游戏的生命arrayindexouttofbounds

Question

我在玩康威的人生游戏。我很确定我快完成了，但是当我运行它时，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

我假设当方法检查数组边缘的邻居时，那里什么都没有，所以它会死。我只是不知道该怎么做才不会发生这种事。有人有什么想法吗?下面的代码。

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。

Answer 1

当r为0时，此为无效:old[r - 1][c]。
这样你就得到了你发布的异常。

我建议你这样简化。

    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;
    }

Answer 2

正如我所看到的，你基本上有两个选择:

1. 应用有限边界，也就是说，对于第一列和最后一列和行中的单元格，在计算"活的"邻居的数量时，您实现了额外的检查。

2. 应用周期边界，即最左边列的单元格和最右边列的单元格被认为是邻居。在模块化算法的帮助下，这些单元不需要与其他单元分开处理。