As has been mentioned, computing a new origin each time means you are subject to error.
// ______ delay _______
// v v
long origin = (long)(System.currentTimeMillis() - System.nanoTime() / 1e6);
// ^
// truncation
If you modify your program so you also compute the origin difference, you'll find out it's very small. About 200ns average I measured which is about right for the time delay.
Using multiplication instead of division (which should be OK without overflow for another couple hundred years) you'll also find that the number of origins computed that fail the equality check is much larger, about 99%. If the reason for error is because of the time delay, they would only pass when the delay happens to be identical to the last one.
A much simpler test is to accumulate elapsed time over some number of subsequent calls to nanoTime and see if it checks out with the first and last calls:
public class SimpleTimeCoherencyTest {
public static void main(String[] args) {
final long anchorNanos = System.nanoTime();
long lastNanoTime = System.nanoTime();
long accumulatedNanos = lastNanoTime - anchorNanos;
long numCallsSinceAnchor = 1L;
for(int i = 0; i < 100; i++) {
TestRun testRun = new TestRun(accumulatedNanos, lastNanoTime);
Thread t = new Thread(testRun);
t.start();
try {
t.join();
} catch(InterruptedException ie) {}
lastNanoTime = testRun.lastNanoTime;
accumulatedNanos = testRun.accumulatedNanos;
numCallsSinceAnchor += testRun.numCallsToNanoTime;
}
System.out.println(numCallsSinceAnchor);
System.out.println(accumulatedNanos);
System.out.println(lastNanoTime - anchorNanos);
}
static class TestRun
implements Runnable {
volatile long accumulatedNanos;
volatile long lastNanoTime;
volatile long numCallsToNanoTime;
TestRun(long acc, long last) {
accumulatedNanos = acc;
lastNanoTime = last;
}
@Override
public void run() {
long lastNanos = lastNanoTime;
long currentNanos;
do {
currentNanos = System.nanoTime();
accumulatedNanos += currentNanos - lastNanos;
lastNanos = currentNanos;
numCallsToNanoTime++;
} while(currentNanos - lastNanoTime <= 100000000L);
lastNanoTime = lastNanos;
}
}
}
That test does indicate the origin is the same (or at least the error is zero-mean).