spejling af 8bit tal

Måske kan nedenstående algoritme bruges (ikke testet *S*):

c = (c & 0x0F) << 4 | (c & 0xF0) >> 4;
c = (c & 0x33) << 2 | (c & 0xCC) >> 2;
c = (c & 0x55) << 1 | (c & 0xAA) >> 1;

revx = reverse_bits[x];

hvor:

int bits_reverse[256] = {
0x00,0x80,0x40,0xC0,0x20,0xA0,0x60,0xE0,
0x10,0x90,0x50,0xD0,0x30,0xB0,0x70,0xF0,
0x08,0x88,0x48,0xC8,0x28,0xA8,0x68,0xE8,
0x18,0x98,0x58,0xD8,0x38,0xB8,0x78,0xF8,
0x04,0x84,0x44,0xC4,0x24,0xA4,0x64,0xE4,
0x14,0x94,0x54,0xD4,0x34,0xB4,0x74,0xF4,
0x0C,0x8C,0x4C,0xCC,0x2C,0xAC,0x6C,0xEC,
0x1C,0x9C,0x5C,0xDC,0x3C,0xBC,0x7C,0xFC,
0x02,0x82,0x42,0xC2,0x22,0xA2,0x62,0xE2,
0x12,0x92,0x52,0xD2,0x32,0xB2,0x72,0xF2,
0x0A,0x8A,0x4A,0xCA,0x2A,0xAA,0x6A,0xEA,
0x1A,0x9A,0x5A,0xDA,0x3A,0xBA,0x7A,0xFA,
0x06,0x86,0x46,0xC6,0x26,0xA6,0x66,0xE6,
0x16,0x96,0x56,0xD6,0x36,0xB6,0x76,0xF6,
0x0E,0x8E,0x4E,0xCE,0x2E,0xAE,0x6E,0xEE,
0x1E,0x9E,0x5E,0xDE,0x3E,0xBE,0x7E,0xFE,
0x01,0x81,0x41,0xC1,0x21,0xA1,0x61,0xE1,
0x11,0x91,0x51,0xD1,0x31,0xB1,0x71,0xF1,
0x09,0x89,0x49,0xC9,0x29,0xA9,0x69,0xE9,
0x19,0x99,0x59,0xD9,0x39,0xB9,0x79,0xF9,
0x05,0x85,0x45,0xC5,0x25,0xA5,0x65,0xE5,
0x15,0x95,0x55,0xD5,0x35,0xB5,0x75,0xF5,
0x0D,0x8D,0x4D,0xCD,0x2D,0xAD,0x6D,0xED,
0x1D,0x9D,0x5D,0xDD,0x3D,0xBD,0x7D,0xFD,
0x03,0x83,0x43,0xC3,0x23,0xA3,0x63,0xE3,
0x13,0x93,0x53,0xD3,0x33,0xB3,0x73,0xF3,
0x0B,0x8B,0x4B,0xCB,0x2B,0xAB,0x6B,0xEB,
0x1B,0x9B,0x5B,0xDB,0x3B,0xBB,0x7B,0xFB,
0x07,0x87,0x47,0xC7,0x27,0xA7,0x67,0xE7,
0x17,0x97,0x57,0xD7,0x37,0xB7,0x77,0xF7,
0x0F,0x8F,0x4F,0xCF,0x2F,0xAF,0x6F,0xEF,
0x1F,0x9F,0x5F,0xDF,0x3F,0xBF,0x7F,0xFF};

koster lidt memory men er både hurtigt og let forståeligt.

Forståeligt. Tjah - det er jo bare maskinen, der skal gøre det. Hvad med:


int main() {
  int i;

  unsigned int rev(unsigned int x) {
    unsigned int s = (x*0x02020202) & 0x84422010;
    unsigned  int t = (x*8) & 0x00000420;
    unsigned  int r = (s+t)%1023;
    return r;
  }

  for (i=0;i<=255;i++) {
    printf("Svar %2X %2X\n",i,rev(i));
  }
}

Ikke kun maskinen - også dem der skal vedligeholde koden !

Jaah, men når nu den funktion virker, skal den jo ikke vedligeholdes ;)

Du kan ikke udlede at fordi kode virker så skal den ikke vedligeholdes. Folk
skal læse og forstå koden for at checke at den virker fordi de leder efter en
fejl. Nogen kommer og vil have den udvidet til 16 bit eller krympet til 4 bit.
Eller 717 andre ting.

Man kan godt bruge en smart algoritme, men så skal man dokumentere
hvordan den virker.