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Flexibility of template alias in C++0x

开发者 https://www.devze.com 2023-01-24 07:13 出处:网络
As I understand, template aliases in C++0x will allow us to do the following: template <typename T>

As I understand, template aliases in C++0x will allow us to do the following:

template <typename T>
using Dictionary = std::map< std::string, T >;

Dictionary<开发者_高级运维int> ints;
ints[ "one" ] = 1;
ints[ "two" ] = 2;

I have two questions:

First, will we be able to do this (bind to any type, or just templates):

template <typename Iter>
using ValueType = std::iterator_traits<Iter>::value_type;

Second, will using the aliases require usage of the typename keyword in templates, e.g.:

template <typename Iter>
typename ValueType<Iter> sum(Iter first, Iter last) { ... }
// ^ required?

Or is it required in the alias declaration?

using ValueType = typename std::iterator_traits<Iter>::value_type;
//                   ^ required?

Or neither?


The syntax is:

template <typename Iter>
using ValueType = typename std::iterator_traits<Iter>::value_type;

as with your second one.

Source: http://www2.research.att.com/~bs/C++0xFAQ.html#template-alias

Their example is:

template<int N>
    using int_exact = typename int_exact_traits<N>::type;  // define alias for convenient notation


typename is required when a member type follows the :: operator and a template-id precedes it.

The usage of typename you mention isn't specific to template aliases nor is it required unless you're aliasing to a member such as ::type, but that is a common use case.

For example, there's no typename when introducing a simple alias name to an existing template.

template< typename x >
class bar;

template< typename x >
using foo = bar< x >; // no typename needed
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