CODEWITHSUNDEEP
javaserializationlambdajava-8
assylias
assylias

Reputation: 328785

Serialization of a lambda after its creation

I can serialize a lambda with the following syntax:

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

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

How can I serialize r below without changing its definition:

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

I have tried to serialize a "derived" object:

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

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

Upvotes: 14

Views: 2914

Answers (1)

Brian Goetz
Brian Goetz

Reputation: 95486

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.

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.)

Upvotes: 16

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