每天学习亿点点day 2, 3 :Bit Twiddling Hacks 位操作经典文章尝试翻译
下面这些方法有些理解起来也很简单，有些自己去想还是比较烦人的.以下的一些算法都是为了在某些不支持if语句或者一些高级功能的机器上运行而优化的, 所以可以一试. 但是平时自己写程序没必要这么写, 可读性太差了. 不怕被打就尽管炫技吧. 以及我会用橙色标注比较常用的条目，有些条目实在是说不上用得很多，或者我这个UE4菜鸡目前还不配用的技法.
1.计算整数的sign:
最后一个式计算结果为32-bit整数的sign=v>>31. 这个操作要比最明显的那个sign=-(v<0)要快. 如果是负数则输出-1,因为数字是负数那就要补1，正数则补0, 所以就输出-1, 其他的均是0. 这个行为是和架构有关的. 有的平台怎么补位的规则可能不一样吧. 其实只要记住任何时候shift都要比那些加法减法快的，能用shift操作解决的问题就不要用加减法,因为因为加法器的延迟比shift操作需要的时钟时间多很多的.
当然如果你需要得到仅有-1和1的结果那么就:
sign = +1 | (v >> (sizeof(int) * CHAR_BIT - 1)); // if v < 0 then -1, else +1
如果你想要-1, 0, 1的结果:
sign = (v != 0) | -(int)((unsigned int)((int)v) >> (sizeof(int) * CHAR_BIT - 1)); // Or, for more speed but less portability: sign = (v != 0) | (v >> (sizeof(int) * CHAR_BIT - 1)); // -1, 0, or +1 // Or, for portability, brevity, and (perhaps) speed: sign = (v > 0) - (v < 0); // -1, 0, or +1
如果只是想知道某个数是否是非负的得到结果是+1或者0则使用:
sign = 1 ^ ((unsigned int)v >> (sizeof(int) * CHAR_BIT - 1)); // if v < 0 then 0, else 1
2. 检测两个整数是否正负符号相反
int x, y; // input values to compare signs bool f = ((x ^ y) < 0); // true iff x and y have opposite signs
3. 计算整数绝对值并且不用条件转移
int v; // we want to find the absolute value of v unsigned int r; // the result goes here int const mask = v >> sizeof(int) * CHAR_BIT - 1; r = (v + mask) ^ mask;
解释一下，32位正数很好理解mask一定是0x00000000，那么v+0然后再和0异或啥也没动. 负数的时候就是v+0x11111111了，这个时候实际上就是-1了, 然后和0x11111111异或就正好把符号位变为0剩余的value位全部取反，就是我们计算补码的方法的逆算法啦, 超级简单(bushi).
4. 计算两个整数的最大和最小且不用条件转移.
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)
r = x ^ ((x ^ y) & -(x < y)); // max(x, y)
如果知道INT_MIN <= x - y <= INT_MAX, 你就可以这么做，这个方法更快因为(x-y)只需要被评估一次.
r = y + ((x - y) & ((x - y) >> (sizeof(int) * CHAR_BIT - 1))); // min(x, y) r = x - ((x - y) & ((x - y) >> (sizeof(int) * CHAR_BIT - 1))); // max(x, y)
因为x<y的时候得到的是-1全为1所以式子变为y^(x^y)&~0=y^x^y=x, 反之以此类推很简单.
5. 确认一个整数是否是2的幂级数
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 & (v - 1)); v&(v-1) v Y 0 2^p=1 1 0 0 0 1 do not care 0 Y=v&&!(v&(v-1))
第一个方法没考虑0所以一般情况下考虑第二个方法，这里我们可以把其中v&(v-1)以及v当作电信号来看待，因为在这个式子也只是当作bool数来运算了，所以这是可行的.
然后利用真值表一写，卧槽，正好是对的.
6. 从一个固定小位宽的数转换到一个大位宽的数字时，sign的扩展.
int x; // convert this from using 5 bits to a full int int r; // resulting sign extended number goes here struct {signed int x:5;} s; r = s.x = x; template <typename T, unsigned B> inline T signextend(const T x) { struct {T x:B;} s; return s.x = x; } int r = signextend<signed int,5>(x); // sign extend 5 bit number x to r
基本思想是利用了struct在存储的时候可以指定每个变量读取时候的bit数. (平时用处不大)
7. 从一个可变小位宽的数转换到一个大位宽的数字的时，sign的扩展
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 = 1U << (b - 1); // mask can be pre-computed if b is fixed x = x & ((1U << b) - 1); // (Skip this if bits in x above position b are already zero.) r = (x ^ m) - m; The code above requires four operations, but when the bitwidth is a constant rather than variable, it requires only two fast operations, assuming the upper bits are already zeroes. A slightly faster but less portable method that doesn't depend on the bits in x above position b being zero is: int const m = CHAR_BIT * sizeof(x) - b; r = (x << m) >> m;
8. 从一个可变小位宽的数转换到一个大位宽的数字的时，sign的扩展仅仅使用三个operation
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) (1U << ((sizeof(x) * CHAR_BIT) - B)) // CHAR_BIT=bits/byte static 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 if using more than 64 bits) static 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) * CHAR_BIT - b; r = (x << s) >> s;
9. 条件设置或者清除某bit位时不使用if语句
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; // OR, for superscalar CPUs: w = (w & ~m) | (-f & m);
f为真的时候把mask位置为1，此时-f是全为1，和w异或之后就是将w取反，式子变为~w&m右边操作其实就是把w的mask位取反其他全置为0，最后再和原来的数字做异或，~b^b必然是1了。也就达到了某位置1的操作. 第二个操作的方法可类似分析.
10. 条件取负时不使用if语句
If you need to negate only when a flag is false, then use the following to avoid branching: bool fDontNegate; // Flag indicating we should not negate v. int v; // Input value to negate if fDontNegate is false. int r; // result = fDontNegate ? v : -v; r = (fDontNegate ^ (fDontNegate - 1)) * v; If you need to negate only when a flag is true, then use this: bool fNegate; // Flag indicating if we should negate v. int v; // Input value to negate if fNegate is true. int r; // result = fNegate ? -v : v; r = (v ^ -fNegate) + fNegate;
11. 根据mask，合并两个数的所有bit位
unsigned int a; // value to merge in non-masked bits unsigned int b; // value to merge in masked bits unsigned int mask; // 1 where bits from b should be selected; 0 where from a. unsigned int r; // result of (a & ~mask) | (b & mask) goes here r = a ^ ((a ^ b) & mask);
简而言之就是根据mask里每位bit的值选择a或者b其中一个当前位的bit值.
12. 计算一个整数的所有bit位里1的个数
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; }
原始方法.
13. 通过查表来数bits
static const unsigned char BitsSetTable256[256] = { # define B2(n) n, n+1, n+1, n+2 # define B4(n) B2(n), B2(n+1), B2(n+1), B2(n+2) # define B6(n) B4(n), B4(n+1), B4(n+1), B4(n+2) B6(0), B6(1), B6(1), B6(2) }; 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]]; // To initially generate the table algorithmically: BitsSetTable256[0] = 0; for (int i = 0; i < 256; i++) { BitsSetTable256[i] = (i & 1) + BitsSetTable256[i / 2]; }
自己体会.
14. 数bits，来自Brian Kernighan的方法
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 }
神来一笔, 理解上你可以先想数字是奇数和偶数的时候这个操作是如何做到消除最低bit上的1的.
15. 使用64位指令数14，24，32位数的bit数
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 14-bit values in v: c = (v * 0x200040008001ULL & 0x111111111111111ULL) % 0xf; // 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;
看看就好.
16. 并行数bits
unsigned int v; // count bits set in this (32-bit value) unsigned int c; // store the total here static const int S[] = {1, 2, 4, 8, 16}; // Magic Binary Numbers static const int B[] = {0x55555555, 0x33333333, 0x0F0F0F0F, 0x00FF00FF, 0x0000FFFF}; c = v - ((v >> 1) & B[0]); c = ((c >> S[1]) & B[1]) + (c & B[1]); c = ((c >> S[2]) + c) & B[2]; c = ((c >> S[3]) + c) & B[3]; c = ((c >> S[4]) + c) & B[4]; The B array, expressed as binary, is: B[0] = 0x55555555 = 01010101 01010101 01010101 01010101 B[1] = 0x33333333 = 00110011 00110011 00110011 00110011 B[2] = 0x0F0F0F0F = 00001111 00001111 00001111 00001111 B[3] = 0x00FF00FF = 00000000 11111111 00000000 11111111 B[4] = 0x0000FFFF = 00000000 00000000 11111111 11111111 We can adjust the method for larger integer 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. For a 32-bit v, 16 operations are used. The best method for counting bits in a 32-bit integer v is the following: v = v - ((v >> 1) & 0x55555555); // reuse input as temporary v = (v & 0x33333333) + ((v >> 2) & 0x33333333); // temp c = ((v + (v >> 4) & 0xF0F0F0F) * 0x1010101) >> 24; // count The best bit counting method takes only 12 operations, which is the same as the lookup-table method, but avoids the memory and potential cache misses of a table. It is a hybrid between the purely parallel method above and the earlier methods using multiplies (in the section on counting bits with 64-bit instructions), though it doesn't use 64-bit instructions. The counts of bits set in the bytes is done in parallel, and the sum total of the bits set in the bytes is computed by multiplying by 0x1010101 and shifting right 24 bits. A generalization of the best bit counting method to integers of bit-widths upto 128 (parameterized by type T) is this: v = v - ((v >> 1) & (T)~(T)0/3); // temp v = (v & (T)~(T)0/15*3) + ((v >> 2) & (T)~(T)0/15*3); // temp v = (v + (v >> 4)) & (T)~(T)0/255*15; // temp c = (T)(v * ((T)~(T)0/255)) >> (sizeof(T) - 1) * CHAR_BIT; // count
看看就好.
17. 从最高的置1的bit位到给定的位置之间所有为1的bit位的sum
uint64_t v; // Compute the rank (bits set) in v from the MSB to pos. unsigned int pos; // Bit position to count bits upto. uint64_t r; // Resulting rank of bit at pos goes here. // Shift out bits after given position. r = v >> (sizeof(v) * CHAR_BIT - pos); // Count set bits in parallel. // r = (r & 0x5555...) + ((r >> 1) & 0x5555...); r = r - ((r >> 1) & ~0UL/3); // r = (r & 0x3333...) + ((r >> 2) & 0x3333...); r = (r & ~0UL/5) + ((r >> 2) & ~0UL/5); // r = (r & 0x0f0f...) + ((r >> 4) & 0x0f0f...); r = (r + (r >> 4)) & ~0UL/17; // r = r % 255; r = (r * (~0UL/255)) >> ((sizeof(v) - 1) * CHAR_BIT);
18. 从最高的值1的bit位，经过n个置1比特位后的位置
uint64_t v; // Input value to find position with rank r. unsigned int r; // Input: bit's desired rank [1-64]. unsigned int s; // Output: Resulting position of bit with rank r [1-64] uint64_t a, b, c, d; // Intermediate temporaries for bit count. unsigned int t; // Bit count temporary. // Do a normal parallel bit count for a 64-bit integer, // but store all intermediate steps. // a = (v & 0x5555...) + ((v >> 1) & 0x5555...); a = v - ((v >> 1) & ~0UL/3); // b = (a & 0x3333...) + ((a >> 2) & 0x3333...); b = (a & ~0UL/5) + ((a >> 2) & ~0UL/5); // c = (b & 0x0f0f...) + ((b >> 4) & 0x0f0f...); c = (b + (b >> 4)) & ~0UL/0x11; // d = (c & 0x00ff...) + ((c >> 8) & 0x00ff...); d = (c + (c >> 8)) & ~0UL/0x101; t = (d >> 32) + (d >> 48); // Now do branchless select! s = 64; // if (r > t) {s -= 32; r -= t;} s -= ((t - r) & 256) >> 3; r -= (t & ((t - r) >> 8)); t = (d >> (s - 16)) & 0xff; // if (r > t) {s -= 16; r -= t;} s -= ((t - r) & 256) >> 4; r -= (t & ((t - r) >> 8)); t = (c >> (s - 8)) & 0xf; // if (r > t) {s -= 8; r -= t;} s -= ((t - r) & 256) >> 5; r -= (t & ((t - r) >> 8)); t = (b >> (s - 4)) & 0x7; // if (r > t) {s -= 4; r -= t;} s -= ((t - r) & 256) >> 6; r -= (t & ((t - r) >> 8)); t = (a >> (s - 2)) & 0x3; // if (r > t) {s -= 2; r -= t;} s -= ((t - r) & 256) >> 7; r -= (t & ((t - r) >> 8)); t = (v >> (s - 1)) & 0x1; // if (r > t) s--; s -= ((t - r) & 256) >> 8; s = 65 - s;
19. 计算bits数的奇偶性
unsigned int v; // word value to compute the parity of bool parity = false; // parity will be the parity of v while (v) { parity = !parity; v = v & (v - 1); }
20. 计算bits数的奇偶性，查表法
static const bool ParityTable256[256] = { # define P2(n) n, n^1, n^1, n # define P4(n) P2(n), P2(n^1), P2(n^1), P2(n) # define P6(n) P4(n), P4(n^1), P4(n^1), P4(n) P6(0), P6(1), P6(1), P6(0) }; unsigned char b; // byte value to compute the parity of bool parity = ParityTable256[b]; // OR, for 32-bit words: unsigned int v; v ^= v >> 16; v ^= v >> 8; bool parity = ParityTable256[v & 0xff]; // Variation: unsigned char * p = (unsigned char *) &v; parity = ParityTable256[p[0] ^ p[1] ^ p[2] ^ p[3]];
21. 使用64位指令中的乘法和模数除法来计算bits数的奇偶性
unsigned char b; // byte value to compute the parity of bool parity = (((b * 0x0101010101010101ULL) & 0x8040201008040201ULL) % 0x1FF) & 1;
22. 用一个乘法来计算bits数的奇偶性
The following method computes the parity of the 32-bit value in only 8 operations using a multiply. unsigned int v; // 32-bit word v ^= v >> 1; v ^= v >> 2; v = (v & 0x11111111U) * 0x11111111U; return (v >> 28) & 1; Also for 64-bits, 8 operations are still enough. unsigned long long v; // 64-bit word v ^= v >> 1; v ^= v >> 2; v = (v & 0x1111111111111111UL) * 0x1111111111111111UL; return (v >> 60) & 1;
23. 并行计算bits数的奇偶性
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;
24. 使用减法和加法来交换数值 (不需要临时变量)
#define SWAP(a, b) ((&(a) == &(b)) || \ \\ 这里的斜杠是换行的意思哈 (((a) -= (b)), ((b) += (a)), ((a) = (b) - (a))))
很好理解也很好用！
25. 使用异或来进行数字交换
#define SWAP(a, b) (((a) ^= (b)), ((b) ^= (a)), ((a) ^= (b)))
老ass的方法了，记不记得y^y^x=x?就是这个原理了.
26. 在一个bit序列里指定两个位置的指定宽度bit序列交换，使用异或.
unsigned int i, j; // positions 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 unsigned int x = ((b >> i) ^ (b >> j)) & ((1U << n) - 1); // XOR temporary r = b ^ ((x << i) | (x << j));
b = 00101111 交换两个宽度位 n = 3 的连续序列 在 i = 1和 j = 5的位置，结果为 r = 11100011.
27. 逆转bits序列
unsigned int v; // input bits to be reversed unsigned int r = v; // r will be reversed bits of v; first get LSB of v int s = sizeof(v) * CHAR_BIT - 1; // extra shift needed at end for (v >>= 1; v; v >>= 1) { r <<= 1; r |= v & 1; s--; } r <<= s; // shift when v's highest bits are zero
典中典算法!
28. 在一个字中逆转bits，通过查表的方法.
static const unsigned char BitReverseTable256[256] = { # define R2(n) n, n + 2*64, n + 1*64, n + 3*64 # define R4(n) R2(n), R2(n + 2*16), R2(n + 1*16), R2(n + 3*16) # define R6(n) R4(n), R4(n + 2*4 ), R4(n + 1*4 ), R4(n + 3*4 ) R6(0), R6(2), R6(1), R6(3) }; 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]];
29. 逆转一个字节的bit序列, 并只使用3个操作, 使用64-bit乘法和模的除法:
unsigned char b; // reverse this (8-bit) byte b = (b * 0x0202020202ULL & 0x010884422010ULL) % 1023;
因为64位很长可以用来做一些变换
30. 逆转一个字节的bit序列, 并只使用4个操作, 只用64-bit乘法:
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.
31. 逆转一个字节的bits只用7个操作 (不用64位操作)
b = ((b * 0x0802LU & 0x22110LU) | (b * 0x8020LU & 0x88440LU)) * 0x10101LU >> 16; Make sure you assign or cast the result to an unsigned char to remove garbage in the higher bits.
32. 逆转一个N-bit 数, 并行时间在5*lg(N)个操作中实现
unsigned int v; // 32-bit word to reverse bit order // swap odd and even bits v = ((v >> 1) & 0x55555555) | ((v & 0x55555555) << 1); // swap consecutive pairs v = ((v >> 2) & 0x33333333) | ((v & 0x33333333) << 2); // swap nibbles ... v = ((v >> 4) & 0x0F0F0F0F) | ((v & 0x0F0F0F0F) << 4); // swap bytes v = ((v >> 8) & 0x00FF00FF) | ((v & 0x00FF00FF) << 8); // swap 2-byte long pairs v = ( v >> 16 ) | ( v << 16); The following variation is also O(lg(N)), however it requires more operations to reverse v. Its virtue is in taking less slightly memory by computing the constants on the fly. unsigned int s = sizeof(v) * CHAR_BIT; // bit size; must be power of 2 unsigned int mask = ~0; while ((s >>= 1) > 0) { mask ^= (mask << s); v = ((v >> s) & mask) | ((v << s) & ~mask); } These methods above are best suited to situations where N is large. If you use the above with 64-bit ints (or larger), then you need to add more lines (following the pattern); otherwise only the lower 32 bits will be reversed and the result will be in the lower 32 bits.
33. 使用1<<s 来计算x%2^n, 不用除号
const unsigned int n; // numerator const unsigned int s; const unsigned int d = 1U << 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);
34. 使用1<<s 来计算x%2^n-1, 不用除号
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;
most时间复杂度是这个O(N * lg(N))
35. 并行使用1<<s 来计算x%2^n-1, 不用除号
// The following is for a word size of 32 bits! static 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 }; static 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} }; static 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 >> s) & M[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, less portably: m = m & -((signed)(m - d) >> s);
36. 找一个整数的最接近的2^p的数里面的p是多少, O(N)复杂度
unsigned int v; // 32-bit word to find the log base 2 of unsigned int r = 0; // r will be lg(v) while (v >>= 1) // unroll for more speed... { r++; }
37. 找一个数的最接近的2^p的数里面的p是多少, 利用64bit的IEEE float数来做
int v; // 32-bit integer to find the log base 2 of int r; // result of log_2(v) goes here union { unsigned int u[2]; double d; } t; // temp t.u[__FLOAT_WORD_ORDER==LITTLE_ENDIAN] = 0x43300000; t.u[__FLOAT_WORD_ORDER!=LITTLE_ENDIAN] = v; t.d -= 4503599627370496.0; r = (t.u[__FLOAT_WORD_ORDER==LITTLE_ENDIAN] >> 20) - 0x3FF;
38. 找一个整数的最接近的2^p的数里面的p是多少, 使用查表法
static const char LogTable256[256] = { #define LT(n) n, n, n, n, n, n, n, n, n, n, n, n, n, n, n, n -1, 0, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, LT(4), LT(5), LT(5), LT(6), LT(6), LT(6), LT(6), LT(7), LT(7), LT(7), LT(7), LT(7), LT(7), LT(7), LT(7) }; unsigned int v; // 32-bit word to find the log of unsigned r; // r will be lg(v) register unsigned int t, tt; // temporaries if (tt = v >> 16) { r = (t = tt >> 8) ? 24 + LogTable256[t] : 16 + LogTable256[tt]; } else { r = (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. Another operation can be trimmed off by using four tables, with the possible additions incorporated into each. Using int table elements may be faster, depending on your architecture. The code above is tuned to uniformly distributed output values. If your inputs are evenly distributed across all 32-bit values, then consider using the following: if (tt = v >> 24) { r = 24 + LogTable256[tt]; } else if (tt = v >> 16) { r = 16 + LogTable256[tt]; } else if (tt = v >> 8) { r = 8 + LogTable256[tt]; } else { r = LogTable256[v]; } To initially generate the log table algorithmically: LogTable256[0] = LogTable256[1] = 0; for (int i = 2; i < 256; i++) { LogTable256[i] = 1 + LogTable256[i / 2]; } LogTable256[0] = -1; // if you want log(0) to return -1
39. 找一个N-bit位的整数的最接近的2^p的数里面的p是多少, 复杂度O(lg(N))
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 r = 0; // result of log2(v) will go here for (i = 4; i >= 0; i--) // unroll for speed... { if (v & b[i]) { v >>= S[i]; r |= S[i]; } } // OR (IF YOUR CPU BRANCHES SLOWLY): unsigned int v; // 32-bit value to find the log2 of register unsigned int r; // result of log2(v) will go here register unsigned int shift; r = (v > 0xFFFF) << 4; v >>= r; shift = (v > 0xFF ) << 3; v >>= shift; r |= shift; shift = (v > 0xF ) << 2; v >>= shift; r |= shift; shift = (v > 0x3 ) << 1; v >>= shift; r |= shift; r |= (v >> 1); // OR (IF YOU KNOW v IS A POWER OF 2): unsigned int v; // 32-bit value to find the log2 of static const unsigned int b[] = {0xAAAAAAAA, 0xCCCCCCCC, 0xF0F0F0F0, 0xFF00FF00, 0xFFFF0000}; register unsigned int r = (v & b[0]) != 0; for (i = 4; i > 0; i--) // unroll for speed... { r |= ((v & b[i]) != 0) << i; }
40. 找一个N-bit位的整数的最接近的2^p的数里面的p是多少, 复杂度O(lg(N)), 使用乘法和查表
uint32_t v; // find the log base 2 of 32-bit v int r; // result goes here static const int MultiplyDeBruijnBitPosition[32] = { 0, 9, 1, 10, 13, 21, 2, 29, 11, 14, 16, 18, 22, 25, 3, 30, 8, 12, 20, 28, 15, 17, 24, 7, 19, 27, 23, 6, 26, 5, 4, 31 }; v |= v >> 1; // first round down to one less than a power of 2 v |= v >> 2; v |= v >> 4; v |= v >> 8; v |= v >> 16; r = MultiplyDeBruijnBitPosition[(uint32_t)(v * 0x07C4ACDDU) >> 27]; The code above computes the log base 2 of a 32-bit integer with a small table lookup and multiply. It requires only 13 operations, compared to (up to) 20 for the previous method. The purely table-based method requires the fewest operations, but this offers a reasonable compromise between table size and speed. If you know that v is a power of 2, then you only need the following: static const int MultiplyDeBruijnBitPosition2[32] = { 0, 1, 28, 2, 29, 14, 24, 3, 30, 22, 20, 15, 25, 17, 4, 8, 31, 27, 13, 23, 21, 19, 16, 7, 26, 12, 18, 6, 11, 5, 10, 9 }; r = MultiplyDeBruijnBitPosition2[(uint32_t)(v * 0x077CB531U) >> 27];
41. 找一个整数的最接近的10^p的数里面的p是多少
unsigned int v; // non-zero 32-bit integer value to compute the log base 10 of int r; // result goes here int t; // temporary static unsigned int const PowersOf10[] = {1, 10, 100, 1000, 10000, 100000, 1000000, 10000000, 100000000, 1000000000}; t = (IntegerLogBase2(v) + 1) * 1233 >> 12; // (use a lg2 method from above) r = t - (v < PowersOf10[t]);
42. 找一个整数的最接近的10^p的数里面的p是多少, 最明显的方法
unsigned int v; // non-zero 32-bit integer value to compute the log base 10 of int r; // result goes here r = (v >= 1000000000) ? 9 : (v >= 100000000) ? 8 : (v >= 10000000) ? 7 : (v >= 1000000) ? 6 : (v >= 100000) ? 5 : (v >= 10000) ? 4 : (v >= 1000) ? 3 : (v >= 100) ? 2 : (v >= 10) ? 1 : 0;
43. 找一个32bit IEEE浮点数的最接近的2^p的数里面的p是多少
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 IEEE 754-compliant architectures utilize subnormal (also called denormal) 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 accomodate 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; } }
44. 找一个32bit IEEE浮点数的最接近的2^（-p）的数里面的p是多少
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;
45. 数最右的0连续序列的长度
unsigned int v; // input to count trailing zero bits int c; // output: c will count v's trailing zero bits, // so if v is 1101000 (base 2), then c will be 3 if (v) { v = (v ^ (v - 1)) >> 1; // Set v's trailing 0s to 1s and zero rest for (c = 0; v; c++) { v >>= 1; } } else { c = CHAR_BIT * sizeof(v); }
46. 数最右的0连续序列的长度, 并行版本
unsigned int v; // 32-bit word input to count zero bits on right unsigned int c = 32; // c will be the number of zero bits on the right v &= -signed(v); if (v) c--; if (v & 0x0000FFFF) c -= 16; if (v & 0x00FF00FF) c -= 8; if (v & 0x0F0F0F0F) c -= 4; if (v & 0x33333333) c -= 2; if (v & 0x55555555) c -= 1;
47. 数最右的0连续序列的长度, 二分查找版本
unsigned int v; // 32-bit word input to count zero bits on right unsigned int c; // c will be the number of zero bits on the right, // so if v is 1101000 (base 2), then c will be 3 // NOTE: if 0 == v, then c = 31. if (v & 0x1) { // special case for odd v (assumed to happen half of the time) c = 0; } else { c = 1; if ((v & 0xffff) == 0) { v >>= 16; c += 16; } if ((v & 0xff) == 0) { v >>= 8; c += 8; } if ((v & 0xf) == 0) { v >>= 4; c += 4; } if ((v & 0x3) == 0) { v >>= 2; c += 2; } c -= v & 0x1; }
48. 数最右的0连续序列的长度, 把它转换成一个浮点数再处理的方法
unsigned int v; // find the number of trailing zeros in v int r; // the result goes here float f = (float)(v & -v); // cast the least significant bit in v to a float r = (*(uint32_t *)&f >> 23) - 0x7f;
49. 数最右的0连续序列的长度, 使用模数除法和查表
unsigned int v; // find the number of trailing zeros in v int r; // put the result in r static const int Mod37BitPosition[] = // map 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];
50. 数最右的0连续序列的长度, 使用乘法和查表
unsigned int v; // find the number of trailing zeros in 32-bit v int r; // result goes here static const int MultiplyDeBruijnBitPosition[32] = { 0, 1, 28, 2, 29, 14, 24, 3, 30, 22, 20, 15, 25, 17, 4, 8, 31, 27, 13, 23, 21, 19, 16, 7, 26, 12, 18, 6, 11, 5, 10, 9 }; r = MultiplyDeBruijnBitPosition[((uint32_t)((v & -v) * 0x077CB531U)) >> 27];
51. 找出下一个比自己大或相同的2的幂级数, 使用浮点转换方法
unsigned int const v; // Round this 32-bit value to the next highest power of 2 unsigned int r; // Put the result here. (So v=3 -> r=4; v=8 -> r=8) if (v > 1) { float f = (float)v; unsigned int const t = 1U << ((*(unsigned int *)&f >> 23) - 0x7f); r = t << (t < v); } else { r = 1; } The code above uses 8 operations, but works on all v <= (1<<31). Quick and dirty version, for domain of 1 < v < (1<<25): float f = (float)(v - 1); r = 1U << ((*(unsigned int*)(&f) >> 23) - 126);
52. 找出下一个比自己大或相同的2的幂级数
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++;
53. 交错杂交两个bits序列
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 Morton Number. for (int i = 0; i < sizeof(x) * CHAR_BIT; i++) // unroll for more speed... { z |= (x & 1U << i) << i | (y & 1U << i) << (i + 1); }
54. 交错杂交两个bits序列, 查表法(空间换时间)
static const unsigned short MortonTable256[256] = { 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 >> 8] << 17 | MortonTable256[x >> 8] << 16 | MortonTable256[y & 0xFF] << 1 | MortonTable256[x & 0xFF];
55. 交错杂交两个bits序列, 使用64-bit乘法
unsigned char x; // Interleave bits of (8-bit) 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;
56. 交错杂交两个bits序列, 使用二元魔法数
static const unsigned int B[] = {0x55555555, 0x33333333, 0x0F0F0F0F, 0x00FF00FF}; static 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 and y must initially be less than 65536. 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);
57. 检查一个word是否有0字节
// 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 #define haszero(v) (((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.
58. 检查一个word是否有一个字节等于某个数n
#define hasvalue(x,n) \ (haszero((x) ^ (~0UL/255 * (n))))
59. 检查一个word是否有一个字节小于某个数n
Test if a word x contains an unsigned byte with value < n. Specifically for n=1, it can be used to find a 0-byte by examining one long at a time, or any byte by XORing x with a mask first. Uses 4 arithmetic/logical operations when n is constant. 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)
60. 检查一个word是否有一个字节大于n
Test if a word x contains an unsigned byte with value > n. Uses 3 arithmetic/logical operations when n is constant. 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)
61. 检查一个word是否有一个字节在m和n之间
When m < n, this technique tests if a word x contains an unsigned byte value, such that m < value < n. It uses 7 arithmetic/logical operations when n and m are constant. 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)
62. 计算当前序列的下一个字典排序的序列
00010011 接下来的序列就是 00010101, 00010110, 00011001,00011010, 00011100, 00100011其实就是一直加1但是保持序列里1的个数不变.
unsigned int v; // current permutation of bits unsigned int w; // next permutation of bits unsigned int t = v | (v - 1); // t gets v's least significant 0 bits set to 1 // Next set to 1 the most significant bit to change, // set to 0 the least significant ones, and add the necessary 1 bits. w = (t + 1) | (((~t & -~t) - 1) >> (__builtin_ctz(v) + 1));
The __builtin_ctz(v) GNU C compiler intrinsic for x86 CPUs returns the number of trailing zeros. If you are using Microsoft compilers for x86, the intrinsic is _BitScanForward. These both emit a bsf instruction, but equivalents may be available for other architectures. If not, then consider using one of the methods for counting the consecutive zero bits mentioned earlier.
Here is another version that tends to be slower because of its division operator, but it does not require counting the trailing zeros.
unsigned int t = (v | (v - 1)) + 1; w = t | ((((t & -t) / (v & -v)) >> 1) - 1);
cnm好累啊，有的还不怎么实用,至少对很多程序员来说...
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