如何用Scanner区分输入字符串+整数与仅字符串的场景？

问题背景 #

我编写了一段数值加减运算的代码，支持输入add 3为结果加3、subtract 7为结果减7，但输入q时程序会卡住等待输入，无法正常退出。原以为kb.hasNextInt()能检测是否有整数输入，实际未达预期。考虑过用split()拆分整行输入，但希望有更简洁的方案。

原代码如下：

private static int result = 0;

public static void main(String[] args){
  Scanner kb = new Scanner(System.in);
  String operation = "";
  int num = 0;
  boolean finished = false;
  
  while(!finished){
    System.out.println("Enter the operation you'd like to perform on " + result + " or type q to quit.");
    operation = kb.next();
    if(kb.hasNextInt()){ num = kb.nextInt(); } //The problem
    
    switch (operation){
      case "add":
        add(num);
        break;
      case "subtract":
        subtract(num);
        break;
      case "q":
        finished = true;
        break;
    }
    System.out.println("New result: " + result);
  }
  System.out.println("Bye!");
}

public static void add(int num){
  result += num;
} 
public static void subtract(int num){
  result -= num;
} 

核心原因 #

Scanner.hasNextInt()是阻塞式方法：它会忽略输入中的空白字符（比如换行），一直等待用户输入非空白内容，直到能判断该内容是否为整数。当输入q回车后，kb.next()读取了q，但hasNextInt()会继续等待下一个输入，导致程序卡住。

解决方案 #

方法1：调整逻辑顺序，优先处理退出指令 #

直接先判断操作是否为q，如果是立即标记退出，跳过后续的整数检测步骤，从根源避免阻塞：

private static int result = 0;

public static void main(String[] args){
  Scanner kb = new Scanner(System.in);
  String operation = "";
  int num = 0;
  boolean finished = false;
  
  while(!finished){
    System.out.println("Enter the operation you'd like to perform on " + result + " or type q to quit.");
    operation = kb.next();
    
    // 先处理退出，不碰整数检测逻辑
    if ("q".equals(operation)) {
      finished = true;
      System.out.println("Bye!");
      break;
    }
    
    // 非退出指令才检查并读取整数
    if(kb.hasNextInt()){ 
      num = kb.nextInt(); 
    }
    
    switch (operation){
      case "add":
        add(num);
        break;
      case "subtract":
        subtract(num);
        break;
    }
    System.out.println("New result: " + result);
  }
}

public static void add(int num){
  result += num;
} 
public static void subtract(int num){
  result -= num;
} 

方法2：整行读取输入（更健壮） #

如果担心用户输入格式多变（比如多空格、误输入非数字），用nextLine()整行读取后拆分的方式反而更可靠，且不会出现阻塞问题：

private static int result = 0;

public static void main(String[] args){
  Scanner kb = new Scanner(System.in);
  String input = "";
  boolean finished = false;
  
  while(!finished){
    System.out.println("Enter the operation you'd like to perform on " + result + " or type q to quit.");
    input = kb.nextLine().trim();
    
    String[] parts = input.split("\\s+");
    if (parts.length == 0) continue;
    
    String operation = parts[0];
    int num = 0;
    
    switch (operation){
      case "add":
      case "subtract":
        if (parts.length > 1) {
          try {
            num = Integer.parseInt(parts[1]);
            if ("add".equals(operation)) add(num);
            else subtract(num);
          } catch (NumberFormatException e) {
            System.out.println("无效数字，请重新输入。");
          }
        } else {
          System.out.println("操作缺少数字，请重新输入。");
        }
        break;
      case "q":
        finished = true;
        break;
      default:
        System.out.println("无效操作，请重新输入。");
    }
    
    if (!finished) System.out.println("New result: " + result);
  }
  System.out.println("Bye!");
}

public static void add(int num){
  result += num;
} 
public static void subtract(int num){
  result -= num;
} 

内容的提问来源于stack exchange，提问作者Babbit