Typename

C++ keyword

"`typename`"[1][2] 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.[3][4] 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 (for example see the difference in source code examples between The Design and Evolution of C++ by Bjarne Stroustrup published in 1994 and the source code examples in The C++ Programming Language: Fourth Edition by Bjarne Stroustrup published in 2013).

## A synonym for "`class`" in template parameters

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In C++'s generic programming feature known as "templates", `typename` can be used for introducing a template parameter:[3][4]

// Define a generic function that returns the greater of its two arguments
template <typename T>
const T& max(const T& x, const T& y)

if (y < x)
return x;
return y;

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

// Define a generic function that returns the greater of its two arguments
template <class T>
const T& max(const T& x, const T& y)

if (y < x)
return x;
return y;

## A method for indicating that a dependent name is a type

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Consider this invalid code:[5]

template <typename T>
void foo(const T& t)

// declares a pointer to an object of type T::bar
T::bar * p;  // error (see text)

struct StructWithBarAsType

typedef int bar;

int main()

StructWithBarAsType x;
foo(x);

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` (which is nowhere to be found) and throw away the result. The fact that in `StructWithBarAsType` the dependent `bar` is in fact a type does not help since `foo()` could be compiled long before `StructWithBarAsType` is seen. Furthermore, if there is also a class like:

struct StructWithBarAsValue

int bar;

then the compiler would be obliged to interpret the `T::bar` in `foo()` as an access to data member `StructWithBarAsValue::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:[3][4]

template <typename T>
void foo(const T& t)

// declares a pointer to an object of type T::bar
typename T::bar * p;

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.

## See also

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* Argument-dependent name lookup – another C++ name lookup rule

## References

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1. ^ Al Stevens (April 2003). "Underdocumented C++". Dr. Dobb's Journal. pp. 72–76.
2. ^ T. L. Veldhuizen (2013). "C++ templates are turing complete" (PDF).
3. ^ a b c "The typename keyword (C++ only)". IBM. Retrieved August 23, 2013.
4. ^ a b c "MSDN - typename". MSDN. Retrieved August 23, 2013.[permanent dead link]
5. ^ "Dependent name lookup for C++ templates". February 6, 2012.