In templates, where and why do I have to put typename and template on dependent names? What exactly are dependent names anyway? I have the following code:

template <typename T, typename Tail> // Tail will be a UnionNode too.

struct UnionNode : public Tail {

    // ...

    template<typename U> struct inUnion {

        // Q: where to add typename/template here?

        typedef Tail::inUnion<U> dummy; 

    };

    template< > struct inUnion<T> {

    };

};

template <typename T> // For the last node Tn.

struct UnionNode<T, void> {

    // ...

    template<typename U> struct inUnion {

        char fail[ -2 + (sizeof(U)%2) ]; // Cannot be instantiated for any U

    };

    template< > struct inUnion<T> {

    };

};

The problem I have is in the typedef Tail::inUnion<U> dummy line. I'm fairly certain that inUnion is a dependent name, and VC++ is quite right in choking on it. I also know that I should be able to add template somewhere to tell the compiler that inUnion is a template-id. But where exactly? And should it then assume that inUnion is a class template, i.e. inUnion<U> names a type and not a function?

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
  
   };
  
  

  
  

C++11

Problem

While the rules in C++03 about when you need typename and template are largely reasonable, there is one annoying disadvantage of its formulation

template<typename T>

struct A {

  typedef int result_type;



  void f() {

    // error, "this" is dependent, "template" keyword needed

    this->g<float>();



    // OK

    g<float>();



    // error, "A<T>" is dependent, "typename" keyword needed

    A<T>::result_type n1;



    // OK

    result_type n2; 

  }



  template<typename U>

  void g();

};

As can be seen, we need the disambiguation keyword even if the compiler could perfectly figure out itself that A::result_type can only be int (and is hence a type), and this->g can only be the member template g declared later (even if A is explicitly specialized somewhere, that would not affect the code within that template, so its meaning cannot be affected by a later specialization of A!).

Current instantiation

To improve the situation, in C++11 the language tracks when a type refers to the enclosing template. To know that, the type must have been formed by using a certain form of name, which is its own name (in the above, A, A<T>, ::A<T>). A type referenced by such a name is known to be the current instantiation. There may be multiple types that are all the current instantiation if the type from which the name is formed is a member/nested class (then, A::NestedClass and A are both current instantiations).

Based on this notion, the language says that CurrentInstantiation::Foo, Foo and CurrentInstantiationTyped->Foo (such as A *a = this; a->Foo) are all member of the current instantiation if they are found to be members of a class that is the current instantiation or one of its non-dependent base classes (by just doing the name lookup immediately).

The keywords typename and template are now not required anymore if the qualifier is a member of the current instantiation. A keypoint here to remember is that A<T> is still a type-dependent name (after all T is also type dependent). But A<T>::result_type is known to be a type - the compiler will "magically" look into this kind of dependent types to figure this out.

struct B {

  typedef int result_type;

};



template<typename T>

struct C { }; // could be specialized!



template<typename T>

struct D : B, C<T> {

  void f() {

    // OK, member of current instantiation!

    // A::result_type is not dependent: int

    D::result_type r1;



    // error, not a member of the current instantiation

    D::questionable_type r2;



    // OK for now - relying on C<T> to provide it

    // But not a member of the current instantiation

    typename D::questionable_type r3;        

  }

};

That's impressive, but can we do better? The language even goes further and requires that an implementation again looks up D::result_type when instantiating D::f (even if it found its meaning already at definition time). When now the lookup result differs or results in ambiguity, the program is ill-formed and a diagnostic must be given. Imagine what happens if we defined C like this

template<>

struct C<int> {

  typedef bool result_type;

  typedef int questionable_type;

};

A compiler is required to catch the error when instantiating D<int>::f. So you get the best of the two worlds: "Delayed" lookup protecting you if you could get in trouble with dependent base classes, and also "Immediate" lookup that frees you from typename and template.

Unknown specializations

In the code of D, the name typename D::questionable_type is not a member of the current instantiation. Instead the language marks it as a member of an unknown specialization. In particular, this is always the case when you are doing DependentTypeName::Foo or DependentTypedName->Foo and either the dependent type is not the current instantiation (in which case the compiler can give up and say "we will look later what Foo is) or it is the current instantiation and the name was not found in it or its non-dependent base classes and there are also dependent base classes.

Imagine what happens if we had a member function h within the above defined A class template

void h() {

  typename A<T>::questionable_type x;

}

In C++03, the language allowed to catch this error because there could never be a valid way to instantiate A<T>::h (whatever argument you give to T). In C++11, the language now has a further check to give more reason for compilers to implement this rule. Since A has no dependent base classes, and A declares no member questionable_type, the name A<T>::questionable_type is neither a member of the current instantiation nor a member of an unknown specialization. In that case, there should be no way that that code could validly compile at instantiation time, so the language forbids a name where the qualifier is the current instantiation to be neither a member of an unknown specialization nor a member of the current instantiation (however, this violation is still not required to be diagnosed).

Examples and trivia

You can try this knowledge on this answer and see whether the above definitions make sense for you on a real-world example (they are repeated slightly less detailed in that answer).

The C++11 rules make the following valid C++03 code ill-formed (which was not intended by the C++ committee, but will probably not be fixed)

struct B { void f(); };

struct A : virtual B { void f(); };



template<typename T>

struct C : virtual B, T {

  void g() { this->f(); }

};



int main() { 

  C<A> c; c.g(); 

}

This valid C++03 code would bind this->f to A::f at instantiation time and everything is fine. C++11 however immediately binds it to B::f and requires a double-check when instantiating, checking whether the lookup still matches. However when instantiating C<A>::g, the Dominance Rule applies and lookup will find A::f instead.