我想通过在C++中使用链表来实现优先级队列。但是,当我运行该程序时,它会在“priorityQLinkedList :: dequeue()”方法内触发一个断点。有人可以告诉我们为什么会出现这种情况,并告诉我如何解决这个问题?我得到一个断点,我不知道为什么
代码:
#include <iostream>
#include <cstring>
#include <iomanip>
using namespace std;
struct DAT
{
int id;
char fullname[50];
double savings;
};
struct NODE
{
DAT data;
NODE *N;
NODE *P;
NODE(const int i, const char *f, const double s)
{
data.id = i;
strcpy_s(data.fullname, f);
data.savings = s;
N = NULL;
P = NULL;
}
};
class priorityQLinkedList
{
private:
NODE *front;
NODE *back;
public:
priorityQLinkedList() { front = NULL; back = NULL; }
~priorityQLinkedList() { destroyList(); }
void enqueue(NODE *);
NODE* dequeue();
void destroyList();
};
void priorityQLinkedList::enqueue(NODE *n)
{
if (front == NULL) {
front = n;
back = n;
}
else {
NODE *temp = front;
if (n->data.id > temp->data.id)
{
front->P = n;
n->N = front;
front = n;
}
else
{
//search for the posistion for the new node.
while (n->data.id < temp->data.id)
{
if (temp->N == NULL) {
break;
}
temp = temp->N;
}
//New node id's smallest then all others
if (temp->N == NULL && n->data.id < temp->data.id)
{
back->N = n;
n->P = back;
back = n;
}
//New node id's is in the medium range.
else {
temp->P->N = n;
n->P = temp->P;
n->N = temp;
temp->P = n;
}
}
}
}
NODE* priorityQLinkedList::dequeue()
{
NODE *temp;
//no nodes
if (back == NULL) {
return NULL;
}
//there is only one node
else if (back->P == NULL) {
NODE *temp2 = back;
temp = temp2;
front = NULL;
back = NULL;
delete temp2;
return temp;
}
//there are more than one node
else {
NODE *temp2 = back;
temp = temp2;
back = back->P;
back->N = NULL;
delete temp2;
return temp;
}
}
void priorityQLinkedList::destroyList()
{
while (front != NULL) {
NODE *temp = front;
front = front->N;
delete temp;
}
}
void disp(NODE *m) {
if (m == NULL) {
cout << "\nQueue is Empty!!!" << endl;
}
else {
cout << "\nID No. : " << m->data.id;
cout << "\nFull Name : " << m->data.fullname;
cout << "\nSalary : " << setprecision(15) << m->data.savings << endl;
}
}
int main() {
priorityQLinkedList *Queue = new priorityQLinkedList();
NODE No1(101, "Qasim Imtiaz", 567000.0000);
NODE No2(102, "Hamad Ahmed", 360200.0000);
NODE No3(103, "Fahad Ahmed", 726000.0000);
NODE No4(104, "Usmaan Arif", 689000.0000);
Queue->enqueue(&No4);
Queue->enqueue(&No3);
Queue->enqueue(&No1);
Queue->enqueue(&No2);
disp(Queue->dequeue());
disp(Queue->dequeue());
disp(Queue->dequeue());
disp(Queue->dequeue());
disp(Queue->dequeue());
delete Queue;
return 0;
}
纠正我,如果我错了,但如果你删除节点出队,然后返回一个指向相同的指针,这不会给调用者带来问题,然后谁会回来一个“死”指针? –
@TimBiegeleisen这就是发生了什么,你应该把它写为答案 –
@AndersK。我尝试了下面的答案。我没有看到处理这个问题的好方法,但我想''dequeue()'应该是删除目标节点。我的答案是复制'NODE'并返回给调用者。但是那时调用者必须在某个时候调用'delete'。 –