java - Java 中的内存问题/错误

Question

下面是我用 Java 制作的冒险游戏的方法:

public static void characterDisperse(){

    int wRand = (int)(Math.random())*wModifiers.length;
    Item[] inv = new Item[1];
    inv[0]=new Item(wModifiers[wRand] + " " + wNames[(int)(Math.random()*wNames.length)],(int)(Math.random()*wModifiers.length)*2,1);
    for(int a=0;a<10;a++){
        for(int b=0;b<10;b++){
            if(Math.random()>0.5){ //density of characters
                charTest[a][b] = new Character(names[(int)(Math.random()*names.length)],(int)(Math.random()*5),(int)(Math.random()*3),(int)(Math.random()*15+5)*10,2,inv); 

                map[a][b].chars[1] = charTest[a][b];
                System.out.println(map[a][b].chars[1]);

            }
            else{
                map[a][b].passable = false;
            }


        }
    }
    for(int d=0;d<10;d++){
        for(int e=0;e<10;e++){
            System.out.println(map[d][e].chars[1]);
        }
    }

}

我遇到的问题是，在第一个双“for”循环中，打印的每个字符都是具有各种不同属性的不同字符。但是，当第二个双“for”循环运行并打印出我刚刚创建的所有字符时，它会打印出所有相同的字符，如下所示:

NAME: Moriah    RACE: 4 ALIGNMENT: 0    HEALTH: 130 SKILL: 2    INVENTORY: 1
NAME: Marge RACE: 1 ALIGNMENT: 0    HEALTH: 160 SKILL: 2    INVENTORY: 1
NAME: Faith RACE: 0 ALIGNMENT: 1    HEALTH: 50  SKILL: 2    INVENTORY: 1
NAME: Morton    RACE: 3 ALIGNMENT: 2    HEALTH: 60  SKILL: 2    INVENTORY: 1
NAME: Sherwood  RACE: 1 ALIGNMENT: 2    HEALTH: 50  SKILL: 2    INVENTORY: 1
NAME: Ezequiel  RACE: 2 ALIGNMENT: 1    HEALTH: 150 SKILL: 2    INVENTORY: 1
NAME: Herschel  RACE: 2 ALIGNMENT: 2    HEALTH: 70  SKILL: 2    INVENTORY: 1
NAME: Lester    RACE: 3 ALIGNMENT: 2    HEALTH: 80  SKILL: 2    INVENTORY: 1
NAME: Corinna   RACE: 2 ALIGNMENT: 0    HEALTH: 190 SKILL: 2    INVENTORY: 1

FIRST FOR LOOP ENDS HERE
SECOND FOR LOOP BEGINS

NAME: Corinna   RACE: 2 ALIGNMENT: 0    HEALTH: 190 SKILL: 2    INVENTORY: 1
NAME: Corinna   RACE: 2 ALIGNMENT: 0    HEALTH: 190 SKILL: 2    INVENTORY: 1
NAME: Corinna   RACE: 2 ALIGNMENT: 0    HEALTH: 190 SKILL: 2    INVENTORY: 1
NAME: Corinna   RACE: 2 ALIGNMENT: 0    HEALTH: 190 SKILL: 2    INVENTORY: 1
NAME: Corinna   RACE: 2 ALIGNMENT: 0    HEALTH: 190 SKILL: 2    INVENTORY: 1
NAME: Corinna   RACE: 2 ALIGNMENT: 0    HEALTH: 190 SKILL: 2    INVENTORY: 1
NAME: Corinna   RACE: 2 ALIGNMENT: 0    HEALTH: 190 SKILL: 2    INVENTORY: 1
NAME: Corinna   RACE: 2 ALIGNMENT: 0    HEALTH: 190 SKILL: 2    INVENTORY: 1

我相信这是某种内存问题，我将map[a][b](位置对象)处的字符设置为charTest[a][b]中的字符。 charTest 已在代码的前面实例化。

map[][]是一个10x10的Location对象数组，charTest[][]是一个10x10的Character对象数组:

位置.java:

import java.awt.*;
public class Location{

Character[] chars;
Feature[] features;
boolean passable;

public Location(Character[] chars, Feature[] features, boolean passable){
    this.chars = chars;
    this.features = features;
    this.passable = passable;
}

public String toString(){
    return "test";
}

public boolean containsChars(Location l){
    if(l.chars.length>0){
        return true;
    }
    else{
        return false;
    }
}

}

字符.java:

public class Character{
String name;  
int race;
int alignment;

int health;

int skill;

Item[] inventory;

public Character(String name,int race,int alignment,int health,int skill,Item[] inventory){
    this.name = name;
    this.race = race;
    this.alignment = alignment;
    this.health = health;
    this.skill = skill;
    this.inventory = inventory;
}


public String toString(){
    String printOut = "NAME: " + name + "\tRACE: " + race + "\tALIGNMENT: " + alignment + "\tHEALTH: " + health + "\tSKILL: " + skill + "\tINVENTORY: " + "1";
    return printOut;
}
}

Answer 1

在此行中，您必须将末尾的 inv 更改为 new Item[1]

charTest[a][b] = new Character(names[(int)(Math.random()*names.length)],(int)(Math.random()*5),(int)(Math.random()*3),(int)(Math.random()*15+5)*10,2,inv); 

所以这一行可以完成这项工作:

charTest[a][b] = new Character(names[(int)(Math.random()*names.length)],(int)(Math.random()*5),(int)(Math.random()*3),(int)(Math.random()*15+5)*10,2,new Item[1]);

原因是，当您使用此 Item[] inv = new Item[1]; 创建数组时，对此数组的引用将分配给变量 inv。然后，如果您将值 inv 分配给另一个变量，则您复制引用。

这意味着在您的情况下，所有 Character 对象都具有相同的对象。

<小时/>

您显然也有同样的问题

map[a][b].chars[1]

当您创建map[a][b]时，您将相同的chars数组分配给map中的所有字段。

然后，每次创建新的 Character 时，都将其放入同一个 chars 数组中，该数组替换旧的并打印它。这就是为什么您在生成新字符时会看到新字符，这就是为什么所有 map 变量中的最后一个字符都是相同的。

<小时/>

当您为 map 生成初始数据时，您必须为每个位置创建新数组，如下所示:

for(int a=0;a<10;a++){
        for(int b=0;b<10;b++){
            map[a][b] = new Location(new Character[5], new Feature[5], true);
        }
}

PS:我强烈建议使用 ArrayList，而不是数组。