Is it possible for Thread 2 to print “x=0”?

Question

Is it possible for Thread 2 to print “x=0”?

int x = 0;
boolean bExit = false;

Thread 1 (not synchronized)
x = 1; 
bExit = true;

Thread 2 (not synchronized)
if (bExit == true) 
System.out.println("x=" + x);

Answer 1

Is it possible for Thread 2 to print “x=0”?

Yes if instructions are reordered by the JIT compiler as:

Thread1:

bExit=true
x=1

Thread2:

 if (bExit == true) System.out.println("x=" + x); //prints 0

When you use synchronized block (or other related constructs) the instrcutions are not reordered by the compiler.

Answer 2

Answer : Yes, It's possible that thread T2 may print x=0.Why? because without any instruction to compiler e.g. synchronized or volatile, bExit=true might come before x=1 in compiler reordering. Also x=1 might not become visible in Thread 2, so Thread 2 will load x=0. Now, how do you fix it? When I asked this question to couple of programmers they answer differently, one suggest to make both thread synchronized on a common mutex, another one said make both variable volatile. Both are correct, as it will prevent reordering and guarantee visibility. But best answer is you just need to make bExit as volatile, then Thread 2 can only print “x=1”. x does not need to be volatile because x cannot be reordered to come after bExit=true when bExit is volatile.