int v; // we want to find the sign of v int sign; // the result goes here sign = v >> (sizeof(int) * 8 - 1); // if v < 0 then -1, else 0The above expression evaluates to sign = v >> 31 for 32-bit integers. This is one operation faster than the obvious way, sign = -(v < 0). This trick works because when integers are shifted right, the value of the far left bit is copied to the other bits. The far left bit is 1 when the value is negative and 0 otherwise; all 1 bits is -1.
Alternatively, if you prefer the result be either -1 or +1, then use:
sign = +1 | (v >> (sizeof(int) * 8 - 1)); // if v < 0 then -1, else +1
Alternatively, if you prefer the result be either -1, 0, or +1, then use:
sign = (v != 0) | (v >> (sizeof(int) * 8 - 1)); // -1, 0, or +1Caveat: On March 7, 2003, Angus Duggan pointed out that the 1989 ANSI C specification leaves the result of signed right-shift implementation-defined, so on some systems this hack might not work.
int v; // we want to find the absolute value of v int r; // the result goes here r = (v ^ (v >> (sizeof(int) * 8 - 1))) - (v >> (sizeof(int) * 8 - 1));Some CPUs don't have an integer absolute value instruction (or the compiler fails to use them). On machines where branching is expensive, the above expression can be faster than the obvious approach, r = (v < 0) ? -v : v, even though the number of operations is the same.
Caveats: On March 7, 2003, Angus Duggan pointed out that the 1989 ANSI C
specification leaves the result of signed right-shift implementation-defined,
so on some systems this hack might not work. I've read that ANSI C does not
require values to be represented as two's complement, so it may not work
for that reason as well (on a diminishingly small number of old machines
that still use one's complement).
On March 14, 2004, Keith H. Duggar sent me the solution above; it is
superior to the one I initially came up with,
r=(+1|(v>>(sizeof(int)*8-1)))*v,
because a multiply is not used.
Unfortunately, this method has been
patented in the USA on June 6, 2000 by Vladimir Yu Volkonsky and
assigned to Sun Microsystems. (For an
unpatented non-multiplying alternative, consider
(v ^ (v >> (sizeof(int) * 8 - 1))) +
((unsigned) v >> (sizeof(int) * 8 - 1)),
suggested by Thorbjørn Willoch on June 21, 2004 or my variation,
(v ^ (v >> (sizeof(int) * 8 - 1))) + (v < 0),
both of which take one more operation than the patented one, though it may
be computed in parallel on modern CPUs.)
int x; // we want to find the minimum of x and y int y; int r; // the result goes here r = y + ((x - y) & -(x < y)); // min(x, y)On machines where branching is expensive, the above expression can be faster than the obvious approach, r = (x < y) ? x : y, even though it involves two more instructions. It works because if x < y, then -(x < y) will be all ones, so r = y + (x - y) & ~0 = y + x - y = x. Otherwise, if x >= y, then -(x < y) will be all zeros, so r = y + (x - y) & 0 = y. On machines like the Pentium, evaluating (x < y) as 0 or 1 requires a branch instruction, so there may be no advantage. MIPS, ARM, and IA64, however, do not require branches.
To find the maximum, use:
r = x - ((x - y) & -(x < y)); // max(x, y)
r = y + ((x - y) & ((x - y) >> (sizeof(int) * 8 - 1))); // min(x, y) r = x - ((x - y) & ((x - y) >> (sizeof(int) * 8 - 1))); // max(x, y)On March 7, 2003, Angus Duggan pointed out the right-shift portability issue. On May 3, 2005, Randal E. Bryant alerted me to the need for the precondition, INT_MIN <= x - y <= INT_MAX, and suggested the non-quick and dirty version as a fix. Both of these issues concern only the quick and dirty version. Nigel Horspoon pointed out on July 6, 2005 that gcc produced the same code on a Pentium as the obvious solution because of how it evaluates (x < y).
unsigned int v; // we want to see if v is a power of 2 bool f; // the result goes here f = (v & (v - 1)) == 0;Note that 0 is incorrectly considered a power of 2 here. To remedy this, use:
f = !(v & (v - 1)) && v;
int x; // convert this from using 5 bits to a full int
int r; // resulting sign extended number goes here
struct {int x:5;} s;
r = s.x = x;
The following is a C++ template function that uses the same
language feature to convert from B bits in one operation (though
the compiler is generating more, of course).
template <typename T, unsigned B>
inline T signextend(const T x)
{
struct {T x:B;} s;
return s.x = x;
}
int r = signextend<int,5>(x); // sign extend 5 bit number x to r
John Byrd caught a typo in the code (attributed to html formatting) on May 2, 2005.
unsigned b; // number of bits representing the number in x int x; // sign extend this b-bit number to r int r; // resulting sign-extended number int const m = 1 << (b - 1); // mask can be pre-computed if b is fixed r = -(x & m) | x;The code above requires five operations. Sean A. Irvine suggested that I add sign extension methods to this page on June 13, 2004, and he provided
m = (1 << (b - 1)) - 1; r = -(x & ~m) | x;
as a starting point from which I optimized.
unsigned b; // number of bits representing the number in x
int x; // sign extend this b-bit number to r
int r; // resulting sign-extended number
#define M(B) (1 << ((sizeof(x) << 3) - B))
int const multipliers[] =
{
0, M(1), M(2), M(3), M(4), M(5), M(6), M(7),
M(8), M(9), M(10), M(11), M(12), M(13), M(14), M(15),
M(16), M(17), M(18), M(19), M(20), M(21), M(22), M(23),
M(24), M(25), M(26), M(27), M(28), M(29), M(30), M(31),
M(32)
}; // (add more for 64 bits)
int const divisors[] =
{
1, ~M(1), M(2), M(3), M(4), M(5), M(6), M(7),
M(8), M(9), M(10), M(11), M(12), M(13), M(14), M(15),
M(16), M(17), M(18), M(19), M(20), M(21), M(22), M(23),
M(24), M(25), M(26), M(27), M(28), M(29), M(30), M(31),
M(32)
}; // (add more for 64 bits)
#undef M
r = (x * multipliers[b]) / divisors[b];
The following variation is not portable,
but on architectures that employ an arithmetic right-shift,
maintaining the sign, it should be fast.
const int s = -b; // OR: sizeof(x) * 8 - b; r = (x << s) >> s;Randal E. Bryant pointed out a bug on May 3, 2005 in an earlier version (that used multipliers[] for divisors[]), where it failed on the case of x=1 and b=1.
bool f; // conditional flag unsigned int m; // the bit mask unsigned int w; // the word to modify: if (f) w |= m; else w &= ~m; w ^= (-f ^ w) & m;On some architectures, the lack of branching can more than make up for what appears to be twice as many operations. For instance, informal speed tests on an AMD Athlon™ XP 2100+ indicated it was 5-10% faster. Glenn Slayden informed me of this expression on December 11, 2003.
unsigned int v; // count the number of bits set in v
unsigned int c; // c accumulates the total bits set in v
for (c = 0; v; v >>= 1)
{
c += v & 1;
}
The naive approach requires one iteration per bit, until no more
bits are set. So on a 32-bit word with only the high set,
it will go through 32 iterations.
const unsigned char BitsSetTable256[] =
{
0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 4,
1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5,
1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5,
2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5,
2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7,
1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5,
2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7,
2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7,
3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7,
4, 5, 5, 6, 5, 6, 6, 7, 5, 6, 6, 7, 6, 7, 7, 8
};
unsigned int v; // count the number of bits set in 32-bit value v
unsigned int c; // c is the total bits set in v
// Option 1:
c = BitsSetTable256[v & 0xff] +
BitsSetTable256[(v >> 8) & 0xff] +
BitsSetTable256[(v >> 16) & 0xff] +
BitsSetTable256[v >> 24];
// Option 2:
unsigned char * p = (unsigned char *) &v;
c = BitsSetTable256[p[0]] +
BitsSetTable256[p[1]] +
BitsSetTable256[p[2]] +
BitsSetTable256[p[3]];
unsigned int v; // count the number of bits set in v
unsigned int c; // c accumulates the total bits set in v
for (c = 0; v; c++)
{
v &= v - 1; // clear the least significant bit set
}
Brian Kernighan's method goes through as many iterations
as there are set bits. So if we have a 32-bit word with only
the high bit set, then it will only go once through the loop.
unsigned int v; // count the number of bits set in v
unsigned int c; // c accumulates the total bits set in v
// option 1, for at most 12-bit values in v:
c = (v * 0x1001001001001ULL & 0x84210842108421ULL) % 0x1f;
// option 2, for at most 24-bit values in v:
c = ((v & 0xfff) * 0x1001001001001ULL & 0x84210842108421ULL) % 0x1f;
c += (((v & 0xfff000) >> 12) * 0x1001001001001ULL & 0x84210842108421ULL)
% 0x1f;
// option 3, for at most 32-bit values in v:
c = ((v & 0xfff) * 0x1001001001001ULL & 0x84210842108421ULL) % 0x1f;
c += (((v & 0xfff000) >> 12) * 0x1001001001001ULL & 0x84210842108421ULL) %
0x1f;
c += ((v >> 24) * 0x1001001001001ULL & 0x84210842108421ULL) % 0x1f;
This method requires a 64-bit CPU with fast modulus division to be efficient.
The first option takes only 3 operations; the second option takes 10; and
the third option takes 15.
Rich Schroeppel originally created a 9-bit version, similiar to option 1; see the Programming Hacks section of Beeler, M., Gosper, R. W., and Schroeppel, R. HAKMEM. MIT AI Memo 239, Feb. 29, 1972. His method was the inspiration for the variants above, devised by Sean Anderson. Randal E. Bryant offered a couple bug fixes on May 3, 2005.
After reading this and pondering it, Roshan James wrote a web page explaining how it works and extending it to 16 bits on March 8, 2004.
unsigned int v; // count bits set in this (32-bit value)
unsigned int c; // store the total here
const int S[] = {1, 2, 4, 8, 16}; // Magic Binary Numbers
const int B[] = {0x55555555, 0x33333333, 0x0F0F0F0F, 0x00FF00FF, 0x0000FFFF};
c = v;
c = ((c >> S[0]) & B[0]) + (c & B[0]);
c = ((c >> S[1]) & B[1]) + (c & B[1]);
c = ((c >> S[2]) & B[2]) + (c & B[2]);
c = ((c >> S[3]) & B[3]) + (c & B[3]);
c = ((c >> S[4]) & B[4]) + (c & B[4]);
The B array, expressed as binary, is:
B[0] = 01010101 01010101 01010101 01010101 B[1] = 00110011 00110011 00110011 00110011 B[2] = 00001111 00001111 00001111 00001111 B[3] = 00000000 11111111 00000000 11111111 B[4] = 00000000 00000000 11111111 11111111We can adjust the method for larger word sizes by continuing with the patterns for the Binary Magic Numbers, B and S. If there are k bits, then we need the arrays S and B to be ceil(lg(k)) elements long, and we must compute the same number of expressions for c as S or B are long.
See Ian Ashdown's nice newsgroup post for more information on counting the number of bits set (also known as sideways addition).
unsigned int v; // word value to compute the parity of
bool parity = false; // parity will be the parity of b
while (v)
{
parity = !parity;
v = v & (v - 1);
}
The above code uses an approach like Brian Kernigan's bit counting, above.
The time it takes is proportional to the number of bits set.
unsigned char b; // byte value to compute the parity of bool parity = (((b * 0x0101010101010101ULL) & 0x8040201008040201ULL) % 0x1FF) & 1;The method above takes around 4 operations, but only works on bytes.
unsigned int v; // word value to compute the parity of v ^= v >> 16; v ^= v >> 8; v ^= v >> 4; v &= 0xf; return (0x6996 >> v) & 1;The method above takes around 9 operations, and works for 32-bit words. It may be optimized to work just on bytes in 5 operations by removing the two lines immediately following "unsigned int v;". The method first shifts and XORs the eight nibbles of the 32-bit value together, leaving the result in the lowest nibble of v. Next, the binary number 0110 1001 1001 0110 (0x6996 in hex) is shifted to the right by the value represented in the lowest nibble of v. This number is like a miniature 16-bit parity-table indexed by the low four bits in v. The result has the parity of v in bit 1, which is masked and returned.
Thanks to Mathew Hendry for pointing out the shift-lookup idea at the end on Dec. 15, 2002. That optimization shaves two operations off using only shifting and XORing to find the parity.
const bool ParityTable[] =
{
0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0,
1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1,
1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1,
0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0,
1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1,
0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0,
0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0,
1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1,
1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1,
0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0,
0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0,
1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1,
0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0,
1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1,
1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1,
0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0
};
unsigned char b; // byte value to compute the parity of
bool parity = parity_table[b];
// OR, for 32-bit words:
unsigned int v; // word value to compute the parity of
bool parity = ParityTable[(v & 0x000000ff)] ^
ParityTable[(v & 0x0000ff00) >> 8] ^
ParityTable[(v & 0x00ff0000) >> 16] ^
ParityTable[(v & 0xff000000) >> 24];
// Variation:
unsigned char * p = (unsigned char *) &v;
bool parity = ParityTable[p[0]] ^
ParityTable[p[1]] ^
ParityTable[p[2]] ^
ParityTable[p[3]];
Randal E. Byrant encouraged the addition of the (admittedly) obvious
last variation with variable p on May 3, 2005.
#define SWAP(a, b) (((a) ^= (b)), ((b) ^= (a)), ((a) ^= (b)))This is an old trick to exchange the values of the variables a and b without using extra space for a temporary variable.
On January 20, 2005, Iain A. Fleming pointed out that the macro above doesn't work when you swap with the same memory location, such as SWAP(a[i], a[j]) with i == j. So if that may occur, consider defining the macro as (((a) == (b)) || (((a) ^= (b)), ((b) ^= (a)), ((a) ^= b))).
unsigned int i, j; // positons of bit sequences to swap unsigned int n; // number of consecutive bits in each sequence unsigned int b; // bits to swap reside in b unsigned int r; // bit-swapped result goes here int x = ((b >> i) ^ (b >> j)) & ((1 << n) - 1); // XOR temporary r = b ^ ((x << i) | (x << j));As an example of swapping ranges of bits suppose we have have b = 00101111 (expressed in binary) and we want to swap the n = 3 consecutive bits starting at i = 1 (the second bit from the right) with the 3 consecutive bits starting at j = 5; the result would be r = 11100011 (binary).
This method of swapping is similar to the general purpose XOR swap trick, but intended for operating on individual bits. The variable x stores the result of XORing the pairs of bit values we want to swap, and then the bits are set to the result of themselves XORed with x. Of course, the result is undefined if the sequences overlap.
unsigned int v; // reverse the bits in this
unsigned int t = v << 1; // t will have the reversed bits of v
int i;
v >>= 1;
for (i = sizeof(v) * 8 - 2; i; i--)
{
t |= v & 1;
t <<= 1;
v >>= 1;
}
t |= v;
On October 15, 2004, Michael Hoisie pointed out a bug in the original version. Randal E. Byrant suggested removing an extra operation on May 3, 2005. Behdad Esfabod suggested a slight change that eliminated one iteration of the loop on May 18, 2005.
const unsigned char BitReverseTable256[] =
{
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
};
unsigned int v; // reverse 32-bit value, 8 bits at time
unsigned int c; // c will get v reversed
// Option 1:
c = (BitReverseTable256[v & 0xff] << 24) |
(BitReverseTable256[(v >> 8) & 0xff] << 16) |
(BitReverseTable256[(v >> 16) & 0xff] << 8) |
(BitReverseTable256[(v >> 24) & 0xff]);
// Option 2:
unsigned char * p = (unsigned char *) &v;
unsigned char * q = (unsigned char *) &c;
q[3] = BitReverseTable256[p[0]];
q[2] = BitReverseTable256[p[1]];
q[1] = BitReverseTable256[p[2]];
q[0] = BitReverseTable256[p[3]];
The first method takes about 17 operations, and the second takes about 8,
assuming your CPU can load and store bytes easily.
unsigned char b; // reverse this byte b = (b * 0x0202020202ULL & 0x010884422010ULL) % 1023;The multiply operation creates five separate copies of the 8-bit byte pattern to fan-out into a 64-bit value. The AND operation selects the bits that are in the correct (reversed) positions, relative to each 10-bit groups of bits. The multiply and the AND operations copy the bits from the original byte so they each appear in only one of the 10-bit sets. The reversed positions of the bits from the original byte coincide with their relative positions within any 10-bit set. The last step, which involves modulus division by 2^10 - 1, has the effect of merging together each set of 10 bits (from positions 0-9, 10-19, 20-29, ...) in the 64-bit value. They do not overlap, so the addition steps underlying the modulus division behave like or operations.
This method was attributed to Rich Schroeppel in the Programming Hacks section of Beeler, M., Gosper, R. W., and Schroeppel, R. HAKMEM. MIT AI Memo 239, Feb. 29, 1972.
unsigned char b; // reverse this byte b = ((b * 0x80200802ULL) & 0x0884422110ULL) * 0x0101010101ULL >> 32;The following shows the flow of the bit values with the boolean variables
a, b, c, d, e, f, g, and h, which
comprise an 8-bit byte. Notice how the first multiply fans out the
bit pattern to multiple copies, while the last multiply combines them
in the fifth byte from the right.
abcd efgh (-> hgfe dcba)
* 1000 0000 0010 0000 0000 1000 0000 0010 (0x80200802)
-------------------------------------------------------------------------------------------------
0abc defg h00a bcde fgh0 0abc defg h00a bcde fgh0
& 0000 1000 1000 0100 0100 0010 0010 0001 0001 0000 (0x0884422110)
-------------------------------------------------------------------------------------------------
0000 d000 h000 0c00 0g00 00b0 00f0 000a 000e 0000
* 0000 0001 0000 0001 0000 0001 0000 0001 0000 0001 (0x0101010101)
-------------------------------------------------------------------------------------------------
0000 d000 h000 0c00 0g00 00b0 00f0 000a 000e 0000
0000 d000 h000 0c00 0g00 00b0 00f0 000a 000e 0000
0000 d000 h000 0c00 0g00 00b0 00f0 000a 000e 0000
0000 d000 h000 0c00 0g00 00b0 00f0 000a 000e 0000
0000 d000 h000 0c00 0g00 00b0 00f0 000a 000e 0000
-------------------------------------------------------------------------------------------------
0000 d000 h000 dc00 hg00 dcb0 hgf0 dcba hgfe dcba hgfe 0cba 0gfe 00ba 00fe 000a 000e 0000
>> 32
-------------------------------------------------------------------------------------------------
0000 d000 h000 dc00 hg00 dcb0 hgf0 dcba hgfe dcba
& 1111 1111
-------------------------------------------------------------------------------------------------
hgfe dcba
Note that the last two steps can be combined on some processors because
the registers can be accessed as bytes;
just multiply so that a register stores the upper 32 bits of the result
and the take the low byte. Thus, it may take only 6 operations.
Devised by Sean Anderson, July 13, 2001.
b = ((b * 0x0802LU & 0x22110LU) | (b * 0x8020LU & 0x88440LU)) * 0x10101LU >> 16Devised by Sean Anderson, July 13, 2001. Typo spotted correction supplied by Mike Keith, January 3, 2002.
unsigned int v; // 32 bit word to reverse bit order
int const S[] = {1, 2, 4, 8, 16}; // Magic Binary Numbers
int const B[] = {0x55555555, 0x33333333, 0x0F0F0F0F, 0x00FF00FF, 0x0000FFFF};
v = ((v >> S[0]) & B[0]) | ((v << S[0]) & ~B[0]); // swap odd and even bits
v = ((v >> S[1]) & B[1]) | ((v << S[1]) & ~B[1]); // swap consecutive pairs
v = ((v >> S[2]) & B[2]) | ((v << S[2]) & ~B[2]); // swap nibbles ...
v = ((v >> S[3]) & B[3]) | ((v << S[3]) & ~B[3]);
v = ((v >> S[4]) & B[4]) | ((v << S[4]) & ~B[4]);
This method is best suited to situations where N is large.
Any reasonable optimizing C compiler should treat the dereferences of B, ~B, and S as constants, requiring no evaluation other than perhaps a load operation for some of the B and ~B references.
See Dr. Dobb's Journal 1983, Edwin Freed's article on Binary Magic Numbers for more information.
const unsigned int n; // numerator const unsigned int s; const unsigned int d = 1 << s; // So d will be one of: 1, 2, 4, 8, 16, 32, ... unsigned int m; // m will be n % d m = n & (d - 1);Most programmers learn this trick early, but it was included for the sake of completeness.
unsigned int n; // numerator
const unsigned int s; // s > 0
const unsigned int d = (1 << s) - 1; // so d is either 1, 3, 7, 15, 31, ...).
unsigned int m; // n % d goes here.
for (m = n; n > d; n = m)
{
for (m = 0; n; n >>= s)
{
m += n & d;
}
}
// Now m is a value from 0 to d, but since with modulus division
// we want m to be 0 when it is d.
m = m == d ? 0 : m;
This method of modulus division by an integer that is one less than a power
of 2 takes at most
5 + (4 + 5 * ceil(N / s)) * ceil(lg(N / s)) operations, where N is the
number of bits in the numerator. In other words, it takes at most
O(N * lg(N)) time.
Devised by Sean Anderson, August 15, 2001. Before Sean A. Irvine corrected me
on June 17, 2004, I mistakenly commented that we could alternatively assign
m = ((m + 1) & d) - 1; at the end. Michael Miller spotted a
typo in the code April 25, 2005.
// The following is for a word size of 32 bits!
const unsigned int M[] =
{
0x00000000, 0x55555555, 0x33333333, 0xc71c71c7,
0x0f0f0f0f, 0xc1f07c1f, 0x3f03f03f, 0xf01fc07f,
0x00ff00ff, 0x07fc01ff, 0x3ff003ff, 0xffc007ff,
0xff000fff, 0xfc001fff, 0xf0003fff, 0xc0007fff,
0x0000ffff, 0x0001ffff, 0x0003ffff, 0x0007ffff,
0x000fffff, 0x001fffff, 0x003fffff, 0x007fffff,
0x00ffffff, 0x01ffffff, 0x03ffffff, 0x07ffffff,
0x0fffffff, 0x1fffffff, 0x3fffffff, 0x7fffffff
};
const unsigned int Q[][6] =
{
{ 0, 0, 0, 0, 0, 0}, {16, 8, 4, 2, 1, 1}, {16, 8, 4, 2, 2, 2},
{15, 6, 3, 3, 3, 3}, {16, 8, 4, 4, 4, 4}, {15, 5, 5, 5, 5, 5},
{12, 6, 6, 6 , 6, 6}, {14, 7, 7, 7, 7, 7}, {16, 8, 8, 8, 8, 8},
{ 9, 9, 9, 9, 9, 9}, {10, 10, 10, 10, 10, 10}, {11, 11, 11, 11, 11, 11},
{12, 12, 12, 12, 12, 12}, {13, 13, 13, 13, 13, 13}, {14, 14, 14, 14, 14, 14},
{15, 15, 15, 15, 15, 15}, {16, 16, 16, 16, 16, 16}, {17, 17, 17, 17, 17, 17},
{18, 18, 18, 18, 18, 18}, {19, 19, 19, 19, 19, 19}, {20, 20, 20, 20, 20, 20},
{21, 21, 21, 21, 21, 21}, {22, 22, 22, 22, 22, 22}, {23, 23, 23, 23, 23, 23},
{24, 24, 24, 24, 24, 24}, {25, 25, 25, 25, 25, 25}, {26, 26, 26, 26, 26, 26},
{27, 27, 27, 27, 27, 27}, {28, 28, 28, 28, 28, 28}, {29, 29, 29, 29, 29, 29},
{30, 30, 30, 30, 30, 30}, {31, 31, 31, 31, 31, 31}
};
const unsigned int R[][6] =
{
{0x00000000, 0x00000000, 0x00000000, 0x00000000, 0x00000000, 0x00000000},
{0x0000ffff, 0x000000ff, 0x0000000f, 0x00000003, 0x00000001, 0x00000001},
{0x0000ffff, 0x000000ff, 0x0000000f, 0x00000003, 0x00000003, 0x00000003},
{0x00007fff, 0x0000003f, 0x00000007, 0x00000007, 0x00000007, 0x00000007},
{0x0000ffff, 0x000000ff, 0x0000000f, 0x0000000f, 0x0000000f, 0x0000000f},
{0x00007fff, 0x0000001f, 0x0000001f, 0x0000001f, 0x0000001f, 0x0000001f},
{0x00000fff, 0x0000003f, 0x0000003f, 0x0000003f, 0x0000003f, 0x0000003f},
{0x00003fff, 0x0000007f, 0x0000007f, 0x0000007f, 0x0000007f, 0x0000007f},
{0x0000ffff, 0x000000ff, 0x000000ff, 0x000000ff, 0x000000ff, 0x000000ff},
{0x000001ff, 0x000001ff, 0x000001ff, 0x000001ff, 0x000001ff, 0x000001ff},
{0x000003ff, 0x000003ff, 0x000003ff, 0x000003ff, 0x000003ff, 0x000003ff},
{0x000007ff, 0x000007ff, 0x000007ff, 0x000007ff, 0x000007ff, 0x000007ff},
{0x00000fff, 0x00000fff, 0x00000fff, 0x00000fff, 0x00000fff, 0x00000fff},
{0x00001fff, 0x00001fff, 0x00001fff, 0x00001fff, 0x00001fff, 0x00001fff},
{0x00003fff, 0x00003fff, 0x00003fff, 0x00003fff, 0x00003fff, 0x00003fff},
{0x00007fff, 0x00007fff, 0x00007fff, 0x00007fff, 0x00007fff, 0x00007fff},
{0x0000ffff, 0x0000ffff, 0x0000ffff, 0x0000ffff, 0x0000ffff, 0x0000ffff},
{0x0001ffff, 0x0001ffff, 0x0001ffff, 0x0001ffff, 0x0001ffff, 0x0001ffff},
{0x0003ffff, 0x0003ffff, 0x0003ffff, 0x0003ffff, 0x0003ffff, 0x0003ffff},
{0x0007ffff, 0x0007ffff, 0x0007ffff, 0x0007ffff, 0x0007ffff, 0x0007ffff},
{0x000fffff, 0x000fffff, 0x000fffff, 0x000fffff, 0x000fffff, 0x000fffff},
{0x001fffff, 0x001fffff, 0x001fffff, 0x001fffff, 0x001fffff, 0x001fffff},
{0x003fffff, 0x003fffff, 0x003fffff, 0x003fffff, 0x003fffff, 0x003fffff},
{0x007fffff, 0x007fffff, 0x007fffff, 0x007fffff, 0x007fffff, 0x007fffff},
{0x00ffffff, 0x00ffffff, 0x00ffffff, 0x00ffffff, 0x00ffffff, 0x00ffffff},
{0x01ffffff, 0x01ffffff, 0x01ffffff, 0x01ffffff, 0x01ffffff, 0x01ffffff},
{0x03ffffff, 0x03ffffff, 0x03ffffff, 0x03ffffff, 0x03ffffff, 0x03ffffff},
{0x07ffffff, 0x07ffffff, 0x07ffffff, 0x07ffffff, 0x07ffffff, 0x07ffffff},
{0x0fffffff, 0x0fffffff, 0x0fffffff, 0x0fffffff, 0x0fffffff, 0x0fffffff},
{0x1fffffff, 0x1fffffff, 0x1fffffff, 0x1fffffff, 0x1fffffff, 0x1fffffff},
{0x3fffffff, 0x3fffffff, 0x3fffffff, 0x3fffffff, 0x3fffffff, 0x3fffffff},
{0x7fffffff, 0x7fffffff, 0x7fffffff, 0x7fffffff, 0x7fffffff, 0x7fffffff}
};
unsigned int n; // numerator
const unsigned int s; // s > 0
const unsigned int d = (1 << s) - 1; // so d is either 1, 3, 7, 15, 31, ...).
unsigned int m; // n % d goes here.
m = (n & M[s]) + ((n & ~M[s]) >> s);
for (const unsigned int * q = &Q[s][0], * r = &R[s][0]; m > d; q++, r++)
{
m = (m >> *q) + (m & *r);
}
m = m == d ? 0 : m; // Or: m = ((m + 1) & d) - 1;
This method of finding modulus division by an integer that is one less
than a power of 2 takes at most O(lg(N)) time, where N is the number of
bits in the numerator (32 bits, for the code above).
The number of operations is at most 12 + 9 * ceil(lg(N)).
The tables may be removed if you know
the denominator at compile time; just extract the few relevent entries
and unroll the loop. It may be easily extended to more bits.
It finds the result by summing the values in base (1 << s) in parallel. First every other base (1 << s) value is added to the previous one. Imagine that the result is written on a piece of paper. Cut the paper in half, so that half the values are on each cut piece. Align the values and sum them onto a new piece of paper. Repeat by cutting this paper in half (which will be a quarter of the size of the previous one) and summing, until you cannot cut further. After performing lg(N/s/2) cuts, we cut no more; just continue to add the values and put the result onto a new piece of paper as before, while there is at least two values.
Devised by Sean Anderson, August 20, 2001. A typo was spotted by Randy E. Bryant on May 3, 2005 (after pasting the code, I had later added "unsinged" to a variable declaration).
unsigned int v; // 32-bit word to find the log base 2 of
unsigned c = 0; // c will be lg(v)
while (v >>= 1) // unroll for more speed...
{
c++;
}
The log base 2 of an integer is the same as the position of the highest
bit set (or most significant bit set, MSB). The following log base 2
methods are faster than this one.
const char LogTable256[] =
{
0, 0, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3,
4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4,
5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5,
5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5,
6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6,
6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6,
6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6,
6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6,
7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7,
7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7,
7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7,
7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7,
7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7,
7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7,
7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7,
7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7
};
unsigned int v; // 32-bit word to find the log of
unsigned c = 0; // c will be lg(v)
register unsigned int t, tt; // temporaries
if (tt = v >> 16)
{
c = (t = v >> 24) ? 24 + LogTable256[t] : 16 + LogTable256[tt & 0xFF];
}
else
{
c = (t = v >> 8) ? 8 + LogTable256[t] : LogTable256[v];
}
The lookup table method takes only about 7 operations to find the log of
a 32-bit value. If extended for 64-bit quantities, it would take roughly
9 operations.
Behdad Esfahbod and I shaved off a fraction of an operation (on average) on May 18, 2005.
unsigned int v; // 32-bit value to find the log2 of
const unsigned int b[] = {0x2, 0xC, 0xF0, 0xFF00, 0xFFFF0000};
const unsigned int S[] = {1, 2, 4, 8, 16};
int i;
register unsigned int c = 0; // result of log2(v) will go here
for (i = 4; i >= 0; i--) // unroll for speed...
{
if (v & b[i])
{
v >>= S[i];
c |= S[i];
}
}
// OR (IF YOU KNOW v IS A POWER OF 2):
const unsigned int b[] = {0xAAAAAAAA, 0xCCCCCCCC, 0xF0F0F0F0, 0xFF00FF00,
0xFFFF0000};
register unsigned int c = (v & b[0]) != 0;
for (i = 4; i > 0; i--) // unroll for speed...
{
c |= ((v & b[i]) != 0) << i;
}
Of course, to extend the code to find the log of a 33- to 64-bit
number, we would append another element, 0xFFFFFFFF00000000, to b,
append 32 to S, and loop from 5 to 0.
The second variation was suggested to me by
John Owens on April 24, 2002; it's faster, but
it is only suitable when the input is known to be a power of 2.
On May 25, 2003, Ken Raeburn suggested improving the general case by
using smaller numbers for b[], which load faster on some architectures
(for instance if the word size is 16 bits, then only one load instruction
may be needed). These values work for the general version, but not for
the special-case version below it, where v is a power of 2; Glenn Slayden
brought this oversight to my attention on December 12, 2003.
const float v; // find int(log2(v)), where v > 0.0 && finite(v) && isnormal(v) int c; // 32-bit int c gets the result; c = *(const int *) &v; // OR, for portability: memcpy(&c, &v, sizeof c); c = (c >> 23) - 127;The above is fast, but many architectures utilize subnormal floating point numbers. These have the exponent bits set to zero (signifying pow(2,-127)), and the mantissa is not normalized, so it contains leading zeros and thus the log2 must be computed from the mantissa. To accomidate for subnormal numbers, use the following:
const float v; // find int(log2(v)), where v > 0.0 && finite(v)
int c; // 32-bit int c gets the result;
int x = *(const int *) &v; // OR, for portability: memcpy(&x, &v, sizeof x);
c = x >> 23;
if (c)
{
c -= 127;
}
else
{ // subnormal, so recompute using mantissa: c = intlog2(x) - 149;
register unsigned int t; // temporary
// Note that LogTable256 was defined earlier
if (t = x >> 16)
{
c = LogTable256[t] - 133;
}
else
{
c = (t = x >> 8) ? LogTable256[t] - 141 : LogTable256[x] - 149;
}
}
On June 20, 2004, Sean A. Irvine suggested that I include code to handle
subnormal numbers. On June 11, 2005, Falk Hüffner pointed out that
ISO C99 6.5/7 specified undefined behavior for the common type punning idiom
*(int *)&, though it has worked on 99.9% of C compilers.
He proposed using memcpy for maximum portability or a union with a float
and an int for better code generation than memcpy on some compilers.
const int r;
const float v; // find int(log2(pow((double) v, 1. / pow(2, r)))),
// where isnormal(v) and v > 0
int c; // 32-bit int c gets the result;
c = *(const int *) &v; // OR, for portability: memcpy(&c, &v, sizeof c);
c = ((((c - 0x3f800000) >> r) + 0x3f800000) >> 23) - 127;
So, if r is 0, for example, we have c = int(log2((double) v)).
If r is 1, then we have c = int(log2(sqrt((double) v))).
If r is 2, then we have c = int(log2(pow((double) v, 1./4))).
On June 11, 2005, Falk Hüffner pointed out that ISO C99 6.5/7 left the type punning idiom *(int *)& undefined, and he suggested using memcpy.
unsigned int v; // 32-bit word input to count zero bits on right
register unsigned int c = 32; // c will be the number of zero bits on the right,
// so if v is 01101000 (base 2), then c will be 3 as output.
const unsigned int B[] = {0x55555555, 0x33333333, 0x0F0F0F0F, 0x00FF00FF, 0x0000FFFF};
const unsigned int S[] = {1, 2, 4, 8, 16}; // Our Magic Binary Numbers
if (v & B[4])
{
v <<= S[4];
c -= S[4];
}
if (v & B[3])
{
v <<= S[3];
c -= S[3];
}
if (v & B[2])
{
v <<= S[2];
c -= S[2];
}
if (v & B[1])
{
v <<= S[1];
c -= S[1];
}
if (v & B[0])
{
v <<= S[0];
c -= S[0];
}
if (v)
{
c--;
}
Here, we are basically doing the same operations as finding the log base 2
in parallel, but the values of b are inverted (in order to
count from the right rather than the left), we shift v up rather than
down, and c starts at the maximum and is decreased. We also have
the additional step at the end, decrementing c if there is anything left in v.
The number of operations is at most 4 * lg(N) + 2, roughly, for N bit words.
unsigned int v; // find the number of trailing zeros in v
int r; // put the result in r
const int Mod37BitPosition[] = // maps a bit value mod 37 to its position
{
32, 0, 1, 26, 2, 23, 27, 0, 3, 16, 24, 30, 28, 11, 0, 13, 4,
7, 17, 0, 25, 22, 31, 15, 29, 10, 12, 6, 0, 21, 14, 9, 5,
20, 8, 19, 18
};
r = Mod37BitPosition[(-v & v) % 37];
The code above finds the number of zeros that are trailing on the right, so
binary 0100 would produce 2. It makes use of the fact that the first 32
bit position values are relatively prime with 37, so performing a modulus
division with 37 gives a unique number from 0 to 36 for each. These numbers
may then be mapped to the number of zeros using a small lookup table.
It uses only 4 operations, however indexing into a table and performing modulus
division may make it unsuitable for some situations.
I came up with this independently and then searched for a subsequence of
the table values, and found it was invented earlier by Reiser, according to
Hacker's Delight.
unsigned int v; // compute the next highest power of 2 of 32-bit v v--; v |= v >> 1; v |= v >> 2; v |= v >> 4; v |= v >> 8; v |= v >> 16; v++;In 12 operations, this code computes the next highest power of 2 for a 32-bit integer. The result may be expressed by the formula 1 << (lg(v - 1) + 1). It would be faster by 2 operations to use the formula and the log base 2 methed that uses a lookup table, but in some situations, lookup tables are not suitable, so the above code may be best. It works by copying the highest set bit to all of the lower bits, and then adding one, which results in carries that set all of the lower bits to 0 and one bit beyond the highest set bit to 1. If the original number was a power of 2, then the decrement will reduce it to one less, so that we round up to the same original value.
Devised by Sean Anderson, Sepember 14, 2001.
unsigned short x; // Interleave bits of x and y, so that all of the
unsigned short y; // bits of x are in the even positions and y in the odd;
unsigned int z = 0; // z gets the resulting 32-bit Morton Number.
for (int i = 0; i < sizeof(x) * 8; i++) // unroll for more speed...
{
z |= (x & 1 << i) << i | (y & 1 << i) << (i + 1);
}
Interleaved bits (aka Morton numbers) are useful for linearizing 2D integer
coordinates, so x and y are combined into a single number that can be
compared easily and has the property that a number is usually close to
another if their x and y values are close.
const unsigned short MortonTable256[] =
{
0x0000, 0x0001, 0x0004, 0x0005, 0x0010, 0x0011, 0x0014, 0x0015,
0x0040, 0x0041, 0x0044, 0x0045, 0x0050, 0x0051, 0x0054, 0x0055,
0x0100, 0x0101, 0x0104, 0x0105, 0x0110, 0x0111, 0x0114, 0x0115,
0x0140, 0x0141, 0x0144, 0x0145, 0x0150, 0x0151, 0x0154, 0x0155,
0x0400, 0x0401, 0x0404, 0x0405, 0x0410, 0x0411, 0x0414, 0x0415,
0x0440, 0x0441, 0x0444, 0x0445, 0x0450, 0x0451, 0x0454, 0x0455,
0x0500, 0x0501, 0x0504, 0x0505, 0x0510, 0x0511, 0x0514, 0x0515,
0x0540, 0x0541, 0x0544, 0x0545, 0x0550, 0x0551, 0x0554, 0x0555,
0x1000, 0x1001, 0x1004, 0x1005, 0x1010, 0x1011, 0x1014, 0x1015,
0x1040, 0x1041, 0x1044, 0x1045, 0x1050, 0x1051, 0x1054, 0x1055,
0x1100, 0x1101, 0x1104, 0x1105, 0x1110, 0x1111, 0x1114, 0x1115,
0x1140, 0x1141, 0x1144, 0x1145, 0x1150, 0x1151, 0x1154, 0x1155,
0x1400, 0x1401, 0x1404, 0x1405, 0x1410, 0x1411, 0x1414, 0x1415,
0x1440, 0x1441, 0x1444, 0x1445, 0x1450, 0x1451, 0x1454, 0x1455,
0x1500, 0x1501, 0x1504, 0x1505, 0x1510, 0x1511, 0x1514, 0x1515,
0x1540, 0x1541, 0x1544, 0x1545, 0x1550, 0x1551, 0x1554, 0x1555,
0x4000, 0x4001, 0x4004, 0x4005, 0x4010, 0x4011, 0x4014, 0x4015,
0x4040, 0x4041, 0x4044, 0x4045, 0x4050, 0x4051, 0x4054, 0x4055,
0x4100, 0x4101, 0x4104, 0x4105, 0x4110, 0x4111, 0x4114, 0x4115,
0x4140, 0x4141, 0x4144, 0x4145, 0x4150, 0x4151, 0x4154, 0x4155,
0x4400, 0x4401, 0x4404, 0x4405, 0x4410, 0x4411, 0x4414, 0x4415,
0x4440, 0x4441, 0x4444, 0x4445, 0x4450, 0x4451, 0x4454, 0x4455,
0x4500, 0x4501, 0x4504, 0x4505, 0x4510, 0x4511, 0x4514, 0x4515,
0x4540, 0x4541, 0x4544, 0x4545, 0x4550, 0x4551, 0x4554, 0x4555,
0x5000, 0x5001, 0x5004, 0x5005, 0x5010, 0x5011, 0x5014, 0x5015,
0x5040, 0x5041, 0x5044, 0x5045, 0x5050, 0x5051, 0x5054, 0x5055,
0x5100, 0x5101, 0x5104, 0x5105, 0x5110, 0x5111, 0x5114, 0x5115,
0x5140, 0x5141, 0x5144, 0x5145, 0x5150, 0x5151, 0x5154, 0x5155,
0x5400, 0x5401, 0x5404, 0x5405, 0x5410, 0x5411, 0x5414, 0x5415,
0x5440, 0x5441, 0x5444, 0x5445, 0x5450, 0x5451, 0x5454, 0x5455,
0x5500, 0x5501, 0x5504, 0x5505, 0x5510, 0x5511, 0x5514, 0x5515,
0x5540, 0x5541, 0x5544, 0x5545, 0x5550, 0x5551, 0x5554, 0x5555
};
unsigned short x; // Interleave bits of x and y, so that all of the
unsigned short y; // bits of x are in the even positions and y in the odd;
unsigned int z; // z gets the resulting 32-bit Morton Number.
z = (MortonTable256[y & 0xFF] << 1) | MortonTable256[x & 0xFF] |
(MortonTable256[y >> 8] << 1 | MortonTable256[x >> 8]) << 16;
unsigned char x; // Interleave bits of x and y, so that all of the
unsigned char y; // bits of x are in the even positions and y in the odd;
unsigned short z; // z gets the resulting 16-bit Morton Number.
z = ((x * 0x0101010101010101ULL & 0x8040201008040201ULL) *
0x0102040810204081ULL >> 49) & 0x5555 |
((y * 0x0101010101010101ULL & 0x8040201008040201ULL) *
0x0102040810204081ULL >> 48) & 0xAAAA;
Holger Bettag was inspired to suggest this technique on
October 10, 2004 after reading the multiply-based bit reversals here.
const unsigned int B[] = {0x55555555, 0x33333333, 0x0F0F0F0F, 0x00FF00FF};
const unsigned int S[] = {1, 2, 4, 8};
unsigned int x; // Interleave lower 16 bits of x and y, so the bits of x
unsigned int y; // are in the even positions and bits from y in the odd;
unsigned int z; // z gets the resulting 32-bit Morton Number.
x = (x | (x << S[3])) & B[3];
x = (x | (x << S[2])) & B[2];
x = (x | (x << S[1])) & B[1];
x = (x | (x << S[0])) & B[0];
y = (y | (y << S[3])) & B[3];
y = (y | (y << S[2])) & B[2];
y = (y | (y << S[1])) & B[1];
y = (y | (y << S[0])) & B[0];
z = x | (y << 1);
// Fewer operations: unsigned int v; // 32-bit word to check if any 8-bit byte in it is 0 bool hasZeroByte = ~((((v & 0x7F7F7F7F) + 0x7F7F7F7F) | v) | 0x7F7F7F7F);The code above may be useful when doing a fast string copy in which a word is copied at a time; it uses 5 operations. On the other hand, testing for a null byte in the obvious ways (which follow) have at least 7 operations (when counted in the most sparing way), and at most 12.
// More operations: bool hasNoZeroByte = ((v & 0xff) && (v & 0xff00) && (v & 0xff0000) && (v & 0xff000000)) // OR: unsigned char * p = (unsigned char *) &v; bool hasNoZeroByte = *p && *(p + 1) && *(p + 2) && *(p + 3);The code at the beginning of this section (labeled "Fewer operations") works by first zeroing the high bits of the 4 bytes in the word. Subsequently, it adds a number that will result in an overflow to the high bit of a byte if any of the low bits were initialy set. Next the high bits of the original word are ORed with these values; thus, the high bit of a byte is set iff any bit in the byte was set. Finally, we determine if any of these high bits are zero by ORing with ones everywhere except the high bits and inverting the result. Extending to 64 bits is trivial; simply increase the constants to be 0x7F7F7F7F7F7F7F7F.
For an additional improvement, a fast pretest that requires only 4 operations may be performed to determine if the word may have a zero byte. The test also returns true if the high byte is 0x80, so there are occasional false positives, but the slower and more reliable version above may then be used on candidates for an overall increase in speed with correct output.
bool hasZeroByte = ((v + 0x7efefeff) ^ ~v) & 0x81010100;
if (hasZeroByte) // or may just have 0x80 in the high byte
{
hasZeroByte = ~((((v & 0x7F7F7F7F) + 0x7F7F7F7F) | v) | 0x7F7F7F7F);
}
There is yet a faster method —
use hasless(v, 1),
which is defined below; it
works in 4 operations and requires no subsquent verification. It simplifies
to
bool hasZeroByte = (v - 0x01010101UL) & ~v & 0x80808080UL;The subexpression (v - 0x01010101UL), evaluates to a high bit set in any byte whenever the corresponding byte in v is zero or greater than 0x80. The sub-expression ~v & 0x80808080UL evaluates to high bits set in bytes where the byte of v doesn't have its high bit set (so the byte was less than 0x80). Finally, by ANDing these two sub-expressions the result is the high bits set where the bytes in v were zero, since the high bits set due to a value greater than 0x80 in the first sub-expression are masked off by the second.
Paul Messmer suggested the fast pretest improvement on October 2, 2004. Juha Järvi later suggested hasless(v, 1) on April 6, 2005, which he found on Paul Hsieh's Assembly Lab; previously it was written in a newsgroup post on April 27, 1987 by Alan Mycroft.
Requirements: x>=0; 0<=n<=128
#define hasless(x,n) (((x)-~0UL/255*(n))&~(x)&~0UL/255*128)To count the number of bytes in x that are less than n in 7 operations, use
#define countless(x,n) \ (((~0UL/255*(127+(n))-((x)&~0UL/255*127))&~(x)&~0UL/255*128)/128%255)
Juha Järvi sent this clever technique to me on April 6, 2005. The
countless macro was added by Sean Anderson on
April 10, 2005, inspired by Juha's countmore, below.
Requirements: x>=0; 0<=n<=127
#define hasmore(x,n) (((x)+~0UL/255*(127-(n))|(x))&~0UL/255*128)To count the number of bytes in x that are more than n in 6 operations, use:
#define countmore(x,n) \ (((((x)&~0UL/255*127)+~0UL/255*(127-(n))|(x))&~0UL/255*128)/128%255)
The macro hasmore was suggested by Juha Järvi on
April 6, 2005, and he added countmore on April 8, 2005.
Note: Bytes that equal n can be reported by likelyhasbetween
as false positives,
so this should be checked by character if a certain result is needed.
Requirements: x>=0; 0<=m<=127; 0<=n<=128
#define likelyhasbetween(x,m,n) \ ((((x)-~0UL/255*(n))&~(x)&((x)&~0UL/255*127)+~0UL/255*(127-(m)))&~0UL/255*128)This technique would be suitable for a fast pretest. A variation that takes one more operation (8 total for constant m and n) but provides the exact answer is:
#define hasbetween(x,m,n) \ ((~0UL/255*(127+(n))-((x)&~0UL/255*127)&~(x)&((x)&~0UL/255*127)+~0UL/255*(127-m))&~0UL/255*128)To count the number of bytes in x that are between m and n (exclusive) in 10 operations, use:
#define countbetween(x,m,n) (hasbetween(x,m,n)/128%255)
Juha Järvi suggested likelyhasbetween on April 6, 2005.
From there,
Sean Anderson created hasbetween and
countbetween on April 10, 2005.
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