CODEWITHSUNDEEP
java
user2662247
user2662247

Reputation: 59

How does JVM know about which run method to be called while executing start() method

In the code given below I have passed the Temp class object's reference id inside the Thread constructor call which would be catched by thread class's constructor in Runnable type reference variable. so I want to ask that is there any code inside that Thread class constructor which tells the JVM that this particular class's run() method is to be executed when a thread is created.

class Temp implements Runnable
{
   public void run()
   {
      System.out.println("Hello from a thread!");
   }

    public static void main(String args[]) 
    {
        (new Thread(new Temp())).start();
    }
}

Upvotes: 1

Views: 363

Answers (5)

Stephen C
Stephen C

Reputation: 719336

What happens (in general) is this.

  • The Thread.start() method allocates the stack, and so on.
  • Then it calls the Thread.run() method.
  • If you have overridden Thread.run(), then your override run() method is called.
    • This does ... basically what you told it to do.
  • Otherwise, the run() method implemented by Thread itself is called.
    • The Thread.run() method checks to see if the Thread instance has a Runnable instance.
    • If it does, then Thread.run() calls the run() method on the Runnable().
    • Otherwise, the Thread.run() method simply returns.
  • Then the Thread.run() method returns (or terminates abnormally), the start() method marks the Thread as no longer alive, and releases the stack and dismantles other stuff associated with the Thread object.

Is there any code inside that Thread class constructor which tells the JVM that this particular class's run() method is to be executed when a thread is created.

Not exactly. The constructor stores the Runnable in a private variable, and the Thread.run() method tests the private variable ... if Thread.run() hasn't been overridden.

Upvotes: 0

Edwin Buck
Edwin Buck

Reputation: 70949

Inside the thread.start() method, the thread calls runnable.run().

A simple way of how this could work but really, it's not done this way due to this example not creating a new thread, is 

public class Thread {

   private Runnable runnable;

   public Thread(Runnable runnable) {
     this.runnable = runnable;
   }

   public void start() {
     if (runnable != null) {
       runnable.run();
     }
   }

}

Then when you call:

new Thread(this).start();

A new Thread will be created, assigning the Runnable this to the inner private member. Later, after the object is created, start() will be called on the object, which will look up the private runnable member, and call it's run() method.

Upvotes: 1

JohnB
JohnB

Reputation: 13733

Thread's start method calls the run method of the object you pass as an argument to the constructor. Since you pass an object of class Temp, Temp's run method will be called.

Upvotes: 0

nneonneo
nneonneo

Reputation: 179592

You passed in a Temp object as the Thread's Runnable. Thread.start will call the run method of its Runnable, so Temp.run is going to get called.

Upvotes: 0

Jeremy Johnson
Jeremy Johnson

Reputation: 469

start() is the command to begin using the thread that you specified in method run(). Whenever start() is called, it will execute code in the run() method.

Upvotes: 0

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