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# Type alias, alias template (since C++11)

From cppreference.com
< cpp‎ | language

Type alias is a name that refers to a previously defined type (similar to typedef)

Alias template is a name that refers to a family of types.

### [edit] Syntax

Alias declarations are block declarations with the following syntax

`using` identifier attr(optional) `=` type-id `;` | (1) |
`template` `<` template-parameter-list `>` `using` identifier attr(optional) `=` type-id `;` | (2) |

attr(C++11) | - | optional sequence of any number of attributes
identifiers | - | the name that is introduced by this declaration, which becomes either a type name (1) or a template name (2)
template-parameter-list | - | template parameter list, as in template declaration
type-id | - | abstract declarator or any other valid type-id. For alias template declaration, type_id cannot directly or indirectly refer to identifier

### [edit] Explanation

1) A type alias declaration introduces a name which can be used as a synonym for the type denoted by type-id. It does not introduce a new type and it cannot change the meaning of an existing type name. There is no difference between a type alias declaration and typedef declaration.
2) An alias template is a template which, when specialized, is equivalent to the result of substituting the template arguments of the alias template for the template parameters in the type-id

    template<class T> struct Alloc {};
    template<class T> using Vec = vector<T, Alloc<T>>; // type-id is vector<T, Alloc<T>>
    Vec<int> v; // Vec<int> is the same as vector<int, Alloc<int>>

The type produced when specializing an alias template is not allowed to directly or indirectly make use of its own type:

    template <class T> struct A;
    template <class T> using B = typename A<T>::U; // type-id is A<T>::U
    template <class T> struct A {
        typedef B<T> U;
    };
    B<short> b; // error: B<short> uses its own type via A<short>::U

Alias templates are never deduced by template argument deduction when deducing a template template parameter.

It is not possible to partially or fully specialize an alias template.

### [edit] Example

Run this code

    #include <string>
    #include <ios>
    #include <vector>
    #include <type_traits>

    // type alias, identical to
    // typedef std::ios_base::fmtflags flags;
    using flags = std::ios_base::fmtflags;
    // the name 'flags' now denotes a type:
    flags fl = std::ios_base::dec;

    // type alias, identical to
    // typedef void (*func)(int, int);
    using func = void (*) (int,int);
    // the name 'func' now denotes a pointer to function:
    void example(int, int) {}
    func fn = example;

    // template type alias
    template<class T> using ptr = T*;
    // the name 'ptr<T>' is now an alias for pointer to T
    ptr<int> x;

    // type alias used to hide a template parameter
    template <class CharT> using mystring =
        std::basic_string<CharT,std::char_traits<CharT>>;
    mystring<char> str;

    // type alias can introduce a member typedef name
    template<typename T>
    struct Container {
        using value_type = T;
    };
    // which can be used in generic programming
    template<typename Container>
    void fn2(const Container& c)
    {
        typename Container::value_type n;
    }

    // type alias used to simplify the syntax of std::enable_if
    template <typename T> using Invoke =
        typename T::type;
    template <typename Condition> using EnableIf =
        Invoke<std::enable_if<Condition::value>>;
    template <typename T, typename = EnableIf<std::is_polymorphic<T>>>
    int fpoly_only(T t) { return 1; }

    struct S { virtual ~S() {} };
    int main()
    {
        Container<int> c;
        fn2(c); // Container::value_type will be int in this function

    //    fpoly_only(c); // error, enable_if prohibits this
        S s;
        fpoly_only(s); // okay, enable_if allows this
    }

### [edit] See also

typedef declaration | creates a synonym for a type[edit]

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