Re: [C]非常郁闷的一个问题

2008/5/29 Thomas X. Iverson <txi@xxxxxxxxxxxxx>:

> 在做一道指针的习题，要求输入任意多个浮点数进行平均数的计算，动态分配内存
> 书上的代码如下：
> /* Exercise 7.1 Calculating a floating-point average using pointers */
> /*********************************************************************
>  * In this solution we allocate a some memory and when it is full    *
>  * allocate a new, larger amount of memory and copy the contents of  *
>  * the old memory to the new. We then free the old memory. This      *
>  * process repeats as often as necessary.                            *
>  *********************************************************************/
> #include <stdio.h>
> #include <stdlib.h>
> #include <ctype.h>
>
> int main(void)
> {
>  double *values = NULL;     /* Pointer to memory holding data values
> */
>  double *temp = NULL;       /* Pointer to newly allocated memory
> */
>  double sum = 0.0;          /* Sum of values
> */
>  int capacity = 0;          /* Maximum number of values that can be stored
> */
>  int increment = 5;         /* Capacity increment for dynamic allocation
> */
>  int count = 0;             /* Number of values read
> */
>  int i = 0;                 /* Index to array
>  */
>  char answer = 'n';
>
>  do
>  {
>    if(count == capacity)    /* Check if there is spare memory */
>    {
>      capacity += increment; /* Increase the capacity of memory by increment
> */
>      temp = (double*)malloc((capacity)*sizeof(double)); /* and allocate it
>  */
>      if(temp == NULL)       /* If memory was not allocated */
>      {                      /* Output a message  and end   */
>        printf("Memory allocation failed. Terminating program.");
>        exit(1);
>      }
>      if(values == NULL)     /* Are there any values?                  */
>        values = temp;       /* No - so just copy address of new memory */
>      else                   /* Yes - so copy data from old to new      */
>      {
>        for(i = 0 ; i<count ; i++)
>          *(temp+i) = *(values+i);
>        free(values);        /* Free the old memory */
>        values = temp;       /* Copy address of new */
>      }
>      temp = NULL;           /* Reset pointer       */
>    }
>
>    printf("Enter a value: ");
>    scanf("%lf", values+count++);
>
>    printf("Do you want to enter another(y or n)? ");
>    scanf(" %c", &answer);
>  }while(tolower(answer) == 'y');
>
>  /* Now sum the values */
>  for(i = 0 ; i<count ; i++)
>    sum += *(values+i);
>
>  /* Output the average */
>  printf("\n The average of the the values you entered is %.2lf.\n",
> sum/count);
>  free(values);     /* We are done - so free the memory */
> }
>
> 在理解了这个算法后，我自己重新实现了一遍(算是默写性质的复习)：
>
> #include<stdio.h>
> #include<stdlib.h>
> #include<ctype.h>
>
> int main(void)
> {
>        double *temp=NULL;
>        double *value=NULL;
>        double sum=0.0;
>        int i;
>        int count=0;
>        int capa=0;
>        int incre=5;
>        char answer='n';
>
>        do
>        {
>                if(count==capa)
>                {
>                        capa+=incre;
>                        temp=(double *)malloc((capa)*sizeof(double));
>                        if(temp==NULL)
>                        {
>                                printf("Failed\n");
>                                return 1;
>                        }
>                        if(value==NULL)
>                                value=temp;
>                        else
>                        {
>                                for(i=0;i<count;i++)
>                                {
>                                        *(temp+i)=*(value+i);
>                                }
>                                free(value);
>                                value=temp;
>                        }
>                        temp=NULL;
>                }
>                printf("Enter a value:");
>                scanf("%lf",value+count++);
>                printf("Enter another?:");
>                scanf(" %c",&answer);
>        }while(tolower(answer)=='y');
>
>        for(i=0;i<count;i++);

狂晕.............检查下这句.......

>
>        {
>                sum+=*(value+i);
>        }
>        printf("%.2lf",sum/count);
>        free(value);
> }
>
> 问题出现了，这两段代码我怎么看都没有觉得有问题，可是就是有两个问题:
> 1.我自己写的这个代码无法很好的处理scanf输入缓冲问题
> 2.我自己写的这个代码计算出的平均值总是0,我后来加了个printf查看sum和count的值，sum总是为0,count没有问题，不知道为什么
>
>
> --
> Keep It Simple Stupid
> http://blog.ghostunix.org
> ghosTM55
>
>

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