Java程序运行时while循环跳过首次迭代是什么原因导致的？

问题根因 #

• Scanner类的nextInt()方法仅读取输入流中的整数部分，不会消耗整数末尾的换行符（\n），该换行符会一直残留在输入缓冲区中。
• 首次加/乘运算结束后，进入下一轮外层while循环时，scan.nextLine()会直接读取缓冲区中残留的空换行，得到空字符串，不会触发用户输入等待逻辑，空字符串不匹配multiply/add分支，直接进入下一次循环，因此会连续打印两次操作提示。

修复方案 #

推荐在所有nextInt()调用后，额外调用一次nextLine()消耗残留的换行符，修正后的代码如下：

import java.util.Scanner;

public class myfun {
    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        int totalNumbers;
        int product = 0;
        int count = 0;
        int sum = 0;
        String choice = "choice";
        System.out.print("You can add up to ten numbers.\n");
        while(! choice.equalsIgnoreCase("stop")){
            System.out.print("Would you like to add or multiply? (Enter stop to be done)");
            choice = scan.nextLine();
            if (choice.equalsIgnoreCase("multiply")) {
                product = 0;
                System.out.print("How many numbers would you like to multiply?");
                totalNumbers = scan.nextInt();
                scan.nextLine(); // 消费数字输入后残留的换行符
                count = 0;
                System.out.print("Please enter " + totalNumbers +" numbers.");
                product +=scan.nextInt();
                count++;
                while (count < totalNumbers) {
                    product *= scan.nextInt();
                    count++;
                }
                scan.nextLine(); // 消费最后一个数字输入后残留的换行符
                System.out.println("The product is " + product);
            }
            else if (choice.equalsIgnoreCase("add")) {
                sum = 0;
                System.out.print("How many numbers would you like to add?");
                totalNumbers = scan.nextInt();
                scan.nextLine(); // 消费数字输入后残留的换行符
                count = 0;
                System.out.print("Please enter " + totalNumbers +" numbers.");
                sum +=scan.nextInt();
                count++;
                    while (count < totalNumbers) {
                        sum += scan.nextInt();
                        count++;
                    }
                scan.nextLine(); // 消费最后一个数字输入后残留的换行符
                System.out.println("The sum is " + sum);
            }
        }
        System.out.print("Come again!");
        scan.close();
    }
}

也可以选择统一用nextLine()读取所有输入，再手动转换为整数类型，从根源上避免不同读取方法的逻辑差异问题。

内容的提问来源于stack exchange，提问作者Jacob Buckingham