The following code:
import java.util.*;
public final class JavaTest {
public static <T extends Comparable<? super T>> int max(List<? extends T> list, int begin, int end) {
return 5;
}
public static void main(String[] args) {
List<scruby<Integer>> List = new ArrayList<>();
JavaTest.<scruby>max(List, 0, 0); //how does this compile?, scruby doesn't meet
//T extends Comparable<? super T>
}
}
class scruby<T> implements Comparable<String>{
public int compareTo(String o) {
return 0;
}
}
How does the statement JavaTest.max(List, 0, 0) compile? How does scruby meet the
T extends Comparable <? super T>
It implements Comparable<String>
which isn't a super type of scruby? If you change it to scruby<Integer>
it won't compile and give the error. So why does it compile now? Why does the raw type compile?
scruby
is a raw type. This suppresses some of the type checks.You should add all required type parameters:
Or just let Java infer them: