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
The diff is %.2f
", 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 }