必须是不可变对象的所有属性都是最终的吗？

问题描述

必须不可变对象的所有属性都是 final ？

Must immutable objects have all properties be final?

据我所知。但我不知道，我是否是对的。

According to me not. But I don't know, whether I am right.

推荐答案

不可变对象（所有属性final）和一个有效的不可变对象（属性不是最终的但不能更改）是安全的发布。

The main difference between an immutable object (all properties final) and an effectively immutable object (properties aren't final but can't be changed) is safe publication.

您可以安全地在多线程上下文中发布不可变对象而无需由于：

You can safely publish an immutable object in a multi threaded context without having to worry about adding synchronization, thanks to the guarantees provided by the Java Memory Model for final fields:

作为旁注，它还可以强制实现不变性（如果你试图在类的未来版本中改变这些字段，因为你忘了它应该是不可变的，它不会编译）。

As a side note, it also enables to enforce immutability (if you try to mutate those fields in a future version of your class because you have forgotten it should be immutable, it won't compile).

澄清

• 使一个对象的所有字段最终都不会使它成为不可变的 - 你还需要确保（i）它的状态不能更改（例如，如果对象包含最终列表，则不得在构建后进行变异操作（添加，删除...）和（ii）您不要


• 一个有效不可变的对象一旦安全发布就是线程安全的
$ b $让在构造期间转义 b

• 不安全发布的示例：

• Making all the fields of an object final does not make it immutable - you also need to make sure that (i) its state can't change (for example, if the object contains a final List, no mutating operations (add, remove...) must be done after construction) and (ii) you don't let this escape during construction
• An effectively immutable object is thread safe once it has been safely published

• Example of unsafe publication:

class EffectivelyImmutable {
    static EffectivelyImmutable unsafe;
    private int i;
    public EffectivelyImmutable (int i) { this.i = i; }
    public int get() { return i; }
}

// in some thread
EffectivelyImmutable.unsafe = new EffectivelyImmutable(1);

//in some other thread
if (EffectivelyImmutable.unsafe != null
    && EffectivelyImmutable.unsafe.get() != 1)
    System.out.println("What???");

这个程序理论上可以打印什么??? 。如果 i 是最终的，那将不是合法的结果。

This program could in theory print What???. If i were final, that would not be a legal outcome.

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