Assuming that it SerializableFunctionis of a type that extends Serializable, the method reference will be serialized. There is nothing special about the particular type of method reference you are asking for.
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, :
import java.io.*;
import java.util.function.Consumer;
public class MyClass {
public void foo() {
System.out.println("Serializable");
}
public static void main(String[] args) throws IOException, ClassNotFoundException {
Consumer<MyClass> consumer = (Consumer<MyClass>&Serializable)MyClass::foo;
byte[] serialized;
try(ByteArrayOutputStream baos=new ByteArrayOutputStream();
ObjectOutputStream oos = new ObjectOutputStream(baos)) {
oos.writeObject(consumer);
oos.flush();
serialized=baos.toByteArray();
}
Consumer<MyClass> deserialized;
try(ByteArrayInputStream bais=new ByteArrayInputStream(serialized);
ObjectInputStream ois=new ObjectInputStream(bais)) {
deserialized = (Consumer<MyClass>)ois.readObject();
}
deserialized.accept(new MyClass());
}
}
, , , ,
import java.io.*;
import java.util.function.Consumer;
public class MyClass {
public void foo() {
System.out.println("Serializable");
}
public static void main(String[] args) throws IOException, ClassNotFoundException {
Runnable runnable = (Runnable&Serializable)new MyClass()::foo;
byte[] serialized;
try(ByteArrayOutputStream baos=new ByteArrayOutputStream();
ObjectOutputStream oos = new ObjectOutputStream(baos)) {
oos.writeObject(runnable);
oos.flush();
serialized=baos.toByteArray();
}
Runnable deserialized;
try(ByteArrayInputStream bais=new ByteArrayInputStream(serialized);
ObjectInputStream ois=new ObjectInputStream(bais)) {
deserialized = (Runnable)ois.readObject();
}
deserialized.run();
}
}
java.io.NotSerializableException: MyClass, MyClass Serializable.