When one thread out of four is executing first line in run() then at the same time what will remaining three threads do?

Question

I have four thread instances and they are executing run() code given below:

public void run (){
   System.out.println("1st line :"+Thread.CurrentThread().getName());
   System.out.println("2nd line :"+Thread.CurrentThread().getName());
   System.out.println("3rd line :"+Thread.CurrentThread().getName());
   System.out.println("4th line :"+Thread.CurrentThread().getName());
   System.out.println("5th line :"+Thread.CurrentThread().getName());
   System.out.println("6th line :"+Thread.CurrentThread().getName());
}

When one thread out of four is executing first line then at the same time what will remaining three threads do?

Either they are waiting for executing first line or will they proceed other lines of code?

Answer 1

Since all the threads run the same method, they'll all print all the lines.

This is like calling run() 4 times, except that instead of waiting for one call to finish before starting the next all the calls are executed at the same time. The order is different, but all the lines are still executed 4 times.

If you want to execute run() only once but in 4 different threads, you'll need to use more complex methods and inevitably restructure it into multiple methods.

Answer 2

Every thread will always start from the first line. No thread will start at the second or third line just because some other thread has executed the first or second; in fact, none of the threads know where any of the other threads are in executing code.

You shouldn't rely on just running multithreaded code to understand what it does, since it is possible for the implementers of the runtime system to change the details of how it works, and the order and length of time each thread runs is unpredictable.

It might be helpful to think of separate threads as separate programs that have shared access to a global memory space.