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：
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| #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;
}
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输入样例：
4 6
输出样例：
The sum is 10.00
The diff is -2.00
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| #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;
}
void sum_diff( float op1, float op2, float *psum, float *pdiff ){
*psum=op1+op2;
*pdiff=op1-op2;
}
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