使用链表从双端队列中删除最后一个节点时避免内存泄漏
Avoid memory leak when removing last node from deque using linkd list
对于我的家庭作业,我需要创建从双端队列添加(追加)和删除(服务)节点的方法。但是,当尝试为最后一个节点提供服务时,我遇到了内存泄漏问题,因为我不知道如何检索系统不再使用的节点。
这是该方法的代码。
void Deque::serveAtRear(int &x)
{
if(empty())
{
cout << "Fila Vazia" << endl;
exit(0);
}
else
{
DequePointer q; //Pointer before p
q = head; //q starts at the deque's head
p = head->nextNode; //Pointer for the next node
if(q->nextNode==NULL) //If there's only one node
{
x = q->entry; //x will receive the value of the removed node to print it afterward
head = tail = NULL;
delete p; //I don't know if I'm using the "delete" right
}
else
{
while(p->nextNode != NULL)
{
p = p->nextNode;
q = q->nextNode;
if(p->nextNode==NULL)
{
q->nextNode=NULL;
}
}
x = p->entry;
delete p->nextNode;
}
delete q->nextNode;
}
}
添加节点我有这个方法:
void Deque::appendAtFront(int x)
{
if(full())
{
cout << "FULL" << endl;
exit(0);
}
else
{
p = new DequeNode;
p->entry=x;
if(p == NULL)
{
cout << "Can't insert" << endl;
exit(0);
}
else if(empty())
{
head = tail = p;
p->nextNode=NULL;
}
else
{
p->nextNode=head;
head = p;
}
}
}
我也无法使用 "deque" 图书馆
如何避免泄漏的一般答案是:避免手动分配内存。
例如,您可以使用智能指针,例如std::unique_ptr
而不是调用 new DequeNode
调用 std::make_unique<DequeNode>()
。然后编译器将指出您的代码需要调整的地方,因为您将限制自己,以便您的每个 DequeNode(s) 将仅由 one unique_ptr 在同时。基于您的 serveAtRear:
的弹出函数示例(其中 head
和 DequeNode.next
是 uniuque_ptrs)
if (!head)
{
return;
}
DequeNode* q; //Pointer before p
DequeNode* p; //will be tail eventually
q = head.get(); //q starts at the deque's head
p = q->next.get(); //Pointer for the next node
if(!p) //If there's only one node
{
x = q->entry; //x will receive the value of the removed node to print it afterward
head.reset(); // hear we release DequeNode pointed by head
}
else
{
while(p->next)
{
q = p;
p = q->next.get();
}
x = p->entry;
q->next.reset(); // here we release tail (i.e. p)
}
当然push的实现也需要采用:).
对于我的家庭作业,我需要创建从双端队列添加(追加)和删除(服务)节点的方法。但是,当尝试为最后一个节点提供服务时,我遇到了内存泄漏问题,因为我不知道如何检索系统不再使用的节点。 这是该方法的代码。
void Deque::serveAtRear(int &x)
{
if(empty())
{
cout << "Fila Vazia" << endl;
exit(0);
}
else
{
DequePointer q; //Pointer before p
q = head; //q starts at the deque's head
p = head->nextNode; //Pointer for the next node
if(q->nextNode==NULL) //If there's only one node
{
x = q->entry; //x will receive the value of the removed node to print it afterward
head = tail = NULL;
delete p; //I don't know if I'm using the "delete" right
}
else
{
while(p->nextNode != NULL)
{
p = p->nextNode;
q = q->nextNode;
if(p->nextNode==NULL)
{
q->nextNode=NULL;
}
}
x = p->entry;
delete p->nextNode;
}
delete q->nextNode;
}
}
添加节点我有这个方法:
void Deque::appendAtFront(int x)
{
if(full())
{
cout << "FULL" << endl;
exit(0);
}
else
{
p = new DequeNode;
p->entry=x;
if(p == NULL)
{
cout << "Can't insert" << endl;
exit(0);
}
else if(empty())
{
head = tail = p;
p->nextNode=NULL;
}
else
{
p->nextNode=head;
head = p;
}
}
}
我也无法使用 "deque" 图书馆
如何避免泄漏的一般答案是:避免手动分配内存。
例如,您可以使用智能指针,例如std::unique_ptr
而不是调用 new DequeNode
调用 std::make_unique<DequeNode>()
。然后编译器将指出您的代码需要调整的地方,因为您将限制自己,以便您的每个 DequeNode(s) 将仅由 one unique_ptr 在同时。基于您的 serveAtRear:
head
和 DequeNode.next
是 uniuque_ptrs)
if (!head)
{
return;
}
DequeNode* q; //Pointer before p
DequeNode* p; //will be tail eventually
q = head.get(); //q starts at the deque's head
p = q->next.get(); //Pointer for the next node
if(!p) //If there's only one node
{
x = q->entry; //x will receive the value of the removed node to print it afterward
head.reset(); // hear we release DequeNode pointed by head
}
else
{
while(p->next)
{
q = p;
p = q->next.get();
}
x = p->entry;
q->next.reset(); // here we release tail (i.e. p)
}
当然push的实现也需要采用:).