在 java 中，我们可以将超类 Object 传递给子类引用吗? [英] In java , can we pass superclass Object to subclass reference?

问题描述

在java中，我们可以将超类Object传递给子类引用吗?

In java, can we pass superclass Object to subclass reference ?

我知道这是一个奇怪的问题/实际上不可行，但我想了解这背后的逻辑为什么在java中不允许.

I know that it is a weird question/practically not viable, but I want to understand the logic behind this Why is it not allowed in java.

class Employee {
    public void met1(){
        System.out.println("met1");
    }
}


class SalesPerson extends Employee 
{
    @Override
    public void met1(){
    System.out.println("new met1");
    }


    public void met2(){
        System.out.println("met2");
    }

}

public class ReferenceTest {
    public static void main(String[] args) {

        SalesPerson sales = new Employee(); // line 1

        sales.met1();  // line 2

        sales.met2();  // line 3
    }
}

如果 Java 允许编译第 1 行会发生什么?问题会出现在哪里?

What would have happened if Java allowed compilation of line 1? Where would the problem arise?

欢迎任何输入/链接.

推荐答案

如果你的 SalesPerson sales = new Employee(); 语句被允许编译，这将违反 多态，这是该语言的特性之一.

If your SalesPerson sales = new Employee(); statement was allowed to compile, this would have broken the principles of Polymorphism, which is one of the features that the language has.

此外，您应该熟悉编译时类型和运行时类型的含义:

Also, you should get familiar with that does compile time type and runtime type mean:

变量的编译时类型是它声明的类型，而运行时类型是变量指向的实际对象的类型.例如:

The compile-time type of a variable is the type it is declared as, while the runtime type is the type of the actual object the variable points to. For example:

Employee sales = new SalesPerson();  

sales 的编译时类型为 Employee，运行时类型为 SalesPerson.编译时类型定义了可以调用哪些方法，而运行时类型定义了实际调用期间会发生什么.

The compile-time type of sales is Employee, and the runtime type will be SalesPerson. The compile-time type defines which methods can be called, while the runtime type defines what happens during the actual call.

让我们假设这个语句是有效的:

Let's suppose for a moment that this statement was valid:

SalesPerson sales = new Employee();

正如我所说，编译时类型定义了可以调用哪些方法，因此 met2() 将有资格调用.同时，Employee 类没有 met2()，因此实际调用是不可能的.

As I said, the compile-time type defines which methods can be called, so met2() would have been eligible for calling. Meanwhile, the Employee class doesn't have a met2() and so the actual call would have been impossible.

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