------- Additional Comments From bonzini at gcc dot gnu dot org 2005-08-23 14:48 ------- Yes, I think that most invocations of next should be inlined, and wrapped in a single synchronized block.
Apart from that, I am pretty sure that here seed = (seed * 0x5DEECE66DL + 0xBL) & ((1L << 48) - 1); return (int) (seed >>> (48 - bits)); it makes no difference if you do the AND or not. So nextDouble could be implemented as: long first; long second; synchronized (this) { seed = (seed * 0x5DEECE66DL + 0xBL); first = (seed & 0x0000FFFFFFC00000L) << 5; seed = (seed * 0x5DEECE66DL + 0xBL); second = (seed >> 21) & 0x7FFFFFF; } return (first | second) / (double) (1L << 53); Similarly, for nextFloat float f; int bits; synchronized (this) { seed = (seed * 0x5DEECE66DL + 0xBL); bits = (int) (seed >> 24); } return bits / 16777216.0f; nextInt int bits; synchronized (this) { seed = (seed * 0x5DEECE66DL + 0xBL); bits = (int) (seed >> 16); } nextLong long first, second; synchronized (this) { seed = (seed * 0x5DEECE66DL + 0xBL); first = (seed << 16) & 0xFFFFFFFF00000000L; seed = (seed * 0x5DEECE66DL + 0xBL); second = (seed >> 16) & 0xFFFFFFFFL; } return first | second; nextInt (n) int bits; synchronized (seed) { seed = (seed * 0x5DEECE66DL + 0xBL); bits = (int) (seed >> 17); } if ((n & -n) == n) // i.e., n is a power of 2 return (int)((n * (long) bits) >> 31); int bits, val; val = bits % n; if (bits - val + n - 1 < 0) synchronized (seed) { do { seed = (seed * 0x5DEECE66DL + 0xBL); bits = (int) (seed >> 17); } while (bits - val + n - 1 < 0); } return val; nextBoolean boolean bit; synchronized (this) { seed = (seed * 0x5DEECE66DL + 0xBL); bit = (seed & 0x800000000000) != 0; } return bit; And I left out nextBytes, I know. Also these are untested, which is why I'm not preparing a full patch. Paolo -- http://gcc.gnu.org/bugzilla/show_bug.cgi?id=23283