Out-of-order writes for Double-checked locking

Question

In the examples mentioned for Out-of-order writes for double-checked locking scenarios (ref: IBM article & Wikipedia Article)

I could not understand the simple reaso

Answer 1

Thread 2 checks to see if the instance is null when Thread 1 is at //3 .

public static Singleton getInstance()
{
if (instance == null)
{
synchronized(Singleton.class) {  //1
  if (instance == null)          //2
    instance = new Singleton();  //3
  }
}
 return instance;//4
}

At this point the memory for instance has been allocated from the heap and the pointer to it is stored in the instance reference, so the "if statement" executed by Thread 2 returns "false". Note that because instance is not null when Thread2 checks it, thread 2 does not enter the synchronized block and instead returns a reference to a " fully constructed, but partially initialized, Singleton object."