斜边程序，在用户输入 2 之前无法弄清楚如何让程序运行

Question

我正在编写一个查找三角形斜边的程序，我需要让程序运行任意次数，直到用户进入2。我无法弄清楚当用户输入2时如何结束程序。

package assignment5a;

import java.util.Scanner;//import Scanner 

public class Assignment5A {

    public static void main(String[] args) {

        Scanner sc = new Scanner(System.in);//new Scanner variable

        int answer;

        double side1, side2, result;

        System.out.println("Enter 1 to calculate the hypotenuse of a triangle or enter 2 to quit.");

        answer = sc.nextInt();

        while(answer < 0 || answer > 2){

            System.err.println("Please enter a valid answer.");

            System.out.println("Enter 1 to calculate the hypotenuse of a triangle or enter 2 to quit.");

            answer = sc.nextInt();

        }

        System.out.println("Enter side 1 of the triangle :");//input for side 1

        side1 = sc.nextDouble();

        System.out.println("Enter side 2 of the triangle :");//input for side 2

        side2 = sc.nextDouble();

        result = hypotenuse(side1, side2);//declares result as the result of the method hypotenuse

        System.out.printf("Hypotenuse of your triangle is: %.2f%n", result);//prints results

    }

    public static double hypotenuse(double s1, double s2){//method for calculating hypotenuse

        double hypot;

        hypot = Math.sqrt((Math.pow(s1, 2) + Math.pow(s2, 2)));

        return hypot;

    }

}

Answer 1

Wilmol的答案和Elliot Frisch的答案/评论是一半的解决方案。另一半是，你需要围绕大多数逻辑的外部循环，这样它就会重复。将大部分内容放在用于启动的循环中，以便它永远循环。main()while (true) {然后使用逻辑...在用户输入 2 时实际突破。if (answer == 2) {

Answer 2

所以我想通了。你的答案帮助很大，但我最终放了两个同时循环。代码如下：public static void main(String[] args) {&nbsp; &nbsp; Scanner sc = new Scanner(System.in);//new Scanner variable&nbsp; &nbsp; int answer;&nbsp; &nbsp; double side1, side2, result;&nbsp; &nbsp; System.out.println("Enter 1 to calculate the hypotenuse of a triangle or enter 2 to quit.");&nbsp; &nbsp; answer = sc.nextInt();&nbsp; &nbsp; while(answer < 0 || answer > 2){&nbsp; &nbsp; &nbsp; &nbsp; System.err.println("Please enter a valid answer.");&nbsp; &nbsp; &nbsp; &nbsp; System.out.println("Enter 1 to calculate the hypotenuse of a triangle or enter 2 to quit.");&nbsp; &nbsp; &nbsp; &nbsp; answer = sc.nextInt();&nbsp; &nbsp; }&nbsp;&nbsp;&nbsp; &nbsp; &nbsp; &nbsp; while(answer == 1){&nbsp; &nbsp; System.out.println("Enter side 1 of the triangle :");//input for side 1&nbsp; &nbsp; side1 = sc.nextDouble();&nbsp; &nbsp; System.out.println("Enter side 2 of the triangle :");//input for side 2&nbsp; &nbsp; side2 = sc.nextDouble();&nbsp; &nbsp; result = hypotenuse(side1, side2);//declares result as the result of the method hypotenuse&nbsp; &nbsp; System.out.printf("Hypotenuse of your triangle is: %.2f%n", result);//prints results&nbsp; &nbsp; System.out.println("Enter 1 to calculate the hypotenuse of a triangle or enter 2 to quit.");&nbsp; &nbsp; answer = sc.nextInt();&nbsp;}} 公共静态双斜边（双 s1、双 s2）{//计算斜边的方法&nbsp; &nbsp; double hypot;&nbsp; &nbsp; hypot = Math.sqrt((Math.pow(s1, 2) + Math.pow(s2, 2)));&nbsp; &nbsp; return hypot;}}

Answer 3

几个选项：if (answer == 2){break;}if (answer == 2){return;}if (answer == 2){System.exit(0);}