Java设计原则1—城堡游戏之消除代码复制

首先看一下城堡游戏的源码

package castle;

public class Room {
public String description;
public Room northExit;
public Room southExit;
public Room eastExit;
public Room westExit;

public Room(String description)
{
this.description = description;
}

public void setExits(Room north, Room east, Room south, Room west)
{
if(north != null)
northExit = north;
if(east != null)
eastExit = east;
if(south != null)
southExit = south;
if(west != null)
westExit = west;
}

@Override
public String toString()
{
return description;
}
}

华丽分割线————————————————————————————————–

package castle;

import java.util.Scanner;

public class Game {
private Room currentRoom;

public Game()
{
createRooms();
}

private void createRooms()
{
Room outside, lobby, pub, study, bedroom;

//  制造房间
outside = new Room("城堡外");
lobby = new Room("大堂");
pub = new Room("小酒吧");
study = new Room("书房");
bedroom = new Room("卧室");

//  初始化房间的出口
outside.setExits(null, lobby, study, pub);
lobby.setExits(null, null, null, outside);
pub.setExits(null, outside, null, null);
study.setExits(outside, bedroom, null, null);
bedroom.setExits(null, null, null, study);

currentRoom = outside;  //  从城堡门外开始
}

private void printWelcome() {
System.out.println();
System.out.println("欢迎来到城堡！");
System.out.println("这是一个超级无聊的游戏。");
System.out.println("如果需要帮助，请输入 'help' 。");
System.out.println();
System.out.println("现在你在" + currentRoom);
System.out.print("出口有：");
if(currentRoom.northExit != null)
System.out.print("north ");
if(currentRoom.eastExit != null)
System.out.print("east ");
if(currentRoom.southExit != null)
System.out.print("south ");
if(currentRoom.westExit != null)
System.out.print("west ");
System.out.println();
}

// 以下为用户命令

private void printHelp()
{
System.out.print("迷路了吗？你可以做的命令有：go bye help");
System.out.println("如：\tgo east");
}

private void goRoom(String direction)
{
Room nextRoom = null;
if(direction.equals("north")) {
nextRoom = currentRoom.northExit;
}
if(direction.equals("east")) {
nextRoom = currentRoom.eastExit;
}
if(direction.equals("south")) {
nextRoom = currentRoom.southExit;
}
if(direction.equals("west")) {
nextRoom = currentRoom.westExit;
}

if (nextRoom == null) {
System.out.println("那里没有门！");
}
else {
currentRoom = nextRoom;
System.out.println("你在" + currentRoom);
System.out.print("出口有: ");
if(currentRoom.northExit != null)
System.out.print("north ");
if(currentRoom.eastExit != null)
System.out.print("east ");
if(currentRoom.southExit != null)
System.out.print("south ");
if(currentRoom.westExit != null)
System.out.print("west ");
System.out.println();
}
}

public static void main(String[] args) {
Scanner in = new Scanner(System.in);
Game game = new Game();
game.printWelcome();

while ( true ) {
String line = in.nextLine();
String[] words = line.split(" ");
if ( words[0].equals("help") ) {
game.printHelp();
} else if (words[0].equals("go") ) {
game.goRoom(words[1]);
} else if ( words[0].equals("bye") ) {
break;
}
}

System.out.println("感谢您的光临。再见！");
in.close();
}

}

以上为城堡游戏的源码。是否发现了一些设计问题？

首先，会发现下面这段代码在Game类中出现了两次，无疑这会使代码看起来并不优雅，重要的是：在程序中存在相似甚至相同的代码块，是非常低级的代码质量问题。

System.out.println("现在你在" + currentRoom);
System.out.print("出口有：");
if(currentRoom.northExit != null)
System.out.print("north ");
if(currentRoom.eastExit != null)
System.out.print("east ");
if(currentRoom.southExit != null)
System.out.print("south ");
if(currentRoom.westExit != null)
System.out.print("west ");

代码复制存在的问题是，如果需要修改一个副本，那么就必须同时修改所有其他的副本，否则就 存在不一致的问题。这增加了维护程序员的工作量，而且存在造成错误的潜在危险。很可能发 生的一种情况是，维护程序员看到一个副本被修改好了，就以为所有要修改的地方都已经改好 了。因为没有任何明显迹象可以表明另外还有一份一样的副本代码存在，所以很可能会遗漏还 没被修改的地方。

我们从消除代码复制开始。消除代码复制的两个基本手段，就是函数和父类。

这里用函数来解决城堡游戏中重复代码的问题：

 public void showPrompt() {
System.out.println("现在你在" + currentRoom);
System.out.print("出口有：");
if(currentRoom.northExit != null)
System.out.print("north ");
if(currentRoom.eastExit != null)
System.out.print("east ");
if(currentRoom.southExit != null)
System.out.print("south ");
if(currentRoom.westExit != null)
System.out.print("west ");
}

删除重复的代码，在重复的地方调用showPrompt() 函数即可。