Exception in thread "main" java.lang.NullPointerException 各位大大帮忙看一下吧

Question

使用netbean运行后，出现如下提示，第65行有问题run:
Enter the width of grid(width>=2): 
Enter the width of grid(width>=2): 20
Enter the height of grid(height>=2): 
Enter the height of grid(height>=2): 20
Enter the number of pixels in each grid square(pixels>=1): 
Enter the number of pixels in each grid square(pixels>=1): 10
Exception in thread "main" java.lang.NullPointerException
        at drawgrid.main(drawgrid.java:65)
/*
 * To change this template, choose Tools | Templates
 * and open the template in the editor.
 *//**
 *
 * @author Administrator
 */import sheffield.*;
import java.util.*;
public class drawgrid { public static void main(String args[] ) {
        int middle2;
        int middle1;
        int midd2;
        int midd1;
        int cross;
        int a;
        int b;
        int c;
        int d;
        int e;
        int f;
EasyReader keyboard = new EasyReader();
int width = keyboard.readInt("Enter the width of grid(width>=2): ");
int height = keyboard.readInt("Enter the height of grid(height>=2): ");
int pix = keyboard.readInt("Enter the number of pixels in each grid square(pixels>=1): ");
EasyGraphics g = new EasyGraphics(width*pix,height*pix);
if (width % 2 == 0)
{
    middle2 = width/2;
}
else
{
 middle2=(width+1)/2;
}
if (height % 2 == 0)
{
 middle1 = height/2;
}
else
{
 middle1=(height+1)/2;
}
        midd2 = middle2 * pix;
        midd1 = middle1 * pix;
int[][] mypoint = new int[width][height];//建立一个整形数据，用来存储每个点的位置 String[][] oldpoint = new String[width][height];//建立一个字符串数组，用于存储x,y 格式的已走过的点
g.moveTo(midd2,middle1 * pix); //移动到中心点
g.plot(midd2,middle1 * pix); //将中心点变黑
int z=0;
        int num=0;
oldpoint[middle2][middle1]=middle2 + "$"+ middle1 ; //将中心点存入走过的点的数组
for (z=0;z<10;z++)
{
String[] crosses = new String[5];//一个可以存5个字符串的
if (z>1){
z=z-1;
}
                if (middle2+1<=height-1 & middle2+1>=0 & middle1<=width-1 &middle1>=0){
                 65行：   if ( oldpoint[middle2+1][middle1].length() == 0 | oldpoint[middle2+1][middle1]==null){                         num = num + 1 ;
                         int temp1=middle2+1;
                        int temp2 = middle1;
                        crosses[num]=temp1+"$"+temp2;
                    }
                }
                if (middle2-1<=height-1 & middle2-1>=0 & middle1<=width-1 &middle1>=0){
                    if (oldpoint[middle2-1][middle1]==null | oldpoint[middle2-1][middle1].length() ==0){
                        num=num+1;
                        int temp1=middle2-1;
                        int temp2=middle1;
                        crosses[num]=temp1+"$"+temp2;
                    }
                }
                if (middle2<=height-1 & middle2>=0 & middle1+1<=width-1 &middle1+1>=0){
                    if (oldpoint[middle2][middle1+1]==null | oldpoint[middle2][middle1+1].length()==0){
                        num=num+1;
                        int temp1=middle2;
                        int temp2=middle1+1;
                        crosses[num]=temp1+"$"+temp2;
                    }
                }
                if (middle2<=height-1 & middle2>=0 & middle1-1<=width-1 &middle1-1>=0){
                    if (oldpoint[middle2][middle1-1]==null | oldpoint[middle2][middle1-1].length()==0){
                        num=num+1;
                        int temp1=middle2;
                        int temp2=middle1-1;
                        crosses[num]=temp1+"$"+temp2;
                    }
                }
if (num>=1){
Random   r=new   Random();
    cross = r.nextInt(num);
                    String temp = new String();
temp = crosses[cross];//获得crosses数组当前随机数下标的字符串值
                    a = temp.indexOf("$");//获得当前字符串中$所在的位置，从最前查
b = temp.lastIndexOf("$");//获得当前字符串中$所在的位置，从最后查
                    e = a-1;
                    f = Integer.parseInt(temp,10);
temp = temp.substring(0,e);
                    c = Integer.parseInt(temp,10);
d = Integer.parseInt(temp.substring(a+1,a+b-1));
                    

oldpoint[c][d]=c + "$"+ d ;
g.lineTo(c*pix,d*pix);
 middle2 = c;
 middle1 = d; }
else
{
break;
} }
}
}

Answer 1

 数组有个 length 属性··没有length()方法吧····

Answer 2

真长，不过确实数组没有length()方法，把所有涉及到length属性的括号去了试试，你把数组的length属性跟字符串的length()方法搞混了

Answer 3

 if ( oldpoint[middle2+1][middle1]==null|| oldpoint[middle2+1][middle1].length() == 0 )
首先判断是不是null 如果是null调用.length()就空指针

Answer 4

null.length()肯定报空指针，检查你的值是否有null