How do I create some variable type alias in Java?

Question

Let's say I have this code

Map<String, String> list = new HashMap<String, String>();
list.put("number1", "one");
list.put("number2", "two");

how can I make some "alias" the type

Map<String, String>

to something that easier to be rewritten like

// may be something like this
theNewType = HashMap<String, String>;

theNewType list = new theNewType();
list.put("number1", "one");
list.put("number2", "two");

Basically, my question is, how to create "alias" to some "type", so I can make it easier to write and easier when need to change the whole program code.

Thanks, and sorry if this is silly question. I'm kinda new in Java.

KARASZI István · Accepted Answer · 2011-04-09 10:17:48Z

85

There are no aliases in Java. You can extend the HashMap class with your class like this:

public class TheNewType extends HashMap<String, String> {
    // default constructor
    public TheNewType() {
        super();
    }
    // you need to implement the other constructors if you need
}

But keep in mind that this will be a class it won't be the same as you type HashMap<String, String>

answered Apr 9, 2011 at 10:17

KARASZI István

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4 Comments

McDowell Over a year ago

@Lee - KARASZI István's caveat about types is important - see ibm.com/developerworks/java/library/j-jtp02216/index.html

Andrey Over a year ago

There is an alternative trick/hack you can do with generics: stackoverflow.com/questions/5604390/…

Niccolò Over a year ago

But because this is a specialization of a more general type, you can't use TheNewType in place of HashMap<String, String> as much as you can't use the latter where a Map<String, String> is returned.

plswork04 Over a year ago

archive of link posted by @McDowell web.archive.org/web/20170206190605/http://www.ibm.com/…

Boann · Accepted Answer · 2015-09-14 08:06:15Z

18

There is no typedef equivalent in Java, and there is no common idiom for aliasing types. I suppose you could do something like

class StringMap extends HashMap<String, String> {}

but this is not common and would not be obvious to a program maintainer.

edited Sep 14, 2015 at 8:06

Boann

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answered Apr 9, 2011 at 10:18

Rom1

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1 Comment

nasukkin Over a year ago

Bear in mind that you should implement the missing details of the Serializable interface for your subclass, should you do this. Else, you may encounter curiosities if you ever run into a situation where you need to (de)serialize your object.

Andrey · Accepted Answer · 2014-12-17 22:53:16Z

13

Although Java doesn't support this, you can use a generics trick to simulate it.

class Test<I extends Integer> {
    <L extends Long> void x(I i, L l) {
        System.out.println(
            i.intValue() + ", " + 
            l.longValue()
        );
    }
}

Source: http://blog.jooq.org/2014/11/03/10-things-you-didnt-know-about-java/

answered Dec 17, 2014 at 22:53

Andrey

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1 Comment

ThomasH Over a year ago

IMO this is the same idea as in the other answers, in that you use the concept of derivation to get a new type name. Here it is I from Integer and L from Long by constraining type variables, where the other answers propose NewType from HashMap<String,String> by constructing a new class. The drawback of the type variable approach is that you can only use it within the scope of the type variables, e.g. the Test class or the x method.

Ingo · Accepted Answer · 2011-04-09 10:19:04Z

4

The closest one could think of is to make a wrapper class like so

class NewType extends HashMap<String, String> {
     public NewType() { }
}

I really wish Java had a sound type aliasing feature.

answered Apr 9, 2011 at 10:19

Ingo

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Comments

Michael Borgwardt · Accepted Answer · 2011-04-09 10:16:44Z

1

Nothing like that exists in Java. You might be able to do something with IDE templates or autocompletion, and look forward to (limited) generics type inference in Java 7.

answered Apr 9, 2011 at 10:16

Michael Borgwardt

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