Java 重写方法后类不可序列化

Java 重写方法后类不可序列化,java,serialization,overriding,Java,Serialization,Overriding,我将测试用例中的createSocket()方法重写为模拟套接字中的pas。执行此操作后,对象不再可序列化 下面是一个不起作用的例子 Foo.java import java.io.Serializable; public class Foo implements Serializable { private static final long serialVersionUID = 3109852436898487119L; public void bar() {

我将测试用例中的
createSocket()
方法重写为模拟套接字中的pas。执行此操作后,对象不再可序列化

下面是一个不起作用的例子

Foo.java

import java.io.Serializable;

public class Foo implements Serializable {
    private static final long serialVersionUID = 3109852436898487119L;

    public void bar() {
        System.out.println("Foo");
    }
}
import java.io.ByteArrayOutputStream;
import java.io.IOException;
import java.io.ObjectOutputStream;

import junit.framework.TestCase;

import org.junit.Test;

public class FooTest extends TestCase {

    // this passes
    @Test
    public void testFooIsSerializable() throws IOException {
        Foo foo = new Foo();

        ByteArrayOutputStream baos = new ByteArrayOutputStream();
        ObjectOutputStream out = new ObjectOutputStream(baos);
        out.writeObject(foo);
    }

    // this throws a java.io.NotSerializableException
    @Test
    public void testFooIsStillSerializableAfterBarIsOverridden()
            throws IOException {

        // Eclipse gives me the warning "The serializable class  does not declare a static final serialVersionUID field of type long"
        // Adding it doesn't help
        Foo foo = new Foo() {
            @Override
            public void bar() {
                System.out.println("Bar");
            }
        };

        ByteArrayOutputStream baos = new ByteArrayOutputStream();
        ObjectOutputStream out = new ObjectOutputStream(baos);
        out.writeObject(foo);
    }
}
FooTest.java

import java.io.Serializable;

public class Foo implements Serializable {
    private static final long serialVersionUID = 3109852436898487119L;

    public void bar() {
        System.out.println("Foo");
    }
}
import java.io.ByteArrayOutputStream;
import java.io.IOException;
import java.io.ObjectOutputStream;

import junit.framework.TestCase;

import org.junit.Test;

public class FooTest extends TestCase {

    // this passes
    @Test
    public void testFooIsSerializable() throws IOException {
        Foo foo = new Foo();

        ByteArrayOutputStream baos = new ByteArrayOutputStream();
        ObjectOutputStream out = new ObjectOutputStream(baos);
        out.writeObject(foo);
    }

    // this throws a java.io.NotSerializableException
    @Test
    public void testFooIsStillSerializableAfterBarIsOverridden()
            throws IOException {

        // Eclipse gives me the warning "The serializable class  does not declare a static final serialVersionUID field of type long"
        // Adding it doesn't help
        Foo foo = new Foo() {
            @Override
            public void bar() {
                System.out.println("Bar");
            }
        };

        ByteArrayOutputStream baos = new ByteArrayOutputStream();
        ObjectOutputStream out = new ObjectOutputStream(baos);
        out.writeObject(foo);
    }
}
使用JUnit运行FooTest时的堆栈跟踪:

java.io.NotSerializableException: FooTest
    at java.io.ObjectOutputStream.writeObject0(ObjectOutputStream.java:1164)
    at java.io.ObjectOutputStream.defaultWriteFields(ObjectOutputStream.java:1518)
    at java.io.ObjectOutputStream.writeSerialData(ObjectOutputStream.java:1483)
    at java.io.ObjectOutputStream.writeOrdinaryObject(ObjectOutputStream.java:1400)
    at java.io.ObjectOutputStream.writeObject0(ObjectOutputStream.java:1158)
    at java.io.ObjectOutputStream.writeObject(ObjectOutputStream.java:330)
    at FooTest.testFooIsStillSerializableAfterBarIsOverridden(FooTest.java:33)
    at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
    at sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:39)
    at sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:25)
    at java.lang.reflect.Method.invoke(Method.java:597)
    at junit.framework.TestCase.runTest(TestCase.java:168)
    at junit.framework.TestCase.runBare(TestCase.java:134)
    at junit.framework.TestResult$1.protect(TestResult.java:110)
    at junit.framework.TestResult.runProtected(TestResult.java:128)
    at junit.framework.TestResult.run(TestResult.java:113)
    at junit.framework.TestCase.run(TestCase.java:124)
    at junit.framework.TestSuite.runTest(TestSuite.java:232)
    at junit.framework.TestSuite.run(TestSuite.java:227)
    at org.junit.internal.runners.JUnit38ClassRunner.run(JUnit38ClassRunner.java:83)
    at org.eclipse.jdt.internal.junit4.runner.JUnit4TestReference.run(JUnit4TestReference.java:49)
    at org.eclipse.jdt.internal.junit.runner.TestExecution.run(TestExecution.java:38)
    at org.eclipse.jdt.internal.junit.runner.RemoteTestRunner.runTests(RemoteTestRunner.java:467)
    at org.eclipse.jdt.internal.junit.runner.RemoteTestRunner.runTests(RemoteTestRunner.java:683)
    at org.eclipse.jdt.internal.junit.runner.RemoteTestRunner.run(RemoteTestRunner.java:390)
    at org.eclipse.jdt.internal.junit.runner.RemoteTestRunner.main(RemoteTestRunner.java:197)
这有什么问题?我必须承认,我没有深入研究Java的可序列化接口,或多或少只是遵循Eclipse的快速修复

具体到我的实现:

java.io.NotSerializableException: FooTest
    at java.io.ObjectOutputStream.writeObject0(ObjectOutputStream.java:1164)
    at java.io.ObjectOutputStream.defaultWriteFields(ObjectOutputStream.java:1518)
    at java.io.ObjectOutputStream.writeSerialData(ObjectOutputStream.java:1483)
    at java.io.ObjectOutputStream.writeOrdinaryObject(ObjectOutputStream.java:1400)
    at java.io.ObjectOutputStream.writeObject0(ObjectOutputStream.java:1158)
    at java.io.ObjectOutputStream.writeObject(ObjectOutputStream.java:330)
    at FooTest.testFooIsStillSerializableAfterBarIsOverridden(FooTest.java:33)
    at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
    at sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:39)
    at sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:25)
    at java.lang.reflect.Method.invoke(Method.java:597)
    at junit.framework.TestCase.runTest(TestCase.java:168)
    at junit.framework.TestCase.runBare(TestCase.java:134)
    at junit.framework.TestResult$1.protect(TestResult.java:110)
    at junit.framework.TestResult.runProtected(TestResult.java:128)
    at junit.framework.TestResult.run(TestResult.java:113)
    at junit.framework.TestCase.run(TestCase.java:124)
    at junit.framework.TestSuite.runTest(TestSuite.java:232)
    at junit.framework.TestSuite.run(TestSuite.java:227)
    at org.junit.internal.runners.JUnit38ClassRunner.run(JUnit38ClassRunner.java:83)
    at org.eclipse.jdt.internal.junit4.runner.JUnit4TestReference.run(JUnit4TestReference.java:49)
    at org.eclipse.jdt.internal.junit.runner.TestExecution.run(TestExecution.java:38)
    at org.eclipse.jdt.internal.junit.runner.RemoteTestRunner.runTests(RemoteTestRunner.java:467)
    at org.eclipse.jdt.internal.junit.runner.RemoteTestRunner.runTests(RemoteTestRunner.java:683)
    at org.eclipse.jdt.internal.junit.runner.RemoteTestRunner.run(RemoteTestRunner.java:390)
    at org.eclipse.jdt.internal.junit.runner.RemoteTestRunner.main(RemoteTestRunner.java:197)
我有一个类,它应该通过ObjectOutputStream发送自身的实例


这是一种根本错误的方法吗?

如果要序列化/反序列化,只需声明serialVersionUID即可:

   Foo foo = new Foo() {
     static final serialVersionUID = 348723598723589723L;
        @Override
        public void bar() {
            System.out.println("Bar");
        }
   };
执行此操作后,对象不再可序列化


你没有任何证据。您所拥有的只是一个Eclipse警告。执行时会发生什么?

您正在创建一个新的注释性内部类,该类通过重写其
bar
方法扩展
Foo
。这个新创建的类没有实现可序列化的接口,因此无法序列化

编辑:

您的子类使用的是
FooTest
,因此该类及其内部类将被序列化。这就是为什么
FooTest
必须是可序列化的

问题是您有一个匿名的内部类,根据定义它包含一个(编译器生成的)对创建它的外部类(
FooTest
)实例的引用。由于序列化在默认情况下包括对象的所有成员,因此运行时也会尝试序列化
FooTest
对象,这是不可序列化的(因为它本来就不应该序列化)。错误消息证实了这一点:

java.io.NotSerializableException: FooTest

因此,如果您将匿名类转换为显式的
静态
内部类,问题应该会消失。

正如我的问题源代码中的一条评论所述,这无助于抛出java.io.NotSerializableException。@Zoran Zaric:命名哪个类?@Zoran Zaric如上面所说,不可序列化的是
FooTest
,而不是
Foo
。原因是构造匿名内部类会创建对包含类的秘密引用,在本例中是
FooTest
,它是不可序列化的。换句话说,测试是无效的。要进行有效的测试,您必须定义一个扩展Foo并重写该方法的静态类。我猜匿名内部类无法实现接口,因此我必须创建一个扩展原始类的内部类?我完全错了,我只是认为这可能是个问题。但由于内部类也将“继承”其超类的接口,因此它当然是可序列化的。我创建了一个与您类似的示例,它的效果与预期的一样