我正在做Conway的生活游戏。我很确定我已接近完成,但当我运行它时,我得到Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: -1
at game.of.life.GameOfLife.generation(GameOfLife.java:77)
at game.of.life.GameOfLife.main(GameOfLife.java:32)
Java Result: 1
我假设当检查阵列边缘的邻居的方法时,那里什么都没有,所以它就会死亡或什么的。我只是不知道如何做到这一点并没有发生。有人有想法吗?代码如下。
package game.of.life;
import java.util.Scanner;
public class GameOfLife {
static boolean[][] current = new boolean[10][10];
static boolean[][] old = new boolean[10][10];
static int population = 10;
public static void main(String[] args) {
String a = " @ ";
String b = " ' ";
int choice = 9;
int gencount = 0;
Scanner input = new Scanner(System.in);
System.out.print("Choose population density. i.e. 10 = 10%: ");
population = input.nextInt();
populate();
copy();
for(int r = 0; r < current.length; r++){
for(int c = 0; c < current[r].length; c++){
if(current[r][c] == true){
System.out.print(a);
}
else
System.out.print(b);
}
System.out.println();
}
System.out.print("Generation " + gencount + ".");
while(choice != 0){
System.out.print("Make a selection: 1 - Advance Generation 0 - Exit");
choice = input.nextInt();
if(choice == 1){
generation();
for(int r = 0; r < current.length; r++){
for(int c = 0; c < current[r].length; c++){
if(current[r][c] == true){
System.out.print(a);
}
else
System.out.print(b);
}
System.out.println();
}
copy();
gencount += 1;
System.out.println("Generation" + gencount + ".");
}
}
}
private static void generation(){
for(int r = 0; r < old.length; r++){
for(int c = 0; c < old[r].length; c++){
if (old[r][c] == true){
int neighbors = 0;
if(old[r + 1][c] == true)
neighbors += 1;
if(old[r - 1][c] == true)
neighbors += 1;
if(old[r][c + 1] == true)
neighbors += 1;
if(old[r][c - 1] == true)
neighbors += 1;
if(old[r + 1][c + 1] == true)
neighbors += 1;
if(old[r + 1][c - 1] == true)
neighbors += 1;
if(old[r - 1][c - 1] == true)
neighbors += 1;
if(old[r - 1][c + 1] == true)
neighbors += 1;
if(neighbors != 3 || neighbors != 2)
current[r][c] = false;
}
else if(old[r][c] == false){
int neighbors = 0;
if(old[r + 1][c] == true)
neighbors += 1;
if(old[r - 1][c] == true)
neighbors += 1;
if(old[r][c + 1] == true)
neighbors += 1;
if(old[r][c - 1] == true)
neighbors += 1;
if(old[r + 1][c + 1] == true)
neighbors += 1;
if(old[r + 1][c - 1] == true)
neighbors += 1;
if(old[r - 1][c - 1] == true)
neighbors += 1;
if(old[r - 1][c + 1] == true)
neighbors += 1;
if(neighbors == 3)
current[r][c] = true;
}
}
}
}
private static void populate(){
for(int r = 0; r < current.length; r++){
for(int c = 0; c < current[r].length; c++){
int q = (int)(Math.random() * 100);
if(q < population){
current[r][c] = true;
}
else{
current[r][c] = false;
}
}
}
}
private static void copy(){
for(int r = 0; r < old.length; r++){
for(int c = 0; c < old[r].length; c++)
old[r][c] = current[r][c];
}
}
}
如果有人能帮助我,我将不胜感激。
答案 0 :(得分:3)
当r
为0
时,此项无效:old[r - 1][c]
这样你就得到了你发布的例外。
我建议你这样简化它。
boolean isValidPosition(int r, int c){
return
0 <= r && r < N &&
0 <= c && c < M;
}
int getNeighboursCount(boolean[][] old, int r, int c){
int neighbors = 0;
for (int i=-1; i<=1; i++){
for (int j=-1; j<=1; j++){
if (i!=0 || j!=0){
if (isValidPosition(r + i, c + j)){
if(old[r + i][c + j])
{
neighbors++;
}
}
}
}
}
return neighbors;
}
答案 1 :(得分:1)
我可以看到,你基本上有两个选择:
应用有限边界,即对于第一列和最后一列和第一行中的单元格,在计算“生活”邻居的数量时执行额外的检查。
应用周期性边界,即最左侧列上的单元格和最右侧列上的单元格被视为邻居。在模块化算术的帮助下,这些单元格不需要与其他单元格分开处理。