以下代码:
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;
}
}
JavaTest.max(List, 0, 0) 语句如何编译?磨砂是如何满足的
T extends Comparable <? super T>
它实现Comparable<String>
了哪些不是超级类型的scruby?如果您将其更改为scruby<Integer>
它将无法编译并给出错误。那为什么现在编译呢?为什么原始类型会编译?