java - 为什么这个乘法整数溢出会导致零？

Question

回答后this question ，我很困惑为什么这段代码中溢出的整数导致 0 而不是负数。很奇怪，为什么会有这么精确的数字？为什么是 0？

public class IntegerOverflow {
  public static void main(String[] args) {
    int x = 10;

    int i = 0;
    for (i = 0; i <= 5; i++)
    {
      x = x * x;
      System.out.println(x);
    }
  }
}

输出:

100
10000
100000000
1874919424
0
0

Answer 1

只有当 x 的起始值为偶数时才会发生这种情况。

根据JLS §15.17.1 :

If an integer multiplication overflows, then the result is the low-order bits of the mathematical product as represented in some sufficiently large two's-complement format. As a result, if overflow occurs, then the sign of the result may not be the same as the sign of the mathematical product of the two operand values.

如果我们以二进制格式而不是十进制格式打印数字，这会变得更加明显:

public class IntegerOverflow {
  public static void main(String[] args) {
    int x = 10;

    int i = 0;
    for (i = 0; i <= 5; i++)
    {
      x *= x;
      System.out.println(Integer.toBinaryString(x));
    }
  }
}

输出:

1100100
10011100010000
101111101011110000100000000
1101111110000010000000000000000
0
0

如您所见，每次平方时，我们都会将零位的数量加倍。由于只保存低阶位，因此每次将零加倍最终将导致零。请注意，如果 x 的起始值为奇数，我们看不到这些尾随零。相反，它会导致看似无关的数字，就像溢出通常那样。

public class IntegerOverflow {
  public static void main(String[] args) {
    int x = 11;

    int i = 0;
    for (i = 0; i <= 5; i++)
    {
      x *= x;
      System.out.format("%-12d\t%s%n", x, Integer.toBinaryString(x));
    }
  }
}

输出:

121             1111001
14641           11100100110001
214358881       1100110001101101101101100001
772479681       101110000010110001101011000001
-1419655807     10101011011000011100010110000001
-1709061375     10011010001000011100101100000001