我想通過在C++中使用鏈表來實現優先級隊列。但是,當我運行該程序時,它會在「priorityQLinkedList :: dequeue()」方法內觸發一個斷點。有人可以告訴我們爲什麼會出現這種情況,並告訴我如何解決這個問題?我得到一個斷點,我不知道爲什麼
代碼:
#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;
}
糾正我,如果我錯了,但如果你刪除節點出隊,然後返回一個指向相同的指針,這不會給調用者帶來問題,然後誰會回來一個「死」指針? –
@TimBiegeleisen這就是發生了什麼,你應該把它寫爲答案 –
@AndersK。我嘗試了下面的答案。我沒有看到處理這個問題的好方法,但我想''dequeue()'應該是刪除目標節點。我的答案是複製'NODE'並返回給調用者。但是那時調用者必須在某個時候調用'delete'。 –