为此,我创建了一个名为InputManager
的单例。这个类处理所有的输入阅读内容。我创建了一个名为ask
的方法,它将回调作为参数。在这个方法中,我创建了一个新线程,并在其中等待使用scanner.hasnextInt
的输入。这个类还有CloseInput
方法,它向上面描述的线程发送中断消息。下面是该类的当前实现:
class InputManager{
private Thread thread;
private InputManager(){}
private static InputManager instance;
private static InputManager getInstance(){
if(instance == null){
instance = new InputManager();
}
return instance;
}
/**
* Ask user to type a number.
* @param onSelected When the user has made his choice, this callback will be executed
*/
public static void ask( Consumer<Integer> onSelected){
getInstance().thread = new Thread(() -> {
System.out.println("Type a number:");
Scanner sc = new Scanner(System.in);
int selection = -1;
while (selection == -1) {
if(Thread.currentThread().isInterrupted()){
return;
}
if(sc.hasNextInt()){
selection = sc.nextInt();
onSelected.accept(selection);
} else {
sc.next();
selection = -1;
}
}
});
getInstance().thread.start();
}
/**
* Reset input stream (?)
*/
public static void closeInput(){
try {
getInstance().thread.interrupt();
} catch(NullPointerException e){
// do nothing
}
}
}
这个代码极不可靠。我一会儿就告诉你我的意思。我制作了一个名为Client的玩具类,在Main
中,我用计时器模拟了TimerEnd
消息收入。
class Client {
/**
* Ask user to type a number and send it to the server
*/
void makeRequest(){
InputManager.closeInput();
InputManager.ask((selected) -> {
System.out.println("Sent message: " + selected);
});
}
public static void main(String[] args) {
Client client = new Client();
client.makeRequest();
// Simulate Server messages
Timer timer = new Timer();
timer.scheduleAtFixedRate(new TimerTask() {
@Override
public void run() {
System.out.println("Message received");
client.makeRequest();
}
}, 5000, 5000);
}
}
以下是它的工作原理:
Type a number:
2
Sent message: 2
Message received
Type a number:
3
Sent message: 3
Message received
Type a number: // Here I don't type anything
Message received
Type a number:
Message received
Type a number:
Message received
Type a number: // Here I can send multiple messages on the same "turn"
1
Sent message: 1
2
Message received
我知道这个问题非常长(也许是不必要的),既然你读了它,让我感谢你花时间。
package com.company;
import java.util.*;
import java.util.function.Consumer;
class InputManager{
private Thread thread;
private InputManager(){}
private static InputManager instance;
private static InputManager getInstance(){
if(instance == null){
instance = new InputManager();
}
return instance;
}
/**
* Ask user to type a number.
* @param onSelected When the user has made his choice, this callback will be executed
*/
public static void ask( Consumer<Integer> onSelected){
getInstance().thread = new Thread(() -> {
System.out.println("Type a number:");
Scanner sc = new Scanner(System.in);
int selection = -1;
while (selection == -1) {
if(Thread.currentThread().isInterrupted()){
return;
}
if(sc.hasNextInt()){
selection = sc.nextInt();
onSelected.accept(selection);
} else {
sc.next();
selection = -1;
}
}
});
getInstance().thread.start();
}
/**
* Reset input stream (?)
*/
public static void closeInput(){
try {
getInstance().thread.interrupt();
} catch(NullPointerException e){
// do nothing
}
}
}
class Client {
/**
* Ask user to type a number and send it to the server
*/
void makeRequest(){
InputManager.closeInput();
InputManager.ask((selected) -> {
System.out.println("Sent message: " + selected);
});
}
public static void main(String[] args) {
Client client = new Client();
client.makeRequest();
Timer timer = new Timer();
timer.scheduleAtFixedRate(new TimerTask() {
@Override
public void run() {
System.out.println("Message received: thread interrupted");
client.makeRequest();
}
}, 5000, 5000);
}
}
在我看来,您可以使用3种类型的线程:
所以我将使用2个生产者和1个消费者如下:
import java.util.Scanner;
public class Main {
private static final Scanner SCAN = new Scanner(System.in);
//This is the Scanner's input Producer:
private static class UserInputProducer extends Thread {
private final UserInputConsumer uInConsumer;
public UserInputProducer(final UserInputConsumer uInConsumer) {
this.uInConsumer = uInConsumer;
}
@Override
public void run() {
while (true) {
final int input = SCAN.nextInt();
SCAN.nextLine(); //Ignore the new line character.
uInConsumer.userInput(input); //Fire user input event (for the current user).
}
}
}
//This is the time out event Producer:
private static class TimeOutEventProducer {
private final UserInputConsumer uInConsumer;
private int validReportId = Integer.MIN_VALUE; //IDs starting from Integer.MIN_VALUE and
//going step by step to Integer.MAX_VALUE, which means about 4 billion resets can be done
//to this Producer before an unhandled overflow occurs.
public TimeOutEventProducer(final UserInputConsumer uInConsumer) {
this.uInConsumer = uInConsumer;
}
public synchronized void reset() {
new TimerOnce(this, ++validReportId).start(); //Start a new TimerOnce. Could be javax.swing.Timer with "setRepeats(false)".
}
/*sleepDone(...) is called by ALL TimerOnce objects... So we need an up-to-date id (the
reportId) to verify that the LAST one TimerOnce finished, rather than any other.*/
public synchronized void sleepDone(final int reportId) {
if (reportId == validReportId) //Only the last one timeout is valid...
uInConsumer.timedOut(); //Fire time out event (for the current user).
}
}
//This is just a "Timer" object which blocks for 5 seconds:
private static class TimerOnce extends Thread {
private final TimeOutEventProducer timeout;
private final int reportId;
public TimerOnce(final TimeOutEventProducer timeout,
final int reportId) {
this.timeout = timeout;
this.reportId = reportId;
}
@Override
public void run() {
try { Thread.sleep(5000); } catch (final InterruptedException ie) {} //Wait.
timeout.sleepDone(reportId); //Report that the time elapsed...
}
}
//This is the Consumer:
private static class UserInputConsumer {
private final String[] names;
private int input;
private boolean timedOut, hasInput;
public UserInputConsumer(final String[] names) {
this.names = names;
}
public synchronized int play() {
new UserInputProducer(this).start(); //Start scanning any user's input...
final TimeOutEventProducer timeout = new TimeOutEventProducer(this);
int i = -1;
do {
i = (i + 1) % names.length;
hasInput = false;
timedOut = false;
timeout.reset(); //Start the input wait timer...
System.out.print("User " + names[i] + " enter a number: "); //Clarify who's player is the turn.
while (!hasInput && !timedOut)
try { wait(); } catch (final InterruptedException ie) {} //Wait for user input or timeout.
//Interpret notification event (either user input, either timeout):
if (timedOut)
System.out.println("Sorry, out of time.");
else if (!hasInput)
throw new UnsupportedOperationException("Probably messed with the flags in the while-condition.");
}
while (input != 5); //Here you test the win/loss condition.
//Lets say, for example, the user that enters number '5' wins...
return i; //Return the winner's index.
}
public synchronized void timedOut() {
timedOut = true;
notify();
}
public synchronized void userInput(final int input) {
this.input = input;
hasInput = true;
notify();
}
}
public static void main(final String[] args) {
System.out.print("Enter number of players: ");
final int numPlayers = SCAN.nextInt();
SCAN.nextLine(); //Ignore the new line character.
final String[] names = new String[numPlayers];
for (int i=0; i<names.length; ++i) {
System.out.print("User " + (i+1) + " enter your name: ");
names[i] = SCAN.nextLine();
}
//Start the consumer (which in turn starts the producers) and start the main logic:
System.out.println(names[new UserInputConsumer(names).play()] + " wins!");
}
}
我对编程相当陌生。目前我正在做一个uni项目,用java创建一个基本的文本游戏。我遇到的问题是,如何实现一个不允许用户输入相同名称的业务规则。我把它设置好了,这样扫描仪就可以读取阵列了。我使用的是Java,这是我第一次使用这个论坛,所以我非常感谢大家提供的帮助,并提前感谢大家!:) 并为糟糕的格式道歉,我不知道如何正确发布。
我正在使用Java的扫描仪读取用户输入。如果我只使用nextLine一次,它可以正常工作。对于两个nextLine,第一个不等待用户输入字符串(第二个)。 输出: X:Y:(等待输入) 我的代码 你知道为什么会发生这种事吗?谢谢
我想在我的代码中手动读取一些值。除了最后一个(lambda2)之外,它对所有值都有效。我可以无限地继续键入值,即使它们不是双精度的,也不会发生任何事情。如果我用任何其他值键入其他值,我会收到一条错误消息,以及第一次输入lambda2时的错误消息。我在另一个代码中以类似的方式(最终只创建了一个不同的对象)进行了操作,并且效果很好。
我这里有两个代码块。一个扫描器正确地等待用户输入,另一个则直接通过它并调用,返回。下面是工作的块: 下面是没有的块: 这两个都是单独的类,并且是从另一个类中的main方法调用的。基本上,调用,后者又调用某个Player类的方法...这是非工作代码所在的地方。程序中是否有不适合接受输入的时候?我的印象是,任何时候我需要用户输入,我都可以使用这种方法。谢谢!
对于作业,我必须编写以下代码: 当我尝试编译它时,它在命令提示符下给了我3个错误,说“无法解析符号,符号:类扫描仪,位置:类单词,扫描仪用户输入=新扫描仪(System.in)”。我不确定错误在哪里。我应该使用BufferedReader作为输入吗?
问题内容: 我正在使用Java的扫描仪读取用户输入。如果我仅使用nextLine一次,那就可以了。使用两个nextLine时,第一个不会等待用户输入字符串(第二个会输入)。 输出: X:Y :(等待输入) 我的密码 任何想法为什么会发生这种情况?谢谢 问题答案: 您可能像以前那样调用方法。这样的程序是这样的: 演示您所看到的行为。 问题是不消耗,因此下一个调用将消耗它,然后等待读取的输入。 您需要