c++ - 使用列表向后打印字母表时缺少“a”

Question

我正在尝试使用链表向后打印字母表，但无法显示“a”。出于某种原因，它跳过了它，我无法弄清楚。这是我的代码:

    int _tmain(int argc, _TCHAR* argv[])
    {
        char s[]="abcdefghijklmnopqrstuvwxyz";

        node *head;
        node *temp;
        node *current;

        head = new node;          // create the head of the linked list
        head->data = s[25];
        head->next = NULL;
        temp = head;   // get ready for the loop - save the head in temp - you are         going to change temp in the loop

        for(size_t i = 25; i >= 1; i--)      // create the rest of the linked list
        {
            current = new node;    // make a new node
            current->data = s[i];  // set it's data member
            current->next = NULL;
            temp->next = current;  // point to the new node
            temp = current;        // make temp point to current node (for next         time through)
        }

        node *ptr = head;    // set a ptr to head, then you are going to "increment"         the pointer

        while (ptr != NULL)
        {
            cout << ptr->data; // print out the linked list
            ptr = ptr->next;   // increment the linked list
        }

        cout << endl;
        system("pause");
        return 0;
    }

有人知道为什么会这样吗？我认为我的 for 循环有问题。谢谢!

Answer 1

问题是您在 for 循环中遗漏了 i=0 的情况。

将您的 for 循环更改为:

    size_t i = 25; // 'z' was already added
    do
    {
        --i;
        current = new node;    // make a new node
        current->data = s[i];  // set it's data member
        current->next = NULL;
        temp->next = current;  // point to the new node
        temp = current;        // make temp point to current node (for next         time through)
    } while ( i != 0 );

你不能简单地执行 for(size_t i = 25; i >= 0; i--) 的原因是因为 i 是无符号的，所以它i >= 0 总是这样，因此循环永远不会终止，或者更有可能出现段错误。