C++从后面删除节点——单链表

Question

为什么我的代码不删除链表的最后一个元素？我创建了一个当前指针来横向穿过我的列表并跳出循环..(下一个是我的结构中名为 Card_Node 的点)。回答起来应该很简单，只是不确定为什么它不会删除列表中的最后一个节点”

    Card_Node *current; 
    current = front;
    while ( current->next->next != NULL){
    {   
        current = current-> next;
    }   
    Card a = current->next->card;
    return a;
    delete current->next;
    current->next = NULL;
}

Answer 1

return current->next->card;   // return !!
delete current->next;         // so this will never be executed
current->next = NULL;

更新

由于下面的评论要求进一步的输入，这里是我尝试保留原始原则的更新。

if (front == nullptr)  // Special handling of empty list
{
    // Nothing to return - add error handling - throw exception perhaps
    // or:
    return ???; // A default card perhaps
}
if (front->next == nullptr)  // Special handling of list with one element
{
    // Only one element
    Card a = front->card;
    delete front;
    front = nullptr;
    return a;
}

Card_Node *current; 
current = front;
while ( current->next->next != NULL)  // Iterate to find last element
{   
    current = current-> next;
}

// Now current->next is last element, i.e. the one to remove   
Card a = current->next->card;
delete current->next;
current->next = NULL;
return a;