预期实现功能为一个简单计算器程序,输入格式为:data1 op data2。其中,data1和data2是参与运算的两个数,op为运算符,它的取值只能是+、-、*、/。
输
参考代码如下:(如有帮助,望采纳!谢谢! 点击我这个回答右上方的【采纳】按钮)
#include <stdio.h>
int main()
{
float a,b;
char op,c;
do {
printf("Please enter the expression:\n");
scanf("%f%1s%f", &a,&op,&b);
switch (op)
{
case '+':
printf("%.2f + %.2f = %.2f\n",a,b,a+b);
break;
case '-':
printf("%.2f - %.2f = %.2f\n",a,b,a-b);
break;
case '*':
printf("%.2f * %.2f = %.2f\n",a,b,a*b);
break;
case '/':
if (b==0)
printf("Division by zero!\n");
else
printf("%.2f / %.2f = %.2f\n",a,b,a/b);
break;
default:
printf("Unknown operator!\n");
break;
}
printf("Do you want to continue(Y/N or y/n)?");
scanf("%1s", &c);
} while (c=='y' || c=='Y');
}