C语言编程 C Language Programming - 0009

编程题0009 (from Programming Teaching Assistant (PTA))

计算两数的和与差

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 

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

Answer - Simple：

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