Question

我正在做一个双向链表，我正在做pop_front函数。当列表中只有一个节点时，删除节点时会遇到一些问题。

int main()
{
    ForwardList<int> l;
    l.push_back(1);
    l.push_back(2);
    l.push_back(3);
    l.push_back(4);
    l.push_back(5);

    l.pop_front();
    l.pop_front();
    l.pop_front();
    l.pop_front();
    l.pop_front();
}

void pop_front()
{
    if (!empty())
    {
        if (this->head == this->tail)
        {
            delete this->head;
            delete this->tail;
            this->head = nullptr;
            this->tail = nullptr;
        }
        else
        {
            this->head = this->head->next;
            delete this->head->prev;
            this->head->prev = nullptr;
        }
    }
}

我收到此错误：

a.out(69846,0x10d5105c0) malloc: *** error for object 0x7fa7a2c02b50: pointer being freed was not allocated
a.out(69846,0x10d5105c0) malloc: *** set a breakpoint in malloc_error_break to debug
[1]    69846 abort      ./a.out

Answer 1

    if (this->head == this->tail)
    {
        delete this->head;
        delete this->tail;
        this->head = nullptr;
        this->tail = nullptr;
    }

看看这些行，因为this-> head == this-> tail，所以delele this-> head和delete this-> tail两次删除了相同的指针。

Answer 2

除了@duyue答案外，更好的设计是将节点取消链接和处置拆分为单独的功能。参见示例here。

在删除我的双向链接列表中的节点时遇到问题

2 个答案: