我昨天在这里提出了一个类似的问题，并纠正了其中的一些问题，但主要问题仍然存在。我将Position对象入队和出队到一个Position队列中。当我入队两个不同的Position对象并将它们都出队时，返回的两个Position对象具有与第二个放入的对象相同的值。当我在enqueue函数内部检查已经入队的值时，它们是正确的。我不明白为什么这样做不起作用，因为我已经解决了逻辑问题，并从一本书中直接使用了出队算法。Position类有三个基于数组的堆栈作为私有成员。
struct Posnode
{
Position *pos;
Posnode *next;
};
class Position
{
private:
Posnode *front,*back,header; //front = back = &header;
Pegs A,B,C;
Position::Position(int num): A(num), B(num), C(num)
{
front = back = &header;
A.assignpeg(num);//assigning 1 to n to each Peg
B.assignpeg(num);
C.assignpeg(num);
}
#include "Pegs.h"
#include "Position.h"
int main ()
{
Position pos(4), intial(3),p,m,n;
intial.geta();//poping Peg A stack
pos.getc();//poping Peg c stack
p.enqueue(intial);
p.enqueue(pos);
p.dequeue(m);//position 'pos' is returned rather than intial
p.dequeue(n);//position 'pos' is returned
cin.get();
return 0;
}
void Position::dequeue(Position&)
{
Position p;
Posnode *ptr=front->next;//front points to an empty node wi
p = *ptr->pos;//assigning the position in ptr to p
front->next = ptr->next;
if (back == ptr) {//if im at the end of the queue
back = front;
}
delete ptr;
return ;
}
void Position::enqueue(Position n)
{
Posnode *ptr = new Posnode;
ptr-> pos = &n;//copying Position n calls Pegs operator =
back->next = ptr;//values that are enqueued check back ok
back = ptr;
return;
}
Pegs& Pegs::operator=(const Pegs & ori)//Pegs copy contructor
{
top=-1;
disks = ori.disks;
peg = new int[disks];
int element=0,g=-1,count=0;
while (count <= ori.top)//copying elements if there are any in ori
{
++count;
element=ori.peg[++g];
push(element);
}
return *this;
}