我正在阅读Unsigned arithmetic in Java，它很好地解释了如何使用以下方法进行无符号的longs
public static boolean isLessThanUnsigned(long n1, long n2) {
return (n1 < n2) ^ ((n1 < 0) != (n2 < 0));
}
但是，我对Guava的实现感到困惑。我希望有人能对此有所启发。
/**
* A (self-inverse) bijection which converts the ordering on unsigned longs to the ordering on
* longs, that is, {@code a <= b} as unsigned longs if and only if {@code flip(a) <= flip(b)} as
* signed longs.
*/
private static long flip(long a) {
return a ^ Long.MIN_VALUE;
}
/**
* Compares the two specified {@code long} values, treating them as unsigned values between
* {@code 0} and {@code 2^64 - 1} inclusive.
*
* @param a the first unsigned {@code long} to compare
* @param b the second unsigned {@code long} to compare
* @return a negative value if {@code a} is less than {@code b}; a positive value if {@code a} is
* greater than {@code b}; or zero if they are equal
*/
public static int compare(long a, long b) {
return Longs.compare(flip(a), flip(b));
}
最佳答案
也许有些图表有帮助。我将使用8位数字来使常量简短，它以明显的方式推广为int和longs。
绝对观点:
Unsigned number line:
[ 0 .. 0x7F ][ 0x80 .. 0xFF]
Signed number line:
[ 0x80 .. 0xFF ][ 0 .. 0x7F]
相对视图:
Unsigned number line:
[ 0 .. 0x7F ][ 0x80 .. 0xFF]
Signed number line:
[ 0x80 .. 0xFF ][ 0 .. 0x7F]
因此，有符号和无符号数字在很大程度上具有相同的相对顺序，不同的是按顺序交换了设置了符号位和未设置符号位的两个范围。当然，反转该位会交换顺序。
x ^ Long.MIN_VALUE反转long的符号位。此技巧适用于仅依赖于相对顺序的任何操作，例如比较和直接相关的操作(例如最小和最大)。它不适用于依赖于数字的绝对大小的运算，例如除法。