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Home/ Questions/Q 6059613
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Editorial Team
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Editorial Team
Asked: May 23, 20262026-05-23T08:43:52+00:00 2026-05-23T08:43:52+00:00

I have an example class defined like below: public class FooBar { void method1(Foo

  • 0

I have an example class defined like below:

public class FooBar {

  void method1(Foo foo){ // Should be overwritten
    ...
  }

}

Later, when I try this:

FooBar fooBar = new FooBar(){
  public String name = null;
  @Override
  void method1(Foo foo){
    ...
  }
};

fooBar.name = "Test";

I get an error saying that the name field does not exist. Why?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-23T08:43:53+00:00Added an answer on May 23, 2026 at 8:43 am

    Because the type of the variable "fooBar" is FooBar (the run-time type of the object in said variable is that of the anonymous class implementing FooBar which is also a subtype of FooBar)…

    …and the type FooBar does not have said member. Hence, a compile error. (Remember, the variable "fooBar" can contain any object conforming to FooBar, even those without name, and thus the compiler rejects the code which is not type-safe.)

    Edit: For one solution, see irreputable’s answer which uses a Local Class Declaration to create a new named type (to replace the anonymous type in the post).

    Java does not support a way to do this (mainly: Java does not support useful type inference), although the following does work, even if not very useful:

    (new foobar(){
      public String name = null;
      @Override
      void method1(Foo foo){
        ...
      }
    }).name = "fred";
    

    Happy coding.


    Both Scala and C# support the required type inference, and thus anonymous type specializations, of local variables. (Although C# does not support extending existing types anonymously). Java, however, does not.

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