gpt4 book ai didi

java - 在 ArrayLists 中使用命令模式进行撤消和重做

转载 作者:塔克拉玛干 更新时间:2023-11-03 04:32:13 25 4
gpt4 key购买 nike

所以我有一个程序,您可以在其中登录并在 friends 数组列表中添加/删除 friend 。我也可以喜欢某个东西,那个东西将被存储到 likes arraylist 中。我被要求为我执行的任何操作设置撤消和重做选项。

所以我想加apple为好友。之后,当我选择撤消选项时,我可以撤消该操作,这样 apple 就不会成为我的 friend 。当输入是我输入要存储到 friends 数组列表中的任何名称或单词时,我如何使用 Command Pattern 来解决这个问题?

我做了一些研究,发现使用命令模式可能是我最好的选择,因为这必须在我已有的 Facebook 类下完成。我假设我将不得不使用两个不同的堆栈,但我有点迷失了主题。

我决定添加我已有的部分内容,以便在我需要做什么以及我的程序做什么方面获得更多帮助。

在驱动程序中

Facebook facebook1 = new Facebook();

if (userInput == 6)
{
System.out.println("Login");
String operand1 = getOperand("What is the Username? ");
String operand2 = getOperand("What is the Password? ");
System.out.println("Enter a friend to be added. ");
String operand3 = getOperand("What is the Username? ");
facebook1.friend(operand3);
}

if (userInput == 7)
{
System.out.println("Login");
String operand1 = getOperand("What is the Username? ");
String operand2 = getOperand("What is the Password? ");
System.out.println("Enter a friend to be removed. ");
String operand3 = getOperand("What is the Username? ");
facebook1.defriend(operand3);
}
if (userInput == 12)
{
System.out.println("Login");
String operand1 = getOperand("What is the Password? ");
facebook1.undo();
}

if (userInput == 13)
{
System.out.println("Login");
String operand1 = getOperand("What is the Password? ");
facebook1.redo();
}

在 Facebook 类里面

ArrayList<FacebookUser> recommendedFriends = new ArrayList<FacebookUser>();

void friend(String newFriend)
{
boolean positiveChecker = false;

for (int i = 0; i < recommendedFriends.size(); i++)
{

if (recommendedFriends.get(i).toString().equalsIgnoreCase(newFriend))
{
System.out.println("Error: This friend already exists.");
positiveChecker = true;
}

}
if (positiveChecker == false)
{
FacebookUser friend = new FacebookUser(newFriend, newFriend );
recommendedFriends.add(friend);
System.out.println(friend + " is now your friend.");
}
positiveChecker = false;
}

void defriend(String formerFriend)
{
boolean positiveChecker = false;

for (int i = 0; i < recommendedFriends.size(); i++)
{

if (recommendedFriends.get(i).toString().equalsIgnoreCase(formerFriend))
{
recommendedFriends.remove(i);
System.out.println(formerFriend + " has been removed from your friends list.");
positiveChecker = true;
}
if (recommendedFriends.size() == (i + 1) && recommendedFriends.get(i).toString() != formerFriend
&& positiveChecker == false)
{
System.out.println("Error: There is no friend with this username.");

}

}
positiveChecker = false;
}

public interface Command
{
public void undo();
public void redo();
}

最佳答案

当你撤消 2 件事然后做一个全新的 Action 时,你需要“忘记”“重做历史”并用新命令替换它,对吧?

例如……

  1. 加好友吉姆
  2. 添加好友Bill
  3. 添加好友 Jill
  4. 删除吉姆
  5. 撤销
  6. 撤销

State 应该是“Jim”和“Bill”。

所以你真的只需要一个列表和一个指向当前“命令”的指针,例如......

// Note: NOT thread safe!
public class CommandStack {
private List<Command> commands = Collections.emptyList();
private int nextPointer = 0;

public void doCommand(Command command) {
List<Command> newList = new ArrayList<>(nextPointer + 1)

for(int k = 0; k < nextPointer; k++) {
newList.add(commands.get(k));
}

newList.add(command);

commands = newList;
nextPointer++;

// Do the command here, or return it to whatever called this to be done, or maybe it has already been done by now or something
// (I can only guess on what your code currently looks like...)
command.execute();
}

public boolean canUndo() {
return nextPointer > 0;
}

public void undo() {
if(canUndo()) {
nextPointer--;
Command commandToUndo = commands.get(nextPointer);
// Undo the command, or return it to whatever called this to be undone, or something
command.undo();
} else {
throw new IllegalStateExcpetion("Cannot undo");
}
}

public boolean canRedo() {
return nextPointer < commands.size();
}

public void redo() {
if(canRedo()) {
commandToDo = commands.get(nextPointer);
nextPointer++;
// Do the command, or return it to whatever called this to be re-done, or something
commandToDo.execute();
} else {
throw new IllegalStateException("Cannot redo");
}
}
}

如果我有...

interface Command { /* execute / undo etc */ }

public class AddFriendCommand implements Command {
private String friendName;

// ... other fields, constructor / getters etc ...

public void execute() {
// Actually do it...
System.out.println("Added friend " + name);
}

public void undo() {
// Undo it...
System.out.println("Removed friend " + name);
}
}

public class RemoveFriendCommand implements Command {
private String friendName;

// ... other fields, constructor / getters etc ...

public void execute() {
// Actually do it, maybe throw exception if friend does not exist?
// (that would have to be a runtime exception unless you want the interface's method to throw stuff);
System.out.println("Removed friend " + name);
}

public void undo() {
// Undo it...
System.out.println("Added friend " + name);
}
}

您可以使用...重复上面的序列

CommandStack stack = new CommandStack();

stack.doCommand(new AddFriendCommand("Jim"));
stack.doCommand(new AddFriendCommand("Bill"));
stack.doCommand(new AddFriendCommand("Jill"));
stack.doCommand(new RemoveFreindCommand("Jim"));

stack.undo();
stack.undo();

如果您现在执行了一个新命令(通过 doCommand),它会忘记您曾经添加过“Jill”或删除过“Jim”,而是现在会记住新命令和未撤消的其余命令历史记录。

希望这对您有所帮助。

关于java - 在 ArrayLists 中使用命令模式进行撤消和重做,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/29660504/

25 4 0
Copyright 2021 - 2024 cfsdn All Rights Reserved 蜀ICP备2022000587号
广告合作:1813099741@qq.com 6ren.com