People sometimes ignore overflow problem because their algorithm is still valid (eg checking if counter has changed within some short period of time). But keep in mind that overflow causes some fundamental math laws don't hold any more. One of them is: when we multiple a few integer numbers we got zero if and only if at least one of the factors is zero. Let's see that in action. Will this loop ever end?

int x = 1;
while(true) {
x *= RandomUtils.nextInt(1, Integer.MAX_VALUE); // positive random number
if (x == 0) break;
}

In practice it ends instantly. Even though we multiply only non-zero integers we got a zero as a result. And, of course, it has nothing to do with rounding precision. When we take a closer look, it becomes obvious that to get zero we just have to produce any number that has a binary representation with 32 (in case of java's int) zeros at the end. And

`x`

accumulates trailing zeros with each multiplication that contains 2 in its prime factors (every second loop on average). So to get zero you can simply multiply two valid ints: (1 << 30) * 4. Same works with any combination of positive and negative numbers: in

java's representation negative powers of two also accumulate trailing zeros.

## 0 komentarze :

## Post a Comment