Typename


is a keyword in the C++ programming language used when writing templates. It is used for specifying that a dependent name in a template definition or declaration is a type. In the original C++ compilers before the first ISO standard was completed, the typename keyword was not part of the C++ language and Bjarne Stroustrup used the class keyword for template arguments instead. While typename is now the preferred keyword, older source code may still use the class keyword instead.

A synonym for class in template parameters

In C++'s generic programming feature known as "templates", typename can be used for introducing a template parameter:

// Define a generic function that returns the greater of its two arguments
template
nodiscard
const T& max noexcept

An alternative and semantically equivalent keyword in this scenario is "class":

// Define a generic function that returns the greater of its two arguments
template
nodiscard
const T& max noexcept

When using concepts, a means of constraining the generic type to satisfy some set of rules, the name of the concept replaces the keyword typename.

import std;
using std::is_base_of_v;
using std::vector;
class Player ;
template
concept ExtendsPlayer = is_base_of_v;
// T is required to be a type whose inheritance upper bound is Player,
// blocking any type that does not inherit from Player
template
void processListOfPlayers

A method for indicating that a dependent name is a type

Consider this invalid code:

template
void foo
struct UsesBar ;
int main

This code looks like it should compile, but it is incorrect because the compiler does not know if T::Bar is a type or a value. The reason it doesn't know is that T::Bar is a "template-parameter dependent name", or "dependent name" for short, which then could represent anything named "Bar" inside a type passed to foo, which could include typedefs, enums, variables, etc.
To resolve this ambiguity, the C++ Language Standard declares:
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.
In short, if the compiler can't tell if a dependent name is a value or a type, then it will assume that it is a value.
In our example, where T::Bar is the dependent name, that means that rather than declaring a pointer to T::Bar named p, the line

T::Bar* p;

will instead multiply the "value" T::Bar by p and throw away the result. The fact that in UsesBar the dependent Bar is in fact a type does not help since foo could be compiled long before UsesBar is seen. Furthermore, if there is also a class like:

struct HoldsBar ;

then the compiler would be obliged to interpret the T::Bar in foo as an access to data member HoldsBar::Bar when instantiated. But since Bar is not a static data member it will flag an error.
The solution to this problem is to explicitly tell the compiler that T::Bar is in fact a type. For this, the typename keyword is used:

template
void foo

Now the compiler knows for sure that T::Bar is a type, and will correctly make p a pointer to an object of that type.