C语言实现简易计算器：使用 switch-case 语句

#include <stdio.h>

int main() {
    double num1, num2, result;
    char operator;

    printf("请输入两个数和一个运算符（+、-、*、/、%）：");
    scanf("%lf %lf %c", &num1, &num2, &operator);

    switch (operator) {
        case '+':
            result = num1 + num2;
            printf("结果为 %.2lf\n", result);
            break;
        case '-':
            result = num1 - num2;
            printf("结果为 %.2lf\n", result);
            break;
        case '*':
            result = num1 * num2;
            printf("结果为 %.2lf\n", result);
            break;
        case '/':
            if (num2 == 0) {
                printf("除数不能为0\n");
            } else {
                result = num1 / num2;
                printf("结果为 %.2lf\n", result);
            }
            break;
        case '%':
            if (num2 == 0) {
                printf("除数不能为0\n");
            } else {
                result = (int)num1 % (int)num2;
                printf("结果为 %.2lf\n", result);
            }
            break;
        default:
            printf("输入的运算符不合法\n");
            break;
    }

    return 0;
}