It is easy to define a method returning the same value as the argument:
class A {
public static void main(String[]args) {
Derived sth = new Derived();
String x = sth.foo("hello");
System.out.println(x);
Derived resultTypeIsKnown = sth.foo(sth); // <==== !!!
System.out.println(""+resultTypeIsKnown);
}
}
class Base {
<T>T foo(T t)
{
return t;
}
}
class Derived extends Base {
}
You see that the compiler knows that although foo() is declared in the Base class, the compiler knows that sth.foo(sth) returns an instance of the Derived class:
Derived derivedRatherThanBase = sth.foo(sth);
How do I declare that the return value is of the same class as the object whose method is called? (In particular, for a method that always returns this?)
Something like:
class Base {
<?ThisClass?> bar() {
return this;
}
}
I can write
<T>T getThis(T t)
{
return (T)this;
}
but x.getThis(x) uses an extra argument and produces a warning.
UPDATE OMG, what they are doing in the "possible duplicate"... But I already have the base class, it's a descendant of Collection. In other words, it is required that class Base extends Collection and knows its class.
And the real life code where I want to know THISCLASS already is very complex.
UPDATE2 As to the XY problem, what I want is:
class MyVerySpecialCollectionBase extends ... {
...
THISCLASS addAll(Collection items) { ... }
}
class MyExtendedVerySpecialCollection extends MyVerySpecialCollectionBase {
...
}
// use:
MyExtendedVerySpecialCollection x =
new MyExtendedVerySpecialCollection(aLotOfArgs)
.addAll(list1)
.addAll(list2);
The proposed solution sounds too complex and even less acceptable than either (1) redefining addAll() in each derived class or (2) making x.addAll(list1).addAll(list2); a separate statement.
