c - 从最后一个节点遍历指针到第一个节点

Question

bn_ptr drive_temp(bn_ptr head,bn_ptr temp,bn_ptr current)
{

    while(temp->next!=current)
    {
        temp=temp->next;
        current=temp;
    }
    temp=head;

    return current; 
}

我有链表，“头”指针保存第一个节点，“当前”指针保存最后一个节点，我想将“当前”一个接一个地带到头，所以我编写了这个函数，但在调试时它给出了段错误程序

Answer 1

temp->next!=current永远不会是真的，除非 temp==temp->next .

试试这个:

bn_ptr drive_temp(bn_ptr head,bn_ptr temp,bn_ptr current)
{

    while(temp->next!=current)
    {
        temp=temp->next;
    }
    current=temp; /* get this out of the loop */
    temp=head;

    return current; 
}

或者更简化:

bn_ptr drive_temp(bn_ptr head,bn_ptr temp,bn_ptr current)
{
    (void)head; /* head isn't used, so put this to avoid warning */
    while(temp->next!=current)
    {
        temp=temp->next;
    }
    /* current and temp will be lost, so assigning to them here is useless */

    return temp; 
}

为了更安全，请确保 temp不是NULL以避免出现 current 时出现运行时错误无效。

bn_ptr drive_temp(bn_ptr head,bn_ptr temp,bn_ptr current)
{
    (void)head; /* head isn't used, so put this to avoid warning */
    while(temp != NULL && temp->next!=current)
    {
        temp=temp->next;
    }

    /* temp will be the previous node of current or NULL */
    return temp; 
}

也许你想要这个:

bn_ptr drive_temp(bn_ptr head,bn_ptr current) /* remove temp from the arguments */
{
    bn_ptr temp = head;
    while(temp != NULL && temp->next!=current)
    {
        temp=temp->next;
    }

    /* temp will be the previous node of current or NULL */
    return temp; 
}