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op1和op2是输入的两个实数，*psum和*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

#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 )
 {
     *psum=op1+op2;
     *pdiff=op1-op2;
 } 

练习8-2 计算两数的和与差（10 分）

原文：https://www.cnblogs.com/2228212230qq/p/9265137.html