任务描述

本关任务：本题要求实现一个计算输入的两数的和与差的简单函数。

函数接口定义： void sum_diff( float op1, float op2, float *psum, float *pdiff ); 其中op1和op2是输入的两个实数，*psum和*pdiff是计算得出的和与差。

主程序样例:

1. #include <stdio.h>
3. void sum_diff( float op1, float op2, float *psum, float *pdiff );
5. int main()
6. {
7. float a, b, sum, diff;
9. scanf("%f %f", &a, &b);
10. sum_diff(a, b, &sum, &diff);
11. printf("The sum is %.2f\nThe diff is %.2f\n", sum, diff);
13. return 0;
14. }
16. /* 你的代码将被嵌在这里 */

编程要求

根据提示，在右侧编辑器补充代码，实现一个计算输入的两数的和与差的简单函数。

测试说明

输入样例：

1. 4 6

输出样例：

1. The sum is 10.00
2. The diff is -2.00

   #include <stdio.h>void sum_diff( float op1, float op2, float *psum, float *pdiff );int main()
   {float a, b, sum, diff;scanf("%f %f", &a, &b);sum_diff(a, b, &sum, &diff);printf("The sum is %.2f\nThe diff is %.2f\n", sum, diff);return 0;
   }/* 你的代码将被嵌在这里 */
   void sum_diff( float op1, float op2, float *psum, float *pdiff )
   {//=======begin=======*psum = op1+op2;*pdiff = op1-op2;//========end========
   }