Can we alias a class name the way we do in namespaces?

Question

For example:

namespace longname{ }
namespace ln = longname;// namespace aliasing

class LONGNAME {};
class LN = LONGNAME; // how to do class name aliasing, if allowed?

`typedef` should do it, unless I'm misunderstanding? – Carl Norum Mar 20 '12 at 16:50 — Carl Norum, Mar 20 '12 at 16:50

Alexandre C. · Accepted Answer · 2012-03-21T08:01:52.673

34

Simple:

typedef LONGNAME LN;

Typedefs are used in C++ a bit like "variables which can store types". Example:

class Car
{
public:
    typedef std::vector<Wheel> WheelCollection;

    WheelCollection wheels;
};

By using Car::WheelCollection everywhere instead of std::vector<Wheel>, you can change the container type in one place and have all your code reflect the changes. This is the C++ way to abstract data types (whereas eg. in C# you'd have a property returning IEnumerable<Wheel>).

edited Mar 21 '12 at 08:01

answered Mar 20 '12 at 16:50

Alexandre C.

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When you posted that as an answer and not comment you should have shown a little more courtesy to explain at least the basics. – Rohit Vipin Mathews Mar 21 '12 at 05:04
@Rohit: What basics ? Typedef is the simplest statement of C++, and there is nothing to know except having seen it once. – Alexandre C. Mar 21 '12 at 07:29
A little theory on how to use it. or what is a typedef, etc would be great answer – Rohit Vipin Mathews Mar 21 '12 at 07:42
1

@Rohit: I added a typical use case. – Alexandre C. Mar 21 '12 at 08:02

score 20 · Answer 2 · edited May 23 '17 at 12:34

20

Beside the answers already provided using the keyword typedef, you also can use the keyword using since C++11. IMHO it looks more consistent regarding the aliasing.

namespace longname{ }
namespace ln = longname;// namespace aliasing

class LONGNAME {};
using LN = LONGNAME; // "class aliasing"

In addition, with using you are able to alias template classes (How to typedef a template class?) by using alias templates.

template<typename T> class LONGNAME {};
template<typename T> using LN = LONGNAME<T>; // alias template

edited May 23 '17 at 12:34

Community

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answered Mar 16 '17 at 11:06

user1810087

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Effective modern C++ 2014, S. Meyers, Item 9 – Petr Javorik Jun 27 '18 at 13:22
This is the right answer for c++11 on-words – Vassilis Apr 13 '22 at 06:49

score 6 · Answer 3 · edited May 23 '17 at 12:26

6

You can use the typedef keyword:

typedef LONGNAME LN;

You can also do something like:

typedef class {

...

} LN;

Edit: You may run into trouble when using templates though. See here for a possible solution.

edited May 23 '17 at 12:26

Community

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answered Mar 20 '12 at 16:51

ApprenticeHacker

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The provided link is unrelated, as it deals with templates, not classes, and those are completely different beasts. – David Rodríguez - dribeas Mar 20 '12 at 18:39
@DavidRodríguez-dribeas I just thought the OP also might be interesting in `typedef`ing template classes. – ApprenticeHacker Mar 21 '12 at 04:56

score 1 · Answer 4 · answered Mar 21 '12 at 05:02

1

typedef int mark;  // for in built data types

class abc
{
};

typedef abc XYZ; // for user written classes.

Typedef allows you to alias a class or datatype name with a more context sensitive name corresponding to the scenario.

answered Mar 21 '12 at 05:02

Rohit Vipin Mathews

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Can we alias a class name the way we do in namespaces?

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