练习：计算器

写一个计算器，要求实现加减乘除功能，并且能够循环接收新的数据，通过用户交互实现。

package com.yanseyutou.method;

import java.util.Scanner;

public class calculator {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);

        String isContinue = "y";

        while(isContinue.equals("y")) {

            double i1 = 0;//第一个数
            double i2 = 0;//第二个数
            String ca = "+";//运算符
            double result = 0;//结果

            System.out.println("请输入第一个数：");
            i1 = scanner.nextDouble();
            System.out.println("请输入运算符：");
            ca = scanner.next();
            System.out.println("请输入第二个数：");
            i2 = scanner.nextDouble();

            switch (ca) {
                case "+":
                    result = add(i1, i2);
                    System.out.println(i1 + ca + i2 + "计算结果为：" + result);
                    break;
                case "-":
                    result = sub(i1, i2);
                    System.out.println(i1 + ca + i2 + "计算结果为：" + result);
                    break;
                case "*":
                    result = mul(i1, i2);
                    System.out.println(i1 + ca + i2 + "计算结果为：" + result);
                    break;
                case "/":
                    if (i2 == 0) {
                        System.out.println("分母不可为零");
                    } else {
                        result = div(i1, i2);
                        System.out.println(i1 + ca + i2 + "计算结果为：" + result);
                    }
                    break;
                default:
                    System.out.println("运算符输入错误");
            }
            System.out.println("进入下一次运算请输入y（不进入请随意输入）：");
            isContinue = scanner.next();
        }

        scanner.close();

    }

    public static double add(double i1,double i2){
        return i1+i2;
    }
    public static double sub(double i1,double i2){
        return i1-i2;
    }
    public static double mul(double i1,double i2){
        return i1*i2;
    }
    public static double div(double i1,double i2){
            return i1/i2;
    }
}