调试用类

Question

这可能是一个很长的镜头，但我需要帮助弄清楚为什么返回语句在它不应该发送空指针时。我的代码是一个可以遍历的二维链表。如果向下移动然后再向上移动会崩溃。 我的主类将调用下一个节点

if((y-dist)>0){ //if not moved out of bounds
    for(int c = 0; c < dist; c++) {//for each step
        node.draw(); //mark current nod as visited
        if(ptr->getup() != NULL){ //another out of bounds check
            ptr = ptr->getup(); //gets the node above it, but returns null
        }
    }
}

我的节点类

#ifndef GRIDNODE_H
#define GRIDNODE_H
#include <cstddef>
#include "gridnode.h"
#include "grid.h"
#include "basenode.h"
class gridnode : public basenode{
public:
    gridnode *l, *r, *u, *d;
    gridnode(){
        l = NULL;
        r = NULL;
        u = NULL;
        d = NULL;
    }
    void draw(){mark = pen;}
}
#endinf

和由四联链表制作的50x50的网格

#include <iostream>
#include "grid.h"
#include "gridnode.h"
#include "basenode.h"
grid::grid(){
gridnode *r1, *r2, *p;
r1 = new gridnode;
head = r1;
for (int i = 2; i <= 50; i++ ){
    r2 = new gridnode;
    r2->l = r1;
    r1->r = r2;
    r1 = r2;
}
r1 = head;
for (int i = 2; i <= 50; i++){
    r2 = new gridnode;
    r2->u = r1;
    r1->d = r2;
    r1 = r2;
    for (int j = 2; j <= 50; j++){
        p = new gridnode;
        p->l = r2;
        r2->r = p;
        p->u = r2->u->r;
        p->u->d = p;
        r2 = p;
    }
}
sethead(head);
}