6-2 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 

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 %.2fnThe diff is %.2fn", sum, diff);

    return 0; 
}

输入样例：

4 6

结尾无空行

输出样例：

The sum is 10.00
The diff is -2.00

结尾无空行

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