Question

我们说我有一个抽象类，一个扩展类和一个这样的接口：

public abstract class SuperClass {
    ...
    public void foo() {...}
    ...
}

public class SubClass extends SuperClass implements MyInterface {...}

public interface MyInterface {
    public void bar();
}

请注意，我有几个实现MyInterface的SuperClass子类，但不是全部。还可以说，我有另一个类，有这样的构造函数：

public class AnotherClass {
    private SuperClass sc;
    public AnotherClass(SuperClass superclass) {
        sc = superclass;
    }
    ...
}

我现在的问题是，如何确保构造函数中的给定对象也实现MyInterface？我需要对象sc才能运行方法foo()和bar()。我怎么能做到这一点？

Answer 1

您可以将generics与交叉类型

一起使用

public class AnotherClass<T extends SuperClass & MyInterface> {
private T sc;
public AnotherClass(T superclass) {
    sc = superclass;
}
void fn(){
 // use both methods
 sc.foo();
 sc.bar();
}
...
}

现在，AnotherClass只允许SuperClass类型并实施MyInterface。

Answer 2

你可以在层次结构中引入另一个类：

public class NonImplementingSubClass extends SuperClass {...}

public class ImplementingSubClass extends SuperClass implements MyInterface {...}

应该实现接口的所有子类将扩展ImplementingSubClass

public class SubClass extends ImplementingSubClass {...}

public class AnotherClass {
    private ImplementingSubClass sc;
    public AnotherClass(ImplementingSubClass superclass) {
        sc = superclass;
    }
    ...
}

Answer 3

public abstract class SuperClass implements MyInterface {
    ...
    public void foo() {...}
    ...
}

public class SubClass extends SuperClass{}

public interface MyInterface {
    public void bar();
}

如何引用既扩展又实现的实例？

3 个答案: