题目
写一个简单的函数实现下面的功能：具有三个参数，完成对两个整型数据的加、减、乘、除四种操作，前两个为操作数，第三个参数为字符型的参数。
解题步骤
（1）定义变量；
（2）接收用户输入；
（3）函数计算；
（4）输出结果；
Java
import java.util.Scanner; public class E20210814 { public static void main(String[] args) { int a1, b; char c; Scanner input = new Scanner(System.in); System.out.print("please enter two whole numbers:"); a1 = input.nextInt(); b = input.nextInt(); System.out.print("please enter your computing type:[a-addition s-subtraction m-multiplication d-division]:"); c = input.next().charAt(0); switch (c) { case 'a' -> System.out.format("%d+%d=%d", a1, b, a1 + b); case 's' -> System.out.format("%d-%d=%d", a1, b, a1 - b); case 'm' -> System.out.format("%d*%d=%d", a1, b, a1 * b); case 'd' -> { if (b == 0) { System.out.println("the divisor is not 0 error!!!"); } else { System.out.format("%d/%d=%d", a1, b, a1 / b); } } default -> System.out.println("input error please try again!!!"); } } }
enhanced switch
switch (c) { case 'a' -> System.out.format("%d+%d=%d", a1, b, a1 + b); case 's' -> System.out.format("%d-%d=%d", a1, b, a1 - b); case 'm' -> System.out.format("%d*%d=%d", a1, b, a1 * b); case 'd' -> { if (b == 0) { System.out.println("the divisor is not 0 error!!!"); } else { System.out.format("%d/%d=%d", a1, b, a1 / b); } } default -> System.out.println("input error please try again!!!"); }
说明
- 注意
switch-case语句中case处的数据类型，因为设定了变量c为char类型，所以需要使用 c = input.next().charAt(0) 语句接收用户键盘上的单个字符输入，charAt()方法用于返回指定索引处的字符，索引范围为从 0 到length() - 1。- Java 中引入增强型
switch结构，给出参考代码。主要特点如下：需要返回值、无需break、使用箭头->、可进行 case 间的合并，以逗号分隔。
C语言
#include <stdio.h> int calculate(int operand1, int operand2, char type) { char a, s, m, d; switch (type) { case 'a': printf("%d+%d=%d", operand1, operand2, operand1 + operand2); break; case 's': printf("%d-%d=%d", operand1, operand2, operand1 - operand2); break; case 'm': printf("%d*%d=%d", operand1, operand2, operand1 * operand2); break; case 'd': { if (operand2 == 0) { printf("the divisor is not 0 the error please try again"); break; } else { printf("%d/%d=%d", operand1, operand2, operand1 / operand2); break; } } default: printf("if the type of operation is not specified please re enter!!!"); } } int main() { int a, b; char c; printf("please enter two integers:"); scanf("%d%d", &a, &b); printf("please enter your computing type:[a-addition s-subtraction m-multiplication d-division]:"); getchar(); scanf("%c", &c); calculate(a, b, c); return 0; }
说明
因为有四种计算类型，所以我们使用
switch-case语句解决，注意除法计算中除数不为 0 的条件判断，且case后需为常量，这里使用字符做判断条件，加上单引号‘’变为字符常量。
