在Java7和Java8中<T>和<? super T>有区别吗？

Question

为什么在调用以下两个方法时在Java8中不报错而在Java7中会报错
public class Test {
    public static void main(String[] args) {

        Animal[] animals = {new Animal(1),new Animal(3),new Animal(2)};
        Dog[] dogs = {new Dog(1),new Dog(3),new Dog(2)};

        min1(animals);//在Java1.7.0_51和Java1.8.0_181编译都没有报错
        min1(dogs);//在Java1.8.0_181编译没有报错 在Java1.7.0_51编译报错
        min2(animals);//在Java1.7.0_51和Java1.8.0_181编译都没有报错
        min2(dogs);//在Java1.7.0_51和Java1.8.0_181编译都没有报错

    }


    //查找出数组中年龄最小的的元素 Comparable<T>中的泛型T没有规定下线
    public static <T extends Comparable<T>> void min1(T[] t){
        T min = t[0];
        for (int i = 1; i < t.length; i++) {
            if (min.compareTo(t[i])<0){
                min = t[i];
            }
        }
        System.out.println(min);
    }

    //查找出数组中年龄最小的的元素 Comparable<T>中的泛型T规定了下线
    public static <T extends Comparable<? super T>> void min2(T[] t){

        T min = t[0];
        for (int i = 1; i < t.length; i++) {
            if (t[i].compareTo(min)<0){
                min = t[i];
            }
        }
        System.out.println(min);
    }

}

//Aminmal实现了Comparable<Animal>
public class Animal implements Comparable<Animal>{

    private int age;

    public Animal(int age) {
        this.age = age;
    }

    @Override
    public int compareTo(Animal o) {
        return age-o.age;
    }

    @Override
    public String toString() {
        return "Animal{" +
                "age=" + age +
                '}';
    }
}

//Dog继承了Animal
public class Dog extends Animal{

    public Dog(int age) {
        super(age);
    }
}

Answer 1

和<T>或者<? super T>没什么关系, Java8比Java7有更好的泛型推断

详见Java generics - Don't allow subclasses