Why does the below code increment value with 2 rather than 1?(New to Java)

Question

In the below code after a thread runs the increment method it prints value 2 to the console.Shouldn't the value be 1 since the method increments it with 1?

class TestSync implements Runnable {
private int balance;

public void run() {

    for(int i = 0; i < 50; i++){

        increment();
        System.out.println(Thread.currentThread().getName() + " balance after increment is " +balance);
    }

}

private synchronized void increment() {

    int i = balance;
    balance = i + 1;
    // System.out.println(balance);
}

}

public class TestSyncTest {

public static void main(String[] args) {

    TestSync job = new TestSync();
    Thread a = new Thread(job);
    Thread b = new Thread(job);
    a.setName("Thread A");
    b.setName("Thread B");

    a.start();
    b.start();

}

}

What if both threads increment the value before any of them gets to print out the value? — tkausl, Oct 28 '17 at 08:00
Try commenting out the existing `System.out.println` and uncommenting the one you now have commented out. — Mick Mnemonic, Oct 28 '17 at 08:07

score 0 · Answer 1 · answered Oct 28 '17 at 08:07

0

Because your code is in the loop .And balance is global data.

In the first loop ,you balance's value is 1 .

for (int i = 0; i < 50; i++) {

      increment();
      System.out.println(Thread.currentThread().getName() + " balance after increment is " + balance);
}

OUTPUT

//          i       balance
//  first   0          1
//  second  1          2
//  third   2          3
//  ...

answered Oct 28 '17 at 08:07

KeLiuyue

8,149
4
25
42

I understand the logic you're trying to explain.The print statement outputs the values starting with 2 rather than 1,that's what is confusing me. – Maaz Shaikh Nov 02 '17 at 16:31
You can debug for it .Make it easier for you to understand – KeLiuyue Nov 03 '17 at 00:21

score 0 · Answer 2 · answered Oct 28 '17 at 08:09

0

Your increment method is synchronized but the balance variable is shared by the threads. After the first thread calls increment, before it can print balance, the other thread can call increment. Put the print statement inside increment method.

answered Oct 28 '17 at 08:09

ssc327

690
1
9
19

score 0 · Answer 3 · answered Oct 28 '17 at 08:09

Both threads share the same object. The only synchronized method increments i's value, but there is no guarantee for the order they'll take when printing the value neither for the state at which they will print it.

If you want each method to print its value after the increment, uncomment the sysout from the synchronized method and remove it from run().

If you expect all threads to finish before you print, you will need to use Thread.join().

Here's a quick example:

class TestSync implements Runnable {
    private int balance;

    public void run() {

        for(int i = 0; i < 50; i++){
            increment();
        }

    }

    private synchronized void increment() {

        int i = balance;
        balance = i + 1;
    }

    public void printBalance() ´{
        System.out.println("Balance: " + balance);
    }
}

public static void main(String[] args) {

    TestSync job = new TestSync();
    Thread a = new Thread(job);
    Thread b = new Thread(job);
    a.setName("Thread A");
    b.setName("Thread B");

    a.start();
    b.start();

    try {
      System.out.println("Wait for threads to finish.");
      a.join();
      b.join();
    } catch (InterruptedException e) {
      System.out.println("Interruption waiting for threads to finish.");
    }
    a.printBalance(); // either method will print the same value.
}

Why does the below code increment value with 2 rather than 1?(New to Java)

3 Answers3