Question

尝试创建一个小应用程序来创建随机密码以获得它的乐趣，而我目前正处于试图让我的用户停止应用程序的阶段。然而，循环继续前进并且无法进入我的扫描仪，但由于某种原因，如果我取消使用扫描仪和同步代码块，这段代码将起作用。

无论如何，这段代码出了什么问题？：

public class Application {

public static void main(String[] args) {

    Scanner stopScanner = new Scanner(System.in);

    final PasswordGenerator pG = new PasswordGenerator();

    Thread t1 = new Thread(new Runnable(){
        @Override
        public void run() {
            try {
                pG.passwordGenerator();



            } catch (InterruptedException e) {
                e.printStackTrace();
            }

        }

    });

    t1.run();

    System.out.println("Press enter to stop!");

    stopScanner.nextLine();
    pG.shutDown();
    }
}

class PasswordGenerator{

private volatile static boolean running = true;
protected Scanner scan = new Scanner(System.in);

public void passwordGenerator() throws InterruptedException{
    synchronized(this){

        System.out.println("Select charecter set:");
        System.out.println(" 1: ABCDEF");
        System.out.println(" 2: GHIJKL");
        System.out.println(" 3: MNOPQR");
        System.out.println(" 4: TUVWXYZ");

        String charecters = scan.nextLine();

        System.out.println("Select a password length");

        int number = scan.nextInt();
        scan.nextLine();

        if(number <= 6){
            System.out.println("Number cannot be smaller or equal to six!");
        } else {
            switch(charecters){
            case "1":
                while(running == true){
                    System.out.println("placeholder");
                    Thread.sleep(1000);
                }
                break;
            case "2":

                break;
            case "3":

                break;
            case "4":

                break;
            default:
                System.out.println("No valid set chosen!");
                break;
            } 
        }
    }

}

public void shutDown(){
    running = false;
}
}

Answer 1

您的running是易失性的，访问volatile成员会在包含它的对象上隐式同步。检查一下：http://www.javamex.com/tutorials/synchronization_volatile.shtml

对（volatile）变量的访问就好像它被包含在一个同步的块中，自身同步

因此，调用pG的线程获取passwordGenerator的锁定。它永远不会释放它。因此，main在调用running

时，shutDown无法访问

data want;
  infile 'myfile.csv' dsd firstobs=2 truncover;
  length var1 $20 var2 8 ... varlast 8 ;
  informat var2 yymmdd10.;
  format var2 yymmdd10.;
  input var1 -- varlast; 
run;

问题与您的扫描仪无关

当我按下输入

1 个答案: