Parameterized typedef possible?

Question

I was wondering if it is possible to have some kind of parameterized typedef.

To illustrate, in my code I use this typedef:

typedef std::queue<std::vector<unsigned char>, std::deque<std::vector<unsigned char> > > UnsignedCharQueue;

As you can see this is a rather unwieldy construct so the typedef makes sense. However, if I want to have queues with other datatypes I need to define them beforehand explizitly.

So I was thinking if it were possible to use a construct like this:

typedef std::queue<std::vector<T>, std::deque<std::vector<T> > > Queue<T>;

private:
    Queue<unsigned char> mMyQueue;

Similar like generics in Java.

possible duplicate of [C++ template typedef](http://stackoverflow.com/questions/2795023/c-template-typedef) — , Apr 30 '13 at 18:40

Andy Prowl · Accepted Answer · 2013-04-30T19:39:54.773

10

In C++11, you can use template aliases, such as in:

template<typename T>
using my_alias = some_class_template<T>;

// ...
my_alias<T> obj; // Same as "some_class_template<T> obj;"

So in your case it would be:

template<typename T>
using Queue = std::queue<std::vector<T>, std::deque<std::vector<T> > >;

Also notice, that in C++11 you do not need to leave a space between closed angle brackets, so the above can be rewritten as follows:

template<typename T>
using Queue = std::queue<std::vector<T>, std::deque<std::vector<T>>>;
//                                                               ^^^

In C++03 you could define a Queue metafunction this way:

template<typename T>
struct Queue
{
    typedef std::queue<std::vector<T>, std::deque<std::vector<T> > > type;
};

Which you would then use this way:

Queue<int>::type obj;

If you are using it in a template with parameter T (as in the following), do not forget the typename disambiguator:

template<typename T>
struct X
{
    typename Queue<T>::type obj;
//  ^^^^^^^^
}

edited Apr 30 '13 at 19:39

answered Apr 30 '13 at 18:39

Andy Prowl

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Damn, you're lightning fast. I was going to post this. – Apr 30 '13 at 18:40
@H2CO3: I was lucky enough to get the new question notification sooner ;) – Andy Prowl Apr 30 '13 at 18:40
@VaughnCato: In the question page... "1 question with new activity" – Andy Prowl Apr 30 '13 at 18:42
1

It may be good to show a metafunction approach for those without C++11 `using`. – ildjarn Apr 30 '13 at 18:43
Thanks! The addtional info helped a lot! The C03 syntax is not perfect but still way nicer then using typedefs. :) – Devolus Apr 30 '13 at 21:32

score 2 · Answer 2 · answered Apr 30 '13 at 18:39

2

Yes, it works like this:

template <typename T> using Queue = std::queue<std::vector<T>, std::deque<std::vector<T> > >;

answered Apr 30 '13 at 18:39

Vaughn Cato

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Is it possible that g++ c – Devolus Apr 30 '13 at 21:14
@Devolus: With g++, make sure to specify -std=c++11 as one of the command-line parameters. – Vaughn Cato Apr 30 '13 at 22:18

Parameterized typedef possible?

2 Answers2