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Lambda Expression throws IllegalAccessError

csh
csh Member Posts: 525
edited Jan 28, 2015 8:18AM in Java 8 Questions

Hi,

I have the following structure and want to use lambda expressions with it, but an exception occurs:

class SuperClass // (package private)


public class SubClass extends SuperClass // in same package as SuperClass

I have a functional interface (a listener), which has one argument of type SubClass.

public interface MyListener() {
    void myMethod(SubClass e);
}

When I use the old-school Java 7 style:

addListener(new MyListener() {
     @Override
     public void myMethod(SubClass e) {
         System.out.println(e);
     }
});

everything works fine!

If I use Java 8 lambda expression instead:

addListener(e -> {
    System.out.println(e);
});

it will throw an exception:

java.lang.IllegalAccessError: tried to access class SuperClass

I wonder if this is a documented shortcoming of Java 8 lambda expressions or a bug?

Can somebody help here?

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Answers

  • dvohra21
    dvohra21 Member Posts: 14,322 Gold Crown

    Is the SuperClass in the default package? Classes from the default package do not get imported. The error should also be in Java 7 if the same package structure is used.

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  • csh
    csh Member Posts: 525

    No, it's in a "real" package, like "com.example". Java 7 does not support Lambdas.

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  • James_D
    James_D Member Posts: 1,496 Gold Trophy

    I tried this, or something similar (under 1.8.0u40 ea) and it worked fine.

    My code:

    SuperClass.java:

    package pri;

class SuperClass { }

    SubClass.java:

    package pri;

public class SubClass extends SuperClass { }

    MyLIstener.java (different package):

    package pub;

import pri.SubClass;

@FunctionalInterface
public interface MyListener {
  void myMethod(SubClass e) ;
}

    Test.java (another package still):

    package test;


import pri.SubClass;
import pub.MyListener;


public class Test {


  public static void main(String[] args) {
      MyListener l = new MyListener() {
          @Override
          public void myMethod(SubClass e) {
              System.out.println(e);    
          }
      };

      MyListener l2 = e -> System.out.println(e);

      MyListener l3 = System.out::println ;

      l.myMethod(new SubClass());
      l2.myMethod(new SubClass());
      l3.myMethod(new SubClass());
  }


}
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  • csh
    csh Member Posts: 525

    Sorry, it turned out, that the problem only exists, when there are generics involved (i.e. class SuperClass<T>). I've got an example at home, if you are interested.

    In the meanwhile I met an Oracle guy (at Devoxx), who said, it looks like a bug in Java 8. I don't know where it's tracked.

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