java协变量返回类型 [英] java covariant return type

问题描述

为什么下面的代码显示"1"?

Why does below code prints "1" ?

class A {
    int x = 1;
}

class B extends A {
    int x = 2;
}

class Base {

    A getObject() {
        System.out.println("Base");
        return new B();
    }
}

public class CovariantReturn extends Base {

B getObject() {
   System.out.println("CovariantReturn");
   return new B(); 
}
/**
 * @param args
 */
public static void main(String[] args) {
    Base test = new CovariantReturn();
    System.out.println(test.getObject() instanceof B);
    System.out.println(test.getObject().x);
}
}

推荐答案

因为您是指不受多态性影响的字段.如果您改用getX()，它将返回2.

Because you are referring to fields, which are not affected by polymorphism. If you instead used getX(), it would've returned 2.

您要问的是，类A中定义的字段x的值(因为Base.getObject()返回A).即使CovariantReturn覆盖了返回B的方法，您也没有将对象称为CovariantReturn.

What you are asking is, the value of field x defined in class A (because Base.getObject() returns A). Even though CovariantReturn overrides the method to return B, you are not referring to your object as CovariantReturn.

为了进一步说明字段如何不受多态性影响-字段访问是在编译时实现的，因此无论编译器看到什么，这就是访问的内容.在您的情况下，该方法定义为返回A，因此可以访问A.x.另一方面，基于运行时类型调用方法.因此，即使您定义返回A但返回B的实例，您调用的方法也会在B上调用.

To expand a bit on how fields are not affected by polymorphism - field access is realized at compile time, so whatever the compiler sees, that's what's accessed. In your case the method defines to return A and so A.x is accessed. On the other hands methods are invoked based on the runtime type. So even if you define to return A but return an instance of B, the method you invoke will be invoked on B.

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