尝试将数组的元素设置为超类对象时发生ArrayStoreException [英] ArrayStoreException when trying to set an element of an array to a superclass object

问题描述

class A {}
class B extends A {}

您好，我正在学习Java，并且试图了解其工作原理:

Hi, I'm learning Java and I'm trying to understand why this works:

A[] tab = new A[10];
tab[0] = new B();

不是这个:

A[] tab = new B[10];
tab[0] = new A();

此行A[] tab = new B[10];表示编译器正在为内存中的B预订10个位置. tab[0] = new A()将tab[0]设置为等于(c)小于B(B extends A)的新A对象.

this line A[] tab = new B[10]; means that the compiler is booking 10 places for B in memory. tab[0] = new A() is setting tab[0] equal to a new A object that is smaller (?) than B (B extends A).

为什么会出现ArrayStoreException: A错误?如何运作?

Why is there an ArrayStoreException: A error ? How does it work ?

推荐答案

您可以在该对象内部存储该对象的子类.即B可以存储在A中，但是A是类型B的超类.基本上，继承链中位于类X之下的任何类都可以称为类型X，但是继承链中位于类X之上的任何类都不能称为类型X.

You can store subclasses of an object, inside that object. ie B can be stored in A, but A is a superclass of type B. Basically, any class below class X in the inheritance chain can be referred to as type X, but any class above class X in the inheritance chain can NOT be referred to as type X.

您的示例

A[] tab = new A[10]; // Perfectly fine. Equal objects.
tab[0] = new B(); // B is a subclass of A, so this is allowed.

A[] tab = new B[10]; // Same principle as above, B is a subclass of A
tab[0] = new A(); // A is a SUPERCLASS of B, so it can not be referred to as A

简而言之，如果X是Y的子类，则只能将X称为Y. (或子类的子类.这是递归定义).

In short, X can only be referred to as Y, if X is a subclass of Y. (Or a subclass of a subclass. It's a recursive definition).

我们使用一些英语术语

让我们使用Item和Book，而不是A和B.

Instead of A and B, let's have Item and Book.

public class Item {}

public class Book extends Item {}

现在，说起来很合理

Item b = new Book(); // All Books are items.

说没有道理

Book b = new Item(); // All items are definitely not books.

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