Java中的矛盾和继承有什么区别？

Question

我有一个父类A，而它的子类B。这两个摘要之间有什么区别：

    public static void main (String[] args) {
        ArrayList<? super A> e = new ArrayList<>();
        B thing = new B();
        e.add(thing);
        System.out.println(e.get(0));
    }

    public static void main (String[] args) {
        ArrayList<A> e = new ArrayList<>();
        B thing = new B();
        e.add(thing);
        System.out.println(e.get(0));
    }

Answer 1

ArrayList<? super A>和ArrayList<A>之间的区别在于，可以为前者分配ArrayList<T>的对象，其中T是A或A本身的超类。

具体来说，这是有效的：

ArrayList<? super A> listOfSuperA = new ArrayList<Object>();

这不是：

ArrayList<A> listOfA = new ArrayList<Object>();

这意味着您可以从listOfA和listOfSuperA中获得不同类型的值：

A a = listOfA.get(0); // valid
Object o = listOfA.get(0); // valid, because A can be converted to Object.
A a2 = listOfSuperA.get(0); // invalid, because listOfSuperA could contain Objects, which are not convertible to A
Object o2 = listOfSuperA.get(0); // valid, because whatever is in listOfSuperA, it can be converted to Object

This might be useful if you want to learn more about where to use ? super T