example f((T)0), f is a dependent name. In the Standard, this is specified at (14.6.2/1).

Additional notes and examples

In enough cases we need both of typename and template. Your code should look like the following

template <typename T, typename Tail>
struct UnionNode : public Tail {
    // ...
    template<typename U> struct inUnion {
        typedef typename Tail::template inUnion<U> dummy;
    };
    // ...
};

The keyword template doesn't always have to appear in the last part of a name. It can appear in the middle before a class name that's used as a scope, like in the following example

typename t::template iterator<int>::value_type v;

In some cases, the keywords are forbidden, as detailed below

On the name of a dependent base class you are not allowed to write typename. It's assumed that the name given is a class type name. This is true for both names in the base-class list and the constructor initializer list:
```
 template <typename T>
 struct derive_from_Has_type : /* typename */ SomeBase<T>::type 
 { };
```

In using-declarations it's not possible to use template after the last ::, and the C++ committee said not to work on a solution.

 template <typename T>
 struct derive_from_Has_type : SomeBase<T> {
    using SomeBase<T>::template type; // error
    using typename SomeBase<T>::type; // typename *is* allowed
 };

Officially, what is typename for?

Following is the quote from Josuttis book:

The keyword typename was introduced to
specify that the identifier that
follows is a type. Consider the
following example:
template <class T>
Class MyClass
{
  typename T::SubType * ptr;
  ...
};
Here, typename is used to clarify that
SubType is a type of class T. Thus,
ptr is a pointer to the type
T::SubType. Without typename, SubType
would be considered a static member.
Thus
T::SubType * ptr
would be a multiplication of value
SubType of type T with ptr.

Why is typename necessary in return type? C++

While parsing the return type (as defined in the question) we are not in the scope of Queue<T> yet. Your reasoning would be correct if you were to write

template<typename T>
auto Queue<T>::test() -> Node* {

}

The nested name qualifier of the fully qualified function name puts us into the scope of the current template specialization. Here unqualified name lookup finds Node* in the current specialization, where it is known to refer to a type.

But while parsing the return type in your question the compiler hasn't yet encountered the "current specialization" where Node* can be unambiguously assumed to name a type. All it sees is us writing a dependent name from some specialization. As such, typename is required. It's no different to us having to write

template<typename T>
typename OtherUnrelatedClass<T>::Node* Queue<T>::test() {

}

Why is the keyword typename needed before qualified dependent names, and not before qualified independent names?

A name in C++ can pertain to three different tiers of entities: Types, values, and templates.

struct Foo
{
    typedef int A;                   // type
    static double B;                 // value
    template <typename T> struct C;  // template
};

The three names Foo::A, Foo::B and Foo::C are examples of all three different tiers.

In the above example, Foo is a complete type, and so the compiler knows already what Foo::A etc. refer to. But now imagine this:

template <typename T> struct Bar
{
    T::A x;
};

Now we are in trouble: what is T::A? if T = Foo, then T::A = int, which is a type, and all is well. But when T = struct { static char A; };, then T::A is a value, which doesn't make sense.

Therefore, the compiler demands that you tell it what T::A and T::B and T::C are supposed to be. If you say nothing, it is assumed to be a value. If you say typename, it is a typename, and if you say template, it is a template:

template <typename T> struct Bar
{
    typename T::A x;    // ah, good, decreed typename

    void foo()
    {
        int a = T::B;   // assumed value, OK

        T::template C<int> z;  // decreed template
        z.gobble(a * x);
    }
};

Secondary checks such as whether T::B is convertible to int, whether a and x can be multiplied, and whether C<int> really has a member function gobble are all postponed until you actually instantiate the template. But the specification whether a name denotes a value, a type or a template is fundamental to the syntactic correctness of the code and must be provided right there during the template definition.

Why is `typename` prefix still required in such a case in C++20?

The answer is speaking loosely about what "context" means. What the feature is doing is finding places where it is grammatically impossible for anything other than a typename to appear in that location. It is those places where typename is no longer necessary.

For example, using name = X;. Grammatically, this is a type alias declaration. As such, whatever X is, the grammar requires that it be a type. If the X happens to be a name dependent on a template parameter, whatever that name is must be a type. So the grammar is unambiguous, and typename is redundant.

The syntax name<X> states that X is a template argument to the template name. But template arguments do not have to be types. A specific template may take a type as a specific argument, but grammatically, X could be an identifier referencing a variable or a template. Therefore, you must disambiguate ambiguous dependent Xs.

Why is repeating typename keyword in function argument list necessary?

You need typename because std::vector::size_type is a depended name.

When Is the "Typename" Keyword Necessary