Why does this code always print nothing after run for a while? [duplicate]










1
















This question already has an answer here:



  • Loop doesn't see value changed by other thread without a print statement

    1 answer



I run the main method after a while, it will always print nothing.



But if I uncomment "System.out.println(111);" and "System.out.println(222);",it will run normal i think.



I fetch the dump of 'soldWorker' thread and 'refundWorker' thread, they stop at the if judge.



public class ProducerAndConsumer2 

public static void main(String args)
SyncTikcet2 syncTikcet = new SyncTikcet2(1);

Runnable syncSoldTicket = () -> syncTikcet.soldTicket();
Runnable syncRefundTicket = () -> syncTikcet.refund();

new Thread(syncSoldTicket, "soldWorker").start();
new Thread(syncRefundTicket, "refundWorker").start();



class SyncTikcet2

private Integer ticketNum;

public SyncTikcet2(Integer ticketNum)
this.ticketNum = ticketNum;



public void soldTicket()
while (true)
if (ticketNum > 0)
ticketNum = ticketNum - 1;
System.out.println("sold ticket,the num is" + ticketNum);
else
// System.out.println(111);




public void refund()
while (true)
if (ticketNum == 0)
ticketNum = ticketNum + 1;
System.out.println("refund ticket,the num is" + ticketNum);
else
// System.out.println(222);




public Integer getTicketNum()
return ticketNum;


public void setTicketNum(Integer ticketNum)
this.ticketNum = ticketNum;











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marked as duplicate by Arnaud, khelwood java
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Nov 14 '18 at 8:46


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.






















    1
















    This question already has an answer here:



    • Loop doesn't see value changed by other thread without a print statement

      1 answer



    I run the main method after a while, it will always print nothing.



    But if I uncomment "System.out.println(111);" and "System.out.println(222);",it will run normal i think.



    I fetch the dump of 'soldWorker' thread and 'refundWorker' thread, they stop at the if judge.



    public class ProducerAndConsumer2 

    public static void main(String args)
    SyncTikcet2 syncTikcet = new SyncTikcet2(1);

    Runnable syncSoldTicket = () -> syncTikcet.soldTicket();
    Runnable syncRefundTicket = () -> syncTikcet.refund();

    new Thread(syncSoldTicket, "soldWorker").start();
    new Thread(syncRefundTicket, "refundWorker").start();



    class SyncTikcet2

    private Integer ticketNum;

    public SyncTikcet2(Integer ticketNum)
    this.ticketNum = ticketNum;



    public void soldTicket()
    while (true)
    if (ticketNum > 0)
    ticketNum = ticketNum - 1;
    System.out.println("sold ticket,the num is" + ticketNum);
    else
    // System.out.println(111);




    public void refund()
    while (true)
    if (ticketNum == 0)
    ticketNum = ticketNum + 1;
    System.out.println("refund ticket,the num is" + ticketNum);
    else
    // System.out.println(222);




    public Integer getTicketNum()
    return ticketNum;


    public void setTicketNum(Integer ticketNum)
    this.ticketNum = ticketNum;











    share|improve this question













    marked as duplicate by Arnaud, khelwood java
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    Nov 14 '18 at 8:46


    This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.




















      1












      1








      1









      This question already has an answer here:



      • Loop doesn't see value changed by other thread without a print statement

        1 answer



      I run the main method after a while, it will always print nothing.



      But if I uncomment "System.out.println(111);" and "System.out.println(222);",it will run normal i think.



      I fetch the dump of 'soldWorker' thread and 'refundWorker' thread, they stop at the if judge.



      public class ProducerAndConsumer2 

      public static void main(String args)
      SyncTikcet2 syncTikcet = new SyncTikcet2(1);

      Runnable syncSoldTicket = () -> syncTikcet.soldTicket();
      Runnable syncRefundTicket = () -> syncTikcet.refund();

      new Thread(syncSoldTicket, "soldWorker").start();
      new Thread(syncRefundTicket, "refundWorker").start();



      class SyncTikcet2

      private Integer ticketNum;

      public SyncTikcet2(Integer ticketNum)
      this.ticketNum = ticketNum;



      public void soldTicket()
      while (true)
      if (ticketNum > 0)
      ticketNum = ticketNum - 1;
      System.out.println("sold ticket,the num is" + ticketNum);
      else
      // System.out.println(111);




      public void refund()
      while (true)
      if (ticketNum == 0)
      ticketNum = ticketNum + 1;
      System.out.println("refund ticket,the num is" + ticketNum);
      else
      // System.out.println(222);




      public Integer getTicketNum()
      return ticketNum;


      public void setTicketNum(Integer ticketNum)
      this.ticketNum = ticketNum;











      share|improve this question















      This question already has an answer here:



      • Loop doesn't see value changed by other thread without a print statement

        1 answer



      I run the main method after a while, it will always print nothing.



      But if I uncomment "System.out.println(111);" and "System.out.println(222);",it will run normal i think.



      I fetch the dump of 'soldWorker' thread and 'refundWorker' thread, they stop at the if judge.



      public class ProducerAndConsumer2 

      public static void main(String args)
      SyncTikcet2 syncTikcet = new SyncTikcet2(1);

      Runnable syncSoldTicket = () -> syncTikcet.soldTicket();
      Runnable syncRefundTicket = () -> syncTikcet.refund();

      new Thread(syncSoldTicket, "soldWorker").start();
      new Thread(syncRefundTicket, "refundWorker").start();



      class SyncTikcet2

      private Integer ticketNum;

      public SyncTikcet2(Integer ticketNum)
      this.ticketNum = ticketNum;



      public void soldTicket()
      while (true)
      if (ticketNum > 0)
      ticketNum = ticketNum - 1;
      System.out.println("sold ticket,the num is" + ticketNum);
      else
      // System.out.println(111);




      public void refund()
      while (true)
      if (ticketNum == 0)
      ticketNum = ticketNum + 1;
      System.out.println("refund ticket,the num is" + ticketNum);
      else
      // System.out.println(222);




      public Integer getTicketNum()
      return ticketNum;


      public void setTicketNum(Integer ticketNum)
      this.ticketNum = ticketNum;






      This question already has an answer here:



      • Loop doesn't see value changed by other thread without a print statement

        1 answer







      java multithreading






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked Nov 14 '18 at 8:34









      wr1ttenyu zhaowr1ttenyu zhao

      163




      163




      marked as duplicate by Arnaud, khelwood java
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      Nov 14 '18 at 8:46


      This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.









      marked as duplicate by Arnaud, khelwood java
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          1 Answer
          1






          active

          oldest

          votes


















          3














          If you haven't made the field volatile, there is no reason for a thread to re-read a value it didn't change.



          Try using private volatile int ticketNum



          I suggest using a primitive e.g. int, unless you have to use an Object reference. It's not only faster but it also makes it clear you don't expect the value to be null.






          share|improve this answer























          • thank you very much, you are so great, your answer guide me to find another question link. This question is the same as mine and it's answer more detailed. Thank you very much once again.

            – wr1ttenyu zhao
            Nov 14 '18 at 9:05






          • 1





            @wr1ttenyuzhao note System.out.println has a synchronized memory barrier, however if it's not called you might not see an update.

            – Peter Lawrey
            Nov 14 '18 at 9:14











          • yeah, this is why it will run normal when i uncomment "System.out.println(111);" and "System.out.println(222);", thank your answer, you are a good man, thanks

            – wr1ttenyu zhao
            Nov 14 '18 at 9:59

















          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          3














          If you haven't made the field volatile, there is no reason for a thread to re-read a value it didn't change.



          Try using private volatile int ticketNum



          I suggest using a primitive e.g. int, unless you have to use an Object reference. It's not only faster but it also makes it clear you don't expect the value to be null.






          share|improve this answer























          • thank you very much, you are so great, your answer guide me to find another question link. This question is the same as mine and it's answer more detailed. Thank you very much once again.

            – wr1ttenyu zhao
            Nov 14 '18 at 9:05






          • 1





            @wr1ttenyuzhao note System.out.println has a synchronized memory barrier, however if it's not called you might not see an update.

            – Peter Lawrey
            Nov 14 '18 at 9:14











          • yeah, this is why it will run normal when i uncomment "System.out.println(111);" and "System.out.println(222);", thank your answer, you are a good man, thanks

            – wr1ttenyu zhao
            Nov 14 '18 at 9:59















          3














          If you haven't made the field volatile, there is no reason for a thread to re-read a value it didn't change.



          Try using private volatile int ticketNum



          I suggest using a primitive e.g. int, unless you have to use an Object reference. It's not only faster but it also makes it clear you don't expect the value to be null.






          share|improve this answer























          • thank you very much, you are so great, your answer guide me to find another question link. This question is the same as mine and it's answer more detailed. Thank you very much once again.

            – wr1ttenyu zhao
            Nov 14 '18 at 9:05






          • 1





            @wr1ttenyuzhao note System.out.println has a synchronized memory barrier, however if it's not called you might not see an update.

            – Peter Lawrey
            Nov 14 '18 at 9:14











          • yeah, this is why it will run normal when i uncomment "System.out.println(111);" and "System.out.println(222);", thank your answer, you are a good man, thanks

            – wr1ttenyu zhao
            Nov 14 '18 at 9:59













          3












          3








          3







          If you haven't made the field volatile, there is no reason for a thread to re-read a value it didn't change.



          Try using private volatile int ticketNum



          I suggest using a primitive e.g. int, unless you have to use an Object reference. It's not only faster but it also makes it clear you don't expect the value to be null.






          share|improve this answer













          If you haven't made the field volatile, there is no reason for a thread to re-read a value it didn't change.



          Try using private volatile int ticketNum



          I suggest using a primitive e.g. int, unless you have to use an Object reference. It's not only faster but it also makes it clear you don't expect the value to be null.







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Nov 14 '18 at 8:38









          Peter LawreyPeter Lawrey

          443k56564967




          443k56564967












          • thank you very much, you are so great, your answer guide me to find another question link. This question is the same as mine and it's answer more detailed. Thank you very much once again.

            – wr1ttenyu zhao
            Nov 14 '18 at 9:05






          • 1





            @wr1ttenyuzhao note System.out.println has a synchronized memory barrier, however if it's not called you might not see an update.

            – Peter Lawrey
            Nov 14 '18 at 9:14











          • yeah, this is why it will run normal when i uncomment "System.out.println(111);" and "System.out.println(222);", thank your answer, you are a good man, thanks

            – wr1ttenyu zhao
            Nov 14 '18 at 9:59

















          • thank you very much, you are so great, your answer guide me to find another question link. This question is the same as mine and it's answer more detailed. Thank you very much once again.

            – wr1ttenyu zhao
            Nov 14 '18 at 9:05






          • 1





            @wr1ttenyuzhao note System.out.println has a synchronized memory barrier, however if it's not called you might not see an update.

            – Peter Lawrey
            Nov 14 '18 at 9:14











          • yeah, this is why it will run normal when i uncomment "System.out.println(111);" and "System.out.println(222);", thank your answer, you are a good man, thanks

            – wr1ttenyu zhao
            Nov 14 '18 at 9:59
















          thank you very much, you are so great, your answer guide me to find another question link. This question is the same as mine and it's answer more detailed. Thank you very much once again.

          – wr1ttenyu zhao
          Nov 14 '18 at 9:05





          thank you very much, you are so great, your answer guide me to find another question link. This question is the same as mine and it's answer more detailed. Thank you very much once again.

          – wr1ttenyu zhao
          Nov 14 '18 at 9:05




          1




          1





          @wr1ttenyuzhao note System.out.println has a synchronized memory barrier, however if it's not called you might not see an update.

          – Peter Lawrey
          Nov 14 '18 at 9:14





          @wr1ttenyuzhao note System.out.println has a synchronized memory barrier, however if it's not called you might not see an update.

          – Peter Lawrey
          Nov 14 '18 at 9:14













          yeah, this is why it will run normal when i uncomment "System.out.println(111);" and "System.out.println(222);", thank your answer, you are a good man, thanks

          – wr1ttenyu zhao
          Nov 14 '18 at 9:59





          yeah, this is why it will run normal when i uncomment "System.out.println(111);" and "System.out.println(222);", thank your answer, you are a good man, thanks

          – wr1ttenyu zhao
          Nov 14 '18 at 9:59



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