返回链表中的值之和 [英] Returning sum of values in linked list

问题描述

我有以下代码:

int sum(LinkedList * list) {

  assert(list!=NULL);

  Node *currentNode = list->head;
  int sum = 0;

  for (currentNode = currentNode->next; currentNode !=NULL; currentNode = currentNode -> next) {
        sum = sum + currentNode->data;

        }
  return sum;

}

我希望它返回链表* list中所有值的总和.但是，我不断遇到细分错误.谁能帮助我发现致命错误?

I want it to return the sum of all the values in the linked list *list. However, I keep getting a segmentation fault. Can anyone help me spot the fatal error?

推荐答案

将循环更改为:

for (currentNode = list->head; currentNode !=NULL; currentNode = currentNode -> next) {
    sum = sum + currentNode->data;
}

这将解决两个问题:

1. 它将检查list->head不是NULL;
2. 计算总和时，它不会跳过列表中的第一个元素.

1. It will check that list->head is not NULL;
2. It won't skip over the first element in the list when calculating the sum.

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