【Java】引用传递问题

 引用传递比较常见，下面是无效的几个示例

public class Test {
    public static void main(String[] args) {
        Test test = new Test();

        // 传null，不管怎么变，结果还是null
        test.test1();

    }

    public void test1() {
        TestPojo testPojo = null;

        setTestPojo(testPojo);// 调用

        System.out.println(JSON.toJSONString(testPojo));// null
    }

    public void setTestPojo(TestPojo testPojo) {
        testPojo = TestPojo.builder()
                .name("新的")
                .build();
    }

}

@Data
@Builder
@NoArgsConstructor
@AllArgsConstructor
public class TestPojo {
    private String name;
}

public class Test {
    public static void main(String[] args) {
        Test test = new Test();

        // 传obj，new后
        test.test2();

    }

    public void test2() {
        TestPojo testPojo = TestPojo.builder()
                .name("原来的")
                .build();

        setTestPojo(testPojo);// 调用

        System.out.println(JSON.toJSONString(testPojo));// {"name":"原来的"}
    }

    public void setTestPojo(TestPojo testPojo) {
        testPojo = TestPojo.builder()
                .name("新的")
                .build();
    }

}

public class Test {
    public static void main(String[] args) {
        Test test = new Test();

        // 传obj，设null
        test.test3();

    }

    public void test3() {
        TestPojo testPojo = TestPojo.builder()
                .name("原来的")
                .build();

        setTestPojoNull(testPojo);// 调用

        System.out.println(JSON.toJSONString(testPojo));// {"name":"改原来的"}
    }

    public void setTestPojoNull(TestPojo testPojo) {
        testPojo.setName("改原来的");

        testPojo = null;
    }

}