该程序从用户处获取算术运算运算符 +, -, *, /和两个操作数。然后，它根据用户输入的运算符对两个操作数执行计算。
使用switch语句的简单计算器
#include
int main() {
char operator;
double first, second;
printf("Enter an operator (+, -, *,): ");
scanf("%c", &operator);
printf("Enter two operands: ");
scanf("%lf %lf", &first, &second);
switch (operator) {
case '+':
printf("%.1lf + %.1lf = %.1lf", first, second, first + second);
break;
case '-':
printf("%.1lf - %.1lf = %.1lf", first, second, first - second);
break;
case '*':
printf("%.1lf * %.1lf = %.1lf", first, second, first * second);
break;
case '/':
printf("%.1lf / %.1lf = %.1lf", first, second, first / second);
break;
// operator doesn't match any case constant
default:
printf("Error! operator is not correct");
}
return 0;
}
输出
Enter an operator (+, -, *,): *
Enter two operands: 1.5
4.5
1.5 * 4.5 = 6.8
用户输入的* 运算符存储在运算符 。并且，两个操作数1.5和4.5分别存储在第一和第二个中 。
由于运算符 *匹配case '*': ：，因此程序的控制跳转到
printf("%.1lf * %.1lf = %.1lf", first, second, first * second);
该语句计算产品并将其显示在屏幕上。
最后， break;语句结束switch语句。