lambda创建后的序列化 [英] Serialization of a lambda after its creation

问题描述

我可以使用以下语法序列化lambda ：

Runnable r = (Runnable & Serializable) () -> System.out.println("");
try (ObjectOutput oo = new ObjectOutputStream(new ByteArrayOutputStream())) {
  oo.writeObject(r);
}

但是，如果我从客户端代码收到lambda并且它还没有被转换恰当地说，我无法将其序列化。

However if I receive the lambda from a client code and it has not been cast appropriately, I can't serialize it.

如何在不更改其定义的情况下序列化 r ：

How can I serialize r below without changing its definition:

Runnable r = () -> System.out.println("");

---

我试图序列化派生对象：

---

I have tried to serialize a "derived" object:

Runnable r1 = (Runnable & Serializable) r::run;
Runnable r2 = (Runnable & Serializable) () -> r.run();

但在每种情况下， oo.writeObject（rxxx）; 失败并带有 NotSerializableException 。

but in each case, oo.writeObject(rxxx); fails with a NotSerializableException.

推荐答案

这是正确的，并按设计。正如您无法在实例化后获取非可序列化对象并使其可序列化一样，一旦创建了lambda，就会设置其可序列化。

This is correct, and by design. Just as you cannot take a non-serializable object and make it serializable after instantiation, once a lambda is created, its serializability is set.

如果lambda的目标类型是可序列化的（并且其捕获的参数是可序列化的），则lambda是可序列化的。您的第一个示例是可序列化的，因为目标类型是交集（Runnable&序列化）。您的两次尝试转换 r 失败，因为在这两种情况下， r 是一个不可序列化的捕获变量，因此生成的lambda表达式/方法引用不可序列化。 （绑定方法引用的接收器充当捕获变量。）

A lambda is serializable if its target type is serializable (and its captured arguments are serializable.) Your first example is serializable because the target type is the intersection (Runnable & Serializable). Your two attempts to convert r fail because in both cases, r is a captured variable that is not serializable, and so the resulting lambda expression / method reference is not serializable. (The receiver for a bound method reference acts as a captured variable.)

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