For the given program, can you explain because I didn't get what I expected:
package javaapplication14;
class Mythread implements Runnable{
String myname;
int myage;
public Mythread( String par_name, int par_age){
myname = par_name;
myage = par_age;
}
public void run(){
try{
for(int i=1; i<=myage; i++) {
System.out.println("Happy birthday "+myname);
Thread.sleep(2000);
}
}catch(Exception e){
System.out.println();
}
}
}
public class JavaApplication14 {
public static void main(String[] args) {
Mythread m = new Mythread("Mutinda ", 2);
Mythread p = new Mythread("Boniface", 2);
Thread k = new Thread(m);
Thread q = new Thread(p);
k.start();
q.start();
Thread t = new Thread(m);
try{
for( int i=1; i<=5; i++){
System.out.println("Main thread executing");
Thread.sleep(1000);
}
}catch(Exception e){
System.out.println("Thread interrupted");
}
}
}
And this was my output:
Main thread executing
Happy birthday Boniface
Happy birthday Mutinda
Main thread executing
Main thread executing
Happy birthday Boniface
Happy birthday Mutinda
Main thread executing
Main thread executing
I expected this:
Main thread executing
Happy birthday Mutinda
Happy birthday Boniface
Main thread executing
Main thread executing
Happy birthday Mutinda
Happy birthday Boniface
Main thread executing
Main thread executing
I need someone to explain to me the priorites set up for the two threads k and q such that the output of q becomes the first one than k, regardless of the sleep time.
My argument: since k.start() was called before the q.start(), is expect my output to start with k, since I called it first and the two takes the same sleep time.
