catch中的return语句的行为,最后

Question

public class J {

     public Integer method(Integer x) 
        {
            Integer val = x;
            try 
            {

                    return val;
            } 
            finally 
            {
                     val = x + x;
            }
        }
        public static void main(String[] args) 
        {
            J littleFuzzy = new J();
            System.out.println(littleFuzzy.method(new Integer(10)));
        }

}

它将返回"10".

现在我只是将Return type Integer替换为StringBuilder并更改了Output.

public class I {

     public StringBuilder method(StringBuilder x) 
        {
            StringBuilder val = x;
            try 
            {

                    return val;
            } 
            finally 
            {
                     val = x.append("aaa");
            }
        }
        public static void main(String[] args) 
        {
            I littleFuzzy = new I();
            System.out.println(littleFuzzy.method(new StringBuilder("abc")));
        }


}

OutPut是"abcaaa"

所以,任何人都可以详细解释我.有什么区别.？

Answer 1

只是因为immutableso after 方法中的整数返回，即使方法中的值发生更改，它也不会反映，并且会反映在StringBuilderObject中

编辑：

public class J {
    public String method(String x) {
        String val = x;
        try {
            return val;
        } finally {
            val = x + x;
        }
    }

    public static void main(String[] args) {
        J littleFuzzy = new J();
        System.out.println(littleFuzzy.method("abc"));
    }
}