Dependent Names


A dependent name is essentially a name that depends on a template parameter. A dependent name can be a type, a non-type, or a template parameter. To express that a dependent name stands for a type or a template, you have to use the keywords typename or template.


Before I write about dependent names, I should write about template parameters.

Template Parameter

Template parameters can be types, non-types, and templates.


Types are the most often used template parameters. Here are a few examples:

std::vector<int> myVec;
std::map<std::string, int> myMap;
std::lock_guard<std::mutex> myLockGuard;


Non-types can be a

  • lvalue reference
  • nullptr
  • pointer
  • enumerator
  • integral types
  • floating-point types (C++20)
  • literal types (C++20)

Non-type template parameters are often just called NTTP.

Integrals are the most used non-types. std::array is the typical example because you have to specify at compile time the size of a std::array:

std::array<int, 3> myArray{1, 2, 3};


With C++20, you can also use two new non-types: floating-point types and literal types.

Literal Types must have the following two properties, essentially:

  • All base classes and non-static data members are public and non-mutable
  • The types of all base classes and non-static data members are structural types or arrays of these

A literal type must have a constexpr constructor.

The following program uses floating-point types and literal types as template parameters.


// nonTypeTemplateParameter.cpp

struct ClassType {
    constexpr ClassType(int) {}  // (1)

template <ClassType cl>          // (2)
auto getClassType() {
    return cl;

template <double d>              // (3)
auto getDouble() {
    return d;

int main() {

    auto c1 = getClassType<ClassType(2020)>();

    auto d1 = getDouble<5.5>();  // (4)
    auto d2 = getDouble<6.5>();  // (4)



ClassType has a constexpr constructor (1) and can be used as a template argument (2). The function template getDouble (3) accepts only doubles. I want to emphasize it explicitly, that each call of the function template getDouble (4) with a new argument triggers the instantiation of a new function template specialization of getDouble.  This means that there are two instantiations for the doubles 5.5 and 6.5 are created.


Templates can be template parameters, and in this case, they are called template template parameters.


// templateTemplateParameters.cpp

#include <iostream>
#include <list>
#include <vector>
#include <string>

template <typename T, template <typename, typename> class Cont >   // (1)
class Matrix{
  explicit Matrix(std::initializer_list<T> inList): data(inList) { // (2)
    for (auto d: data) std::cout << d << " ";
  int getSize() const{
    return data.size();

  Cont<T, std::allocator<T>> data;                                 // (3)                               


int main() {

  std::cout << '\n';

                                                                    // (4)
  Matrix<int, std::vector> myIntVec{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}; 
  std::cout << std::endl;
  std::cout << "myIntVec.getSize(): " << myIntVec.getSize() << '\n';

  std::cout << std::endl;

  Matrix<double, std::vector> myDoubleVec{1.1, 2.2, 3.3, 4.4, 5.5}; // (5)
  std::cout << std::endl;
  std::cout << "myDoubleVec.getSize(): "  << myDoubleVec.getSize() << '\n';

  std::cout << std::endl;
                                                                    // (6)
  Matrix<std::string, std::list> myStringList{"one", "two", "three", "four"};  
  std::cout << std::endl;
  std::cout << "myStringList.getSize(): " << myStringList.getSize() << '\n';

  std::cout << '\n';



Matrix is a simple class template that can be initialized by a std::initializer_list (line 2). A Matrix can be used with a std::vector (line 4 and line 5), or a std::list (line 6) to hold its values. So far, nothing special. 


Line 1 declares a class template that has two template parameters. The first parameter is the type of the elements, and the second parameter stands for the container. Let's have a closer look at the second parameter: template <typename, typename> class Cont>. This means the second template argument should be a template requiring two template parameters. The first template parameter is the type of elements the container stores, and the second template parameter is the defaulted allocator a container of the standard template library has. The allocator has a default value, such as in the case of a std::vector. The allocator depends on the type of elements.

    typename T,
    typename Allocator = std::allocator<T>
> class vector;


Line 3 shows the usage of the allocator in this internally used container. The matrix can be instantiated with all containers that are of the kind: container< type of the elements, allocator of the elements>. This is true for the sequence containers such as std::vector, std::deque, or std::liststd::array, and std::forward_list would fail because std::array needs an additional non-type for specifying its size at compile-time and std::forward_list does not support the size method.

Now, I can write about dependent names.

Dependent Names

What is a dependent name? A dependent name is essentially a name that depends on a template parameter. Here are a few examples based on


template<typename T>
struct X : B<T>                 // "B<T>" is dependent on T
    typename T::A* pa;          // "T::A" is dependent on T
    void f(B<T>* pb) {
        static int i = B<T>::i; // "B<T>::i" is dependent on T
        pb->j++;                // "pb->j" is dependent on T


Now, the fun starts. A dependent name can be a type, a non-type, or a template parameter. The name lookup is the big difference between non-dependent and dependent names.

  • Non-dependent names are looked up at the point of the template definition.
  • Dependent names are looked up at the point of the template instantiation.

When you use a dependent name in a template declaration, the compiler has no idea, whether this name refers to a type, a non-type, or a template parameter. In this case, the compiler assumes that the dependent name refers to a non-type, which may be wrong. This is the case in which you have to give the compiler a hint.

Before I show you two examples, I have to write about an exception to this rule. You can skip my next few words if you want to get a general idea and jump directly to the next section. The exception is: if the name refers in the template definition to the current instantiation, the compiler can deduce the name at the point of the template definition. Here are a few examples:


template <typename T> 
class A { A* p1; // A is the current instantiation A<T>* p2; // A<T> is the current instantiation ::A<T>* p4; // ::A<T> is the current instantiation A<T*> p3; // A<T*> is not the current instantiation }; template <typename T>
class A<T*> { A<T*>* p1; // A<T*> is the current instantiation A<T>* p2; // A<T> is not the current instantiation }; template <int I>
struct B { static const int my_I = I; static const int my_I2 = I + 0; static const int my_I3 = my_I; B<my_I>* b3; // B<my_I> is the current instantiation B<my_I2>* b4; // B<my_I2> is not the current instantiation B<my_I3>* b5; // B<my_I3> is the current instantiation };


Finally, I come to the critical idea of my post. If a dependent name could be a type, a non-type, or a template, you have to give the compiler a hint.

Use typename if the Dependent Name is a Type

After such a long introduction, the following program snippet makes it pretty clear.


template <typename T>
void test(){
    std::vector<T>::const_iterator* p1;          // (1)
    typename std::vector<T>::const_iterator* p2; // (2)


Without the typename keyword in line 2, the name std::vector<T>::const_iterator in line 2 would be interpreted as a non-type and, consequently, the * stands in this case for multiplication and not for a pointer declaration. Exactly this is happening in line (1).

Similarly, if your dependent name should be a template, you have to give the compiler a hint.

Use .template if the Dependent Name is a Template

Honestly, this syntax looks a bit weird.


template<typename T>
struct S {
    template <typename U> void func(){}
template<typename T>
void func2(){
    S<T> s;
    s.func<T>();             // (1)
    s.template func<T>();    // (2)


Same story as before. Compare lines 1 and 2. When the compiler reads the name s.func (line 1) it decides to interpret it as non-type. This means the < sign stands for the comparison operator, but not opening square bracket of the template argument of the generic method func. In this case, you have to specify that s.func is a template such as in line 2: s.template func

Here is the essence of this post in one sentence: When you have a dependent name, use typename to specify that the dependent name is a type or use .template to specify that the dependent name is a template.

What's next?

 In my next post, I present automatic returns types. They are often a life saver when it comes to function templates.


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0 #1 Marlon 2021-10-11 08:52
Thank you for your posts on templates. I find them very clear and easy to follow. Please keep them coming !

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