How to avoid calling this.field for every field in a java class

Question

Is there a way to avoid calling this.field for every field in a class ?

public class Test {
    private String name;
    private String email;

    public Test(String name, String email) {
        // I want to avoid this
        this.name = name;
        this.email = email;
    }

    public Test(Test test) {
        // Something like this would be perfect, setting both name and email to test
        this(test);
    }
}

Answer 1

The use of this is only required in cases of name collisions, to resolve the ambiguity.

Some programmers like me prefer using the this. prefix routinely, whereas other use only where necessary.

See Answer by Wasserman for an example of how to avoid naming collision.

Use the IDE, Luke

Your IDE will generate constructors, accessors (getters/setters), equals & hashCode, toString, and so on. So you need not type this.; let the machine do the typing.

Use custom settings to control whether you want the IDE to include or omit this. prefixes.

record

You may be interested in using the records feature, new in Java 16+. A record is a brief way to write a class whose main purpose is to communicate data transparently and immutably.

With a record, by default, the compiler implicitly writes the constructor, getters, equals & hashCode, and toString. The implicitly created constructor populates each and every member field on your behalf. You write none of that code.

Here is your entire example class when written as a record. No this required. All your member fields are automatically assigned.

public record Test ( String name , String email ) {}

Be cautious in using records. The reason for their invention was not writing less code. The reason was to provide an explicit mechanism for transmitting immutable data transparently, a “nominal tuple” in academic-speak. Less boilerplate coding is merely a nice side-effect. I highly recommend reading JEP 395 for more explanation.

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Tip: You can combine the two points of this Answer. Ask your IDE to generate a full-blown class by starting with a record.

1. Write a record with all your member fields listed in the parentheses.
2. Invoke your IDE to convert from a record to a class.

Voilà, you have a complete class with constructor, accessors, equals & hashCode, and toString all written out with an absolute minimum of typing by you.

For example, in IntelliJ 2022, choosing Convert record to class from the light-bulb icon menu turns this:

public record Test ( String name , String email ) {}

… into this:

package work.basil.example.recs;

import java.util.Objects;

public final class Test
{
    private final String name;
    private final String email;

    public Test ( String name , String email )
    {
        this.name = name;
        this.email = email;
    }

    public String name ( ) { return name; }

    public String email ( ) { return email; }

    @Override
    public boolean equals ( Object obj )
    {
        if ( obj == this ) { return true; }
        if ( obj == null || obj.getClass() != this.getClass() ) { return false; }
        var that = ( Test ) obj;
        return Objects.equals( this.name , that.name ) &&
                Objects.equals( this.email , that.email );
    }

    @Override
    public int hashCode ( )
    {
        return Objects.hash( name , email );
    }

    @Override
    public String toString ( )
    {
        return "Test[" +
                "name=" + name + ", " +
                "email=" + email + ']';
    }
}

Caveat: That result may not be the default. I may have altered the settings in IntelliJ.