问题
For example, in PHP, how would I reverse the bits of the byte 11011111 to 11111011?
回答1:
The straight forward approach is to perform 8 masks, 8 rotates, and 7 additions:
$blah = $blah & 128 >> 7 + $blah & 64 >> 5 + $blah & 32 >> 3 + $blah & 16 >> 1 + $blah & 8 << 1 + $blah & 4 << 3 + $blah & 2 << 5 + $blah & 1 << 7;
回答2:
If you have already the bits in the form of a string, use strrev.
If not, convert first the byte to its binary representation by using decbin, then reverse using strrev, then go back to byte (if necessary) by using bindec.
回答3:
Check the section on reversing bit sequences in Bit Twiddling Hacks. Should be easy to adapt one of the techniques into PHP.
While probably not practical for PHP, there's a particularly fascinating one using 3 64bit operations:
unsigned char b; // reverse this (8-bit) byte
b = (b * 0x0202020202ULL & 0x010884422010ULL) % 1023;
回答4:
The quickest way, but also the one requiring more space is a lookup, whereby each possible value of a byte (256 if you go for the whole range), is associated with its "reversed" equivalent.
If you only have a few such bytes to handle, bit-wise operators will do but that will be slower, maybe something like:
function reverseBits($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 $out;
}
回答5:
This is O(n) with the bit length. Just think of the input as a stack and write to the output stack.
My attempt at writing this in PHP.
function bitrev ($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 $inBits;
}
回答6:
Some people have been suggesting a lookup table, while I have been making one:
[
0x00, 0x80, 0x40, 0xC0, 0x20, 0xA0, 0x60, 0xE0, 0x10, 0x90, 0x50, 0xD0, 0x30, 0xB0, 0x70, 0xF0,
0x08, 0x88, 0x48, 0xC8, 0x28, 0xA8, 0x68, 0xE8, 0x18, 0x98, 0x58, 0xD8, 0x38, 0xB8, 0x78, 0xF8,
0x04, 0x84, 0x44, 0xC4, 0x24, 0xA4, 0x64, 0xE4, 0x14, 0x94, 0x54, 0xD4, 0x34, 0xB4, 0x74, 0xF4,
0x0C, 0x8C, 0x4C, 0xCC, 0x2C, 0xAC, 0x6C, 0xEC, 0x1C, 0x9C, 0x5C, 0xDC, 0x3C, 0xBC, 0x7C, 0xFC,
0x02, 0x82, 0x42, 0xC2, 0x22, 0xA2, 0x62, 0xE2, 0x12, 0x92, 0x52, 0xD2, 0x32, 0xB2, 0x72, 0xF2,
0x0A, 0x8A, 0x4A, 0xCA, 0x2A, 0xAA, 0x6A, 0xEA, 0x1A, 0x9A, 0x5A, 0xDA, 0x3A, 0xBA, 0x7A, 0xFA,
0x06, 0x86, 0x46, 0xC6, 0x26, 0xA6, 0x66, 0xE6, 0x16, 0x96, 0x56, 0xD6, 0x36, 0xB6, 0x76, 0xF6,
0x0E, 0x8E, 0x4E, 0xCE, 0x2E, 0xAE, 0x6E, 0xEE, 0x1E, 0x9E, 0x5E, 0xDE, 0x3E, 0xBE, 0x7E, 0xFE,
0x01, 0x81, 0x41, 0xC1, 0x21, 0xA1, 0x61, 0xE1, 0x11, 0x91, 0x51, 0xD1, 0x31, 0xB1, 0x71, 0xF1,
0x09, 0x89, 0x49, 0xC9, 0x29, 0xA9, 0x69, 0xE9, 0x19, 0x99, 0x59, 0xD9, 0x39, 0xB9, 0x79, 0xF9,
0x05, 0x85, 0x45, 0xC5, 0x25, 0xA5, 0x65, 0xE5, 0x15, 0x95, 0x55, 0xD5, 0x35, 0xB5, 0x75, 0xF5,
0x0D, 0x8D, 0x4D, 0xCD, 0x2D, 0xAD, 0x6D, 0xED, 0x1D, 0x9D, 0x5D, 0xDD, 0x3D, 0xBD, 0x7D, 0xFD,
0x03, 0x83, 0x43, 0xC3, 0x23, 0xA3, 0x63, 0xE3, 0x13, 0x93, 0x53, 0xD3, 0x33, 0xB3, 0x73, 0xF3,
0x0B, 0x8B, 0x4B, 0xCB, 0x2B, 0xAB, 0x6B, 0xEB, 0x1B, 0x9B, 0x5B, 0xDB, 0x3B, 0xBB, 0x7B, 0xFB,
0x07, 0x87, 0x47, 0xC7, 0x27, 0xA7, 0x67, 0xE7, 0x17, 0x97, 0x57, 0xD7, 0x37, 0xB7, 0x77, 0xF7,
0x0F, 0x8F, 0x4F, 0xCF, 0x2F, 0xAF, 0x6F, 0xEF, 0x1F, 0x9F, 0x5F, 0xDF, 0x3F, 0xBF, 0x7F, 0xFF,
][$byte]
And here's a character version:
[
"\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($byte)];
回答7:
Try to get this book, there is whole chapter about bits reversion: Hacker's Delight. But please check content first if this suits you.
回答8:
I disagree with using a look up table as (for larger integers) the amount of time necessary to load it into memory trumps processing performance.
I also use a bitwise masking approach for a O(logn) solution, which looks like:
MASK = onescompliment of 0
while SIZE is greater than 0
SIZE = SIZE shiftRight 1
MASK = MASK xor (MASK shiftLeft SIZE)
output = ((output shiftRight SIZE) bitwiseAnd MASK) bitwiseOR ((onescompliment of MASK) bitwiseAnd (output shfitLeft SIZE))
The advantage of this approach is it handles the size of your integer as an argument
in php this might look like:
function bitrev($bitstring, $size){
$mask = ~0;
while ($size > 0){
$size = $size >> 1;
$mask = $mask ^ ($mask << $size);
$bitstring = (($bitstring >> $size) & $mask) | ((~$mask) & ($bitstring << $size));
}
}
unless I screwed up my php somewhere :(
来源：https://stackoverflow.com/questions/1688532/how-to-reverse-bits-of-a-byte