我正在构建一个程序来询问乘法,我想设置一个计时器来强制该人在给定时间内给出答案:
I'm building a programm to ask multiplication and I want to set up a timer to force the person to give its answer in a given time :
目前,案例 1 可以完成,但 2 不是,我正在考虑一种方法来返回; 从方法内部就像一个Thread或者什么东西,我不知道怎么样
For the moment, case 1 can be done, but 2 not, I was thinking about a way to return; from the method within like a Thread or something, bu I don't know how
所以我遇到问题,如果扫描仪打开,等待输入,如何停止?我已经尝试将它放在一个Thread和 interrupt()中,或者使用 boolean 作为标志,但它没有' t停止扫描仪
So I'm facing a problem, if a Scanner is open, waiting for input, how to stop it ? I've tried putting it in a Thread and interrupt() it or using boolean as flags, but it doesn't stop the Scanner
class Multiplication extends Calcul { Multiplication() { super((nb1, nb2) -> nb1 * nb2); } @Override public String toString() { return getNb1() + "*" + getNb2(); } } abstract class Calcul { private int nb1, nb2; private boolean valid; private boolean inTime = true; private boolean answered = false; private BiFunction<Integer, Integer, Integer> function; Calcul(BiFunction<Integer, Integer, Integer> f) { this.nb1 = new Random().nextInt(11); this.nb2 = new Random().nextInt(11); this.function = f; } void start() { Scanner sc = new Scanner(System.in); System.out.println("What much is " + this + " ?"); Timer timer = new Timer(); timer.schedule(new TimerTask() { @Override public void run() { if (!answered) { inTime = false; } } }, 5 * 1000); int answer = Integer.parseInt(sc.nextLine()); if (inTime) { checkAnswer(answer); timer.cancel(); } } private void checkAnswer(int answer) { System.out.println("You said " + answer); valid = (function.apply(nb1, nb2) == answer) && inTime; answered = true; } int getNb1() { return nb1; } int getNb2() { return nb2; } boolean isValid() { return valid; } public static void main(String[] args) { List<Calcul> l = Arrays.asList(new Multiplication(), new Multiplication(), new Multiplication()); l.forEach(Calcul::start); } }推荐答案
你可以检查 System.in.available()> 0 查看是否有要读取的行。只有在返回true时才调用 sc.nextLine()来实际接收输入。
You can check for System.in.available() > 0 to see if there is a line to read. Only if this returns true call the sc.nextLine() to actually receive the input.
示例:
Scanner sc = new Scanner(System.in); long sTime = System.currentTimeMillis(); while (System.currentTimeMillis() - sTime < 5000) { if (System.in.available() > 0) { System.out.println(sc.nextLine()); } } sc.close();如果有东西需要阅读并再打印出来,这将从控制台读取5秒钟。 注意:实际使用时,你可能会在循环中抛出 sleep ,而不是拥抱许多系统资源。
This reads from the console for 5 seconds if there is something to read and just prints it out again. Note: When actually using this you would probably throw a sleep in the loop to not hug to many system resources.
请注意,这是一个可以工作解决方案: available()往往是一个不可靠的方法,做一些估计可能是错误的。我可能不会在时间要求严格的系统等中依赖它。
Please note that this is a can work solution: available() tends to be an unreliable method that does some estimation and can be in the wrong. I would probably not rely on it in a time-critical system, etc.
此外,为了进一步扩展,这种方法依赖于控制台以的方式工作控制台工作(换句话说:我所知道的所有控制台):只有当用户输入换行符时(例如按下输入),该行实际上被赋予 System.in 处理。否则只有一个字符被输入时, available()将返回true。
Also to further expand, this approach relies on the console to work the way most consoles work (in other words: All consoles that I know of): Only when the user enters a newline (by e.g. pressing enter) the line is actually given to System.in to process. Else available() would already return true when only one character gets typed.
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