java.lang.ArrayStoreException

When creating an array, remember that it will be not of the type declared for the variable, but of the type decided in the initializer:

public class ArrayStoreExceptinTest {
 public static void main(String[] args) {
  System.out.println("starting");

  // test 1: all ok

  Object[] myarray1 = new Object[2];
  System.out.println("myarray1 " + myarray1.getClass().getName());
  myarray1[0] = new Integer(0);
  myarray1[1] = new String("");
  myarray1[1] = new Integer(0);

  // test 2: fails with java.lang.ArrayStoreException
  try {
   Object[] myarray2 = new String[2];
   System.out.println("myarray2 " + myarray2.getClass().getName());
   myarray2[0] = new Integer(0);
  } catch (Exception e) {
   e.printStackTrace();
  }

  // test 3: all ok
  Animal[] myarray3 = new Animal[] { new Cat(), new Dog() };
  System.out.println("myarray3 " + myarray3.getClass().getName());
  myarray3[0] = new Dog();

  try {
   // test 4: fails with java.lang.ArrayStoreException
   Animal[] myarray4 = new Cat[] { new Cat(), new Cat() };
   System.out.println("myarray4 " + myarray4.getClass().getName());
   myarray4[0] = new Dog();
  } catch (Exception e) {
   e.printStackTrace();
  }

 }
}

interface Animal {

}

class Dog implements Animal {

}

class Cat implements Animal {

}

this is the output:

starting
myarray1 [Ljava.lang.Object;
myarray2 [Ljava.lang.String;
java.lang.ArrayStoreException: java.lang.Integer
 at ArrayStoreExceptinTest.main(ArrayStoreExceptinTest.java:18)
myarray3 [LAnimal;
myarray4 [LCat;
java.lang.ArrayStoreException: Dog
 at ArrayStoreExceptinTest.main(ArrayStoreExceptinTest.java:32)

so if you initialize an Animal[] as a Cat[], you can't assign to a member a variable of type Dog, even if a Dog is an Animal...