我的程序，但是验证的结果却是： 我用n来查看scanf是否把数据读进去了？可结果是n=2，我请请问各位这个问题怎么解决，我实在不知道我的错误出现在什么地方！请大家畅所欲言！谢谢，急！！！！！！！！！

/*-------------------------------
programmer:
version:
functional description:
--------------------------------*/



#include <stdio.h>

//function  declaration
void sum_n_avg(double  *n1,double  *n2,double  *n3, double  *sump, double  *avgp);


int
main()
{
    //function  declaration
    double  one, two, three, sum_of_3, avg_of_3;
    // Verify return value of scanf
    int n = 1;                  
   
    printf("please input three numbers>");
    n = scanf("%.2lf,%.2lf,%.2lf\n",&one,&two,&three);
   
    sum_n_avg(&one, &two, &three, &sum_of_3, &avg_of_3);
   
    //print the result
    printf("sum_of_3 = %6.2f\n",sum_of_3);
    printf("avg_of_3 = %6.2f\n",avg_of_3);
    printf("n = %d",n);
}



void
sum_n_avg(double  *n1,double  *n2,double  *n3, double  *sump, double  *avgp)
{   
   
    *sump = *n1 + *n2 + *n3;
    *avgp = (*sump) / 3;
}

怎么没有人给回呀，我很急！！！

怎么回事？

#include <stdio.h>

//function  declaration
void sum_n_avg(double  *n1,double  *n2,double  *n3, double  *sump, double  *avgp);


int
main()
{
    //function  declaration
    double  one, two, three, sum_of_3, avg_of_3;
    // Verify return value of scanf
    int n = 1;                 
  
    printf("please input three numbers>");
    n = scanf("%lf,%lf,%lf",&one,&two,&three);
  
    sum_n_avg(&one, &two, &three, &sum_of_3, &avg_of_3);
  
    //print the result
    printf("sum_of_3 = %6.2f\n",sum_of_3);
    printf("avg_of_3 = %6.2f\n",avg_of_3);
    printf("n = %d",n);
}



void
sum_n_avg(double  *n1,double  *n2,double  *n3, double  *sump, double  *avgp)
{  
  
    *sump = *n1 + *n2 + *n3;
    *avgp = (*sump) / 3;
}

看红色的改动

n = scanf("%lf,%lf,%lf",&one,&two,&three);

晕，代码里不能加红色
看这个

厨房，我没看到你改的呀

谢谢！厨房，我的问题解决了！谢谢，非常感谢！