Raw type. References to generic types should be parameterized

Question

I have a Cage class:

public class Cage {
// the construtor takes in an integer as an explicit parameter
...
}

I am

Answer 1

For other java newbie like me.

• Code is look like this:

public class ContinuousAddressBuilder<T> extends VariableLengthPacket {
  ...

  /* T=int/float/double */
  private ArrayList<T> informosomes;

  ...

  public ContinuousAddressBuilder builderCon(int con) {
    ...
  }
}

• Solution:

Add <T> after your class:

change from

public ContinuousAddressBuilder builderCon(int con)

to

public ContinuousAddressBuilder<T> builderCon(int con)

Answer 2

Cage<T> is a generic type, so you need to specify a type parameter, like so (assuming that there is a class Dog extends Animal):

private Cage<Dog> cage5 = new Cage<Dog>(5);

You can use any type that extends Animal (or even Animal itself).

If you omit the type parameter then what you wind up with in this case is essentially Cage<Animal>. However, you should still explicitly state the type parameter even if this is what you want.