Question

If I have a variable int x = 1, say, and I declare a runnable in the main thread, and I want to pass x to the runnable’s run() method, it must be declared final. Why?

final int x = 0;//<----must be final...
private class myRun implements Runnable {

    @Override
    public void run() {
        x++;//
    }

}

Answer 1

Because if they are able to be changed, it could cause a lot of problems, consider this:

public void count()
{
    int x;

    new Thread(new Runnable()
    {
        public void run()
        {
            while(x < 100)
            {
                x++;
                try
                {
                    Thread.sleep(1000);
                }catch(Exception e){}
            }
        }
     }).start();

     // do some more code...

     for(x = 0;x < 5;x++)
         for(int y = 0;y < 10;y++)
             System.out.println(myArrayElement[x][y]);
 }

This is a rough example but you can see where a lot of unexplained errors could occur. This is why the variables must be final. Here is a simple fix for the problem above:

public void count()
{
    int x;

    final int w = x;

    new Thread(new Runnable()
    {
        public void run()
        {
            int z = w;

            while(z < 100)
            {
                z++;
                try
                {
                    Thread.sleep(1000);
                }catch(Exception e){}
            }
        }
     }).start();

     // do some more code...

     for(x = 0;x < 5;x++)
         for(int y = 0;y < 10;y++)
             System.out.println(myArrayElement[x][y]);
 } 

If you want a more full explanation, it is sort of like synchronized. Java wants to prevent you from referencing one Object from multiple Threads. Here is a little bit about synchronization:

• http://docs.oracle.com/javase/tutorial/essential/concurrency/sync.html

Hope this helped!