在Java中的getter和修改列表组合 [英] Combination of getter and list modification in Java

问题描述

今天我处理了一个Java问题，真正让我感到困惑。
我有以下的code：

today i dealt with a Java problem that really confused me. I have the following code:

List<ObjectXY> someList = obj.getListOfObjectsXY(); // getter returns 2 elements
someList.add(new ObjectXY());

obj.getListOfObjectsXY(); // getter now returns 3 elements

当我一个元素添加到列表，吸气得到某种覆盖。这是因为 someList 就像在这种情况下，吸气结果的参考？还是什么别的原因导致这种效果？

When i add an element to a list, the getter gets some kind of overwritten. Is this because someList acts like a reference on the result of the getter in this case? Or what else causes this effect?

我用另一个列表解决了以下code中的问题：

I solved the problem with the following code by using another list:

List<ObjectXY> someList = obj.getListOfObjectsXY(); // result: 2 elements

List<ObjectXY> someOtherList = new ArrayList<ObjectXY>();
someOtherList.addAll(someList);
someOtherList.add(new ObjectXY());

obj.getListOfObjectsXY(); // result: 2 elements

但我还是某种困惑，因为我没有想到Java来表现这种方式。
谁能给我解释一下我做错了什么，为什么会这样？

But i am still some kind of confused because i didn't expect Java to behave this way. Can anyone explain to me what i did wrong and why it is so?

在此先感谢！

推荐答案

返回的结果确实只是一个相同的对象的引用作为您内部使用的副本。望着调用者不能修改的对象是容易出错。

The returned result is indeed just a copy of a reference to the same object as you are using internally. Counting on the caller to not modify the object is error-prone.

一个解决方案是一个参考返回一个不可修改的列表包装列表。参见 Col​​lections.unmodifiableList（）。 getter方法​​调用者将无法修改列表。

One solution is to return a reference to an unmodifiable list wrapping your list. See Collections.unmodifiableList(). The getter caller will be unable to modify your list.

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