Question

最好的PHP，

例如，

11011111 ==＆gt; 11111011

Answer 1

直接的方法是执行8个掩码，8个旋转和7个添加：

$blah = $blah & 128 >> 7 + $blah & 64 >> 5 + $blah & 32 >> 3 + $blah & 16 >> 1 + $blah & 8 << 1 + $blah & 4 << 3 + $blah & 2 << 5 + $blah & 1 << 7;

Answer 2

如果您已经拥有字符串形式的位，请使用strrev。

如果没有，请先使用decbin将字节转换为二进制表示，然后使用strrev反转，然后使用bindec返回字节（如有必要）。

Answer 3

检查Bit Twiddling Hacks中反转位序列的部分。应该很容易将其中一种技术改编成PHP。

尽管对PHP来说可能不实用，但使用3 64位操作有一个特别吸引人的地方：

unsigned char b; // reverse this (8-bit) byte
b = (b * 0x0202020202ULL & 0x010884422010ULL) % 1023;

Answer 4

最快的方式，也是需要更多空间的方法是查找，其中一个字节的每个可能值（如果你去整个范围，则为256）与其“反向”等价物相关联。

如果你只有几个这样的字节要处理，逐位运算符会做，但会慢一些，可能是这样的：

function reverseBits($in) {
  $out = 0;

  if ($in & 0x01) $out |= 0x80;
  if ($in & 0x02) $out |= 0x40;
  if ($in & 0x04) $out |= 0x20;
  if ($in & 0x08) $out |= 0x10;
  if ($in & 0x10) $out |= 0x08;
  if ($in & 0x20) $out |= 0x04;
  if ($in & 0x40) $out |= 0x02;
  if ($in & 0x80) $out |= 0x01;

  return $out;
}

Answer 5

这是O（n）的位长。只需将输入视为堆栈并写入输出堆栈即可。

我尝试用PHP编写这个。

function bitrev ($inBits, $bitlen){
   $cloneBits=$inBits;
   $inBits=0;
   $count=0;

   while ($count < $bitlen){
      $count=$count+1;
      $inBits=$inBits<<1;
      $inBits=$inBits|($cloneBits & 0x1);
      $cloneBits=$cloneBits>>1;
   }

    return $inBits;
}

Answer 6

试着拿到这本书，关于比特反转有整章： Hacker's Delight。但如果这适合你，请先检查内容。

Answer 7

有些人一直在建议查找表，而我一直在制作一个：

[
        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,
][$byte]

这是一个角色版本：

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

Answer 8

我不同意使用查找表（对于更大的整数）将加载到内存中所需的时间量超过处理性能。

我还对O（logn）解决方案使用按位掩码方法，如下所示：

MASK = onescompliment of 0    
while SIZE is greater than 0
  SIZE = SIZE shiftRight 1
  MASK = MASK xor (MASK shiftLeft SIZE)
  output = ((output shiftRight  SIZE) bitwiseAnd MASK) bitwiseOR ((onescompliment of MASK) bitwiseAnd (output shfitLeft SIZE))

这种方法的优点是它将整数的大小作为参数处理

php中的

可能看起来像：

function bitrev($bitstring, $size){

  $mask = ~0;
  while ($size > 0){

    $size = $size >> 1;
    $mask = $mask ^ ($mask << $size);
    $bitstring = (($bitstring >> $size) & $mask) | ((~$mask) & ($bitstring << $size));
  }
}

除非我搞砸了我的php :(

Answer 9

1984 年，我想出了这个解决方案（在 Commodore Vic20 上，当时内存很贵）。所以我在 BASIC 中的两行计算成功了。简单、快速且不占用内存的表。

从 =11001111=207 开始，在 W 中输入。

程序返回V = 243... = 11110011

程序：

input w: v=0: for x=0 to 7
if w>(2^(7-x))-1 then v=v+(2^x): w=w-(2^(7-x))
next
print v

如何反转一个字节的位？

9 个答案: