c - 我的 'NumberSize' 函数中的二进制操作数无效

Question

我创建了一个函数，该函数应该获取链接列表中的数字并确定哪个是最小的数字，哪个是最大的数字。我创建了一个 while 循环，该循环应该获取每个数字并将其与下一个数字进行比较。 “newNumber”是当前号码，“next”应该是列表中的下一个号码。我无法理解如何调用当前号码和下一个号码。我的 while 循环也是无限的，是因为我在确定中再次引用数字吗？

double NumberSize(NewNumber *start){

NewNumber *determine = start;
double SecondNumber =0;
double FirstNumber = 0;
while(determine != NULL){
    FirstNumber = determine->newNum;
    SecondNumber = determine->next;

    if(FirstNumber < SecondNumber){
        printf("The biggest number is:\n", SecondNumber);
    }else{
        printf("The smallest number is:\n", FirstNumber);
    }
}

Answer 1

您忘记单步浏览列表。使用以下解决方案求平均值:

void NumberSize(NewNumber * start)
{
  double num = 0.0;

  double biggest;
  double smallest;
  int flag = 0;

  NewNumber *temp = start;
  // set start values
  if(tmp != NULL){
     biggest = temp->newNum;
     smallest = temp->newNum;
     tmp = tmp->next;
     flag = 1;
  }

  // while temp is not NULL
  while (temp) {
    // get number from current node
    num = temp->newNum;
    // if that number is bigger than "biggest"
    if (num > biggest) {
      // exchange "biggest" with "num"
      biggest = num;
    }
    // do it the same way for the smallest number
    else if (num < smallest) {
      smallest = num;
    }
    // here is the forgotten line:
    // go to the next node
    temp = temp->next;
  }
  if(flag){
    // no returns, just printing
    printf("The smallest number is: \n", smallest);
    printf("The biggest number is: \n", biggest);
  } else {
    puts("list empty");
  }
}