调用自己的方法时，旧字符串值会留下

Question

当我收到不需要的输入时，我再次调用该方法，但在第二次迭代时，字符串上仍然留有一个字符。为什么它不创建一个全新的字符串并替换旧的值？

所以首先我在扫描仪中输入“mmmm”，这会触发 if 语句中的条件并调用createSocial()，然后重新运行循环，然后我输入新的输入 5555。现在这是我没有得到的：当我打印出新的字符串时，它们显示为 5555，但是在循环中打印它们时留下了“m”，为什么？

    public class test {

    public static String createSocial() {

            Scanner sc = new Scanner(System.in);

            String social = sc.nextLine();

            String digit = social; 

            System.out.println(social + "     SOCIAL");

            System.out.println(digit + "     DIGIT");

            if (social.length() != 4) {

                System.out.println("You did not type 4 digits, try again");  

                createSocial(); 

            }

            //check non-integers        

            while(digit.length() > 0) {

                System.out.println( digit.charAt(0) + "<---");  

                if(Character.isDigit( digit.charAt(0) ) == false) {

                    System.out.println("You did not type your last 4 digits correctly, try again"); 

                    createSocial();  

                }

                digit = digit.substring(1); 

            }   

            return social; 

        }

}

这是我的输出：

    enter the last 4 digits of your Social Security number

    mmmm    //(This is my first input for the scanner)

    mmmm     SOCIAL

    mmmm     DIGIT

    m<---

    You did not type your last 4 digits correctly, try again

    5555   //(this is my input for the scanner the second time around)

    5555     SOCIAL

    5555     DIGIT

    5<---

    5<---

    5<---

    5<---

    m<---

    You did not type your last 4 digits correctly, try again

Answer 1

您看到m 的原因如下：5<---5<---5<---5<---m<---仅仅是因为您在开始递归调用之前没有清除社交字符串变量。实际上，当进行此递归调用时，您应该这样做：social = createSocial();这样，您就可以保证Social将始终保留该递归调用的真实值，而不是先前递归调用的值。如果您不从createSocial()方法获取返回值，您将永远不会获得新结果，因为社交变量的范围对于方法本身而言是本地的。您需要接受返回值，就像第一次初始调用createSocial()方法时所做的那样。但这还不是全部，您还需要在每次递归调用（ ）之后更新数字digit = social;变量内容。在我看来，这比必要的更复杂。我理解为什么您添加了数字变量，但是如果您仔细查看代码，您可以通过继续利用社交变量来进行数字细分来完成同样的事情。您只需在该方法返回的变量中进行数值细分之前，将原始值保存在social中，例如：private static int methodCalls = 0; // Class Global Member Variablepublic static String createSocial() {&nbsp; &nbsp; methodCalls++;&nbsp; &nbsp; Scanner sc = new Scanner(System.in);&nbsp; &nbsp; String social = sc.nextLine();&nbsp; &nbsp; System.out.println(social + "&nbsp; &nbsp; &nbsp;SOCIAL");&nbsp; &nbsp; if (social.length() != 4) {&nbsp; &nbsp; &nbsp; &nbsp; System.out.println("You did not type 4 digits, try again");&nbsp; &nbsp; &nbsp; &nbsp; social = createSocial();&nbsp; &nbsp; }&nbsp; &nbsp; // returnResult will eventually will hold the&nbsp;&nbsp; &nbsp; // valid result to return.&nbsp; &nbsp; String returnResult = social;&nbsp; &nbsp; //check non-integers&nbsp; &nbsp; &nbsp;&nbsp;&nbsp; &nbsp; while (social.length() > 0) {&nbsp; &nbsp; &nbsp; &nbsp; System.out.println(social.charAt(0) + " <--- Method call: " + methodCalls);&nbsp; &nbsp; &nbsp; &nbsp; if (Character.isDigit(social.charAt(0)) == false) {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; System.out.println("You did not type your last 4 digits correctly, try again");&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; social = createSocial();&nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; &nbsp; &nbsp; social = social.substring(1);&nbsp; &nbsp; }&nbsp; &nbsp; methodCalls--;&nbsp; &nbsp; return returnResult;}使用您发布的示例参数（mmmmm 和 5555），您可以使用上面的代码，如下所示：System.out.println(createSocial());输入mmmmm和5555后，控制台的输出结果将如下所示：mmmmmmmmmm&nbsp; &nbsp; &nbsp;SOCIALYou did not type 4 digits, try again55555555&nbsp; &nbsp; &nbsp;SOCIAL5 <--- Method call: 25 <--- Method call: 25 <--- Method call: 25 <--- Method call: 25 <--- Method call: 15 <--- Method call: 15 <--- Method call: 15 <--- Method call: 15555为什么都是5 <---？因为单次递归调用。标记为 5 的方法调用：1是对createSocial () 方法的初始调用。任何超过 1 的方法调用都是递归调用。每次调用时5 <---都会列出四个。这一切都可以以更简单的方式完成，并且不需要递归来执行此任务，可以使用循环来代替，例如：public static String createSocial() {&nbsp; &nbsp; String ls = System.lineSeparator();&nbsp; &nbsp; Scanner sc = new Scanner(System.in);&nbsp; &nbsp; String social = "";&nbsp; &nbsp; while (social.equals("")) {&nbsp; &nbsp; &nbsp; &nbsp; System.out.print("Please enter the last four digits of " + ls&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; + "your Social Security Number: --> ");&nbsp; &nbsp; &nbsp; &nbsp; social = sc.nextLine();&nbsp; &nbsp; &nbsp; &nbsp; if (!social.matches("\\d{4}")) {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; System.err.println("Invalid Entry! Please Try Again..." + ls);&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; social = "";&nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; }&nbsp; &nbsp; return social;}但如果需要递归，那么你可以这样做：public static String createSocial() {&nbsp; &nbsp; String ls = System.lineSeparator();&nbsp; &nbsp; Scanner sc = new Scanner(System.in);&nbsp; &nbsp; String social = "";&nbsp; &nbsp; System.out.print("Please enter the last four digits of " + ls&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; + "your Social Security Number: --> ");&nbsp; &nbsp; social = sc.nextLine();&nbsp; &nbsp; if (!social.matches("\\d{4}")) {&nbsp; &nbsp; &nbsp; &nbsp; System.err.println("Invalid Entry! Please Try Again..." + ls);&nbsp; &nbsp; &nbsp; &nbsp; social = createSocial();&nbsp; &nbsp; }&nbsp; &nbsp; return social;}使用 Java 的String#matches()方法和一个小的正则表达式来检查数据输入的有效性，消除了对大量代码的需要，并使内容更容易阅读。上面两个代码示例中IF语句条件的matches()方法中使用的正则表达式 (RegEx)if (!social.matches("\\d{4}")) {基本上意味着：如果用户输入字符串不是四位整数值，则显示“无效输入！” 信息。