Question

In the “Effective Java”, the author mentioned that

while (!done) i++;

can be optimized by HotSpot into

if (!done) {
    while (true) i++;
}

I am very confused about it. The variable done is usually not a const, why can compiler optimize that way?

Answer 1

The author assumes there that the variable done is a local variable, which does not have any requirements in the Java Memory Model to expose its value to other threads without synchronization primitives. Or said another way: the value of done won’t be changed or viewed by any code other than what’s shown here.

In that case, since the loop doesn’t change the value of done, its value can be effectively ignored, and the compiler can hoist the evaluation of that variable outside the loop, preventing it from being evaluated in the “hot” part of the loop. This makes the loop run faster because it has to do less work.

This works in more complicated expressions too, such as the length of an array:

int[] array = new int[10000];
for (int i = 0; i < array.length; ++i) {
    array[i] = Random.nextInt();
}

In this case, the naive implementation would evaluate the length of the array 10,000 times, but since the variable array is never assigned and the length of the array will never change, the evaluation can change to:

int[] array = new int[10000];
for (int i = 0, $l = array.length; i < $l; ++i) {
    array[i] = Random.nextInt();
}

Other optimizations also apply here unrelated to hoisting.

Hope that helps.