CODEWITHSUNDEEP
java
Sree Ram
Sree Ram

Reputation: 859

java static variables

i have two java files file1 and file2 as follows in package pak

file1:

package pak;
public  class file1 {
  public  static int x=432;

  public static void main(String[]  args){
    System.out.println("y is "+file2.y);
    while(x==432) {
      System.out.println("x is "+file1.x);
    }

  }
}

file 2:

package pak;
public  class file2 {
  public  static int y=46;
  public static void main(String[]  args){

   System.out.println("x is "+file1.x);
   ++file1.x;
   System.out.println("x is "+file1.x);

  }
}

i will run first file1 and it will be running and when i run file 2 from another shell it should increment x value and it should come out of loop in file 1 can some one pls help me ??????

Upvotes: 0

Views: 278

Answers (3)

Peter Lawrey
Peter Lawrey

Reputation: 533870

i will run killjava that kills the process in iostat.java and just before killing i need to perform one action

You can do this with a shutdown hook.

public static void main(String... args) throws InterruptedException {
    Runtime.getRuntime().addShutdownHook(new Thread(new Runnable() {
        @Override
        public void run() {
            System.out.println("Final stats");
        }
    }));
    System.out.println("PID " + ManagementFactory.getRuntimeMXBean().getName());
    while (true) {
        System.out.println("stats");
        Thread.sleep(2000);
    }
}

prints

PID 29490@plus-dev-01
stats
stats
stats
stats
stats
Final stats

The last line occurs after I run kill 29490 in another window.

how will i do it for two different processes

This is a complex answer with too many possible solution to mention. (Shared memory, JMS, RMI, signals, Sockets, IPC etc)

Generally, this is done as a high level concept like "stop process" rather than "increment x" even if the result is the same.

The problem is that as its an advanced topic, you would need to have a good understand of interprocess communication to understand the full answer.

The simple solution is; don't do it, find another way to do what you want.

Upvotes: 0

Péter Török
Péter Török

Reputation: 116306

It won't work in such a simple way. Either you

  • run the code currently within your two main methods from two distinct threads within the same process (i.e. Java app), (and you also need to declare file1.x volatile for this to work), or
  • run the two distinct processes as described above, but then you need to make them communicate with each other, which is far from trivial. Otherwise the two processes each will have their own copies of file1.x so modifications in one won't be visible for the other.

Upvotes: 2

Azodious
Azodious

Reputation: 13882

You are invoking two different jvm processes. Both the processes will have it's own copy of file1 and file2. Change in static variable within one process will not be seen by other process.

Upvotes: 0

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