练习8-2 计算两数的和与差 (10分)

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op1op2是输入的两个实数,*psum*pdiff是计算得出的和与差。

裁判测试程序样例:

#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 */
 

输入样例:

4 6
 

输出样例:

The sum is 10.00
The diff is -2.00


1 void sum_diff( float op1, float op2, float *psum, float *pdiff ){
2     *psum=op1+op2;
3     *pdiff=op1-op2;
4 }

 

posted @ 2020-06-30 10:24  行行行行行行行  阅读(531)  评论(0编辑  收藏  举报