检测一个无符号数是不为2^n-1(^为幂)： x&(x+1)将最右侧0位改为1位： x | (x+1)二进制补码运算公式：-x = ~x + 1 = ~(x-1)~x = -x-1 -(~x) = x+1~(-x) = x-1x+y = x - ~y - 1 = (x|y)+(x&y) x-y = x + ~y + 1 = (x|~y)-(~x&y) x^y = (x|y)-(x&y)x|y = (x&~y)+yx&y = (~x|y)-~xx==y: ~(x-y|y-x)x!=y: x-y|y-xx< y: (x-y)^((x^y)&((x-y)^x))x<=y: (x|~y)&((x^y)|~(y-x))x< y: (~x&y)|((~x|y)&(x-y))//无符号x,y比较x<=y: (~x|y)&((x^y)|~(y-x))//无符号x,y比较使用位运算的无分支代码：计算绝对值int abs( int x ) { int y ; y = x >> 31 ; return (x^y)-y ;//or: (x+y)^y}符号函数：sign(x) = -1, x<0; 0, x == 0 ; 1, x > 0int sign(int x){ return (x>>31) | (unsigned(-x))>>31 ;//x=-2^31时失败(^为幂)}三值比较：cmp(x,y) = -1, x<y; 0, x==y; 1, x > yint cmp( int x, int y ){ return (x>y)-(x-y) ;}doz=x-y, x>=y; 0, x<yint doz(int x, int y ){ int d ; d = x-y ; return d & ((~(d^((x^y)&(d^x))))>>31) ;}int max(int x, int y ) { int m ; m = (x-y)>>31 ; return y & m | x & ~m ;}不使用第三方交换x,y:1.x ^= y ; y ^= x ; x ^= y ;2.x = x+y ; y = x-y ; x = x-y ;3.x = x-y ; y = y+x ; x = y-x ;4.x = y-x ; x = y-x ; x = x+y ; 双值交换:x = a, x==b; b, x==a//常规编码为x = x==a ? b :a ;1.x = a+b-x ;2.x = a^b^x ;下舍入到2的k次方的倍数:1.x & ((-1)<<k)2.(((unsigned)x)>>k)<<k上舍入：1. t = (1<<k)-1 ; x = (x+t)&~t ;2.t = (-1)<<k ; x = (x-t-1)&t ;位计数,统计1位的数量：1.int pop(unsigned x){ x = x-((x>>1)&0x55555555) ; x = (x&0x33333333) + ((x>>2) & 0x33333333 ) ; x = (x+(x>>4)) & 0x0f0f0f0f ; x = x + (x>>8) ; x = x + (x>>16) ; return x & 0x0000003f ;}2.int pop(unsigned x) { static char table[256] = { 0,1,1,2, 1,2,2,3, ...., 6,7,7,8 } ; return table[x&0xff]+table[(x>>8)&0xff]+table[(x>>16)&0xff]+table[(x>>24)] ;}奇偶性计算:x = x ^ ( x>>1 ) ;x = x ^ ( x>>2 ) ;x = x ^ ( x>>4 ) ;x = x ^ ( x>>8 ) ;x = x ^ ( x>>16 ) ;结果中位于x最低位，对无符号x,结果的第i位是原数第i位到最左侧位的奇偶性位反转：unsigned rev(unsigned x){ x = (x & 0x55555555) << 1 | (x>>1) & 0x55555555 ; x = (x & 0x33333333) << 2 | (x>>2) & 0x33333333 ; x = (x & 0x0f0f0f0f) << 4 | (x>>4) & 0x0f0f0f0f ; x = (x<<24) | ((x&0xff00)<<8) | ((x>>8) & 0xff00) | (x>>24) ; return x ;}递增位反转后的数：unsigned inc_r(unsigned x){ unsigned m = 0x80000000 ; x ^= m ; if( (int)x >= 0 ) do { m >>= 1 ; x ^= m ; } while( x < m ) ; return x ;}混选位：abcd efgh ijkl mnop ABCD EFGH IJKL MNOP->aAbB cCdD eEfF gGhH iIjJ kKlL mMnN oOpPunsigned ps(unsigned x){ unsigned t ; t = (x ^ (x>>8)) & 0x0000ff00; x = x ^ t ^ (t<<8) ; t = (x ^ (x>>4)) & 0x00f000f0; x = x ^ t ^ (t<<4) ; t = (x ^ (x>>2)) & 0x0c0c0c0c; x = x ^ t ^ (t<<2) ; t = (x ^ (x>>1)) & 0x22222222; x = x ^ t ^ (t<<1) ; return x ;}位压缩：选择并右移字x中对应于掩码m的1位的位,如：compress(abcdefgh,01010101)=0000bdfhcompress_left(x,m)操作与此类似，但结果位在左边: bdfh0000.unsigned compress(unsigned x, unsigned m){ unsigned mk, mp, mv, t ; int i ; x &= m ; mk = ~m << 1 ; for( i = 0 ; i < 5 ; ++i ) { mp = mk ^ ( mk << 1) ; mp ^= ( mp << 2 ) ; mp ^= ( mp << 4 ) ; mp ^= ( mp << 8 ) ; mp ^= ( mp << 16 ) ; mv = mp & m ; m = m ^ mv | (mv >> (1<<i) ) ; t = x & mv ; x = x ^ t | ( t >> ( 1<<i) ) ; mk = mk & ~mp ; } return x ;}位置换：用32个5位数表示从最低位开始的位的目标位置，结果是一个32*5的位矩阵，将该矩阵沿次对角线转置后用5个32位字p[5]存放。SAG(x,m) = compress_left(x,m) | compress(x,~m) ;准备工作：void init( unsigned *p ) { p[1] = SAG( p[1], p[0] ) ; p[2] = SAG( SAG( p[2], p[0]), p[1] ) ; p[3] = SAG( SAG( SAG( p[3], p[0] ), p[1]), p[2] ) ; p[4] = SAG( SAG( SAG( SAG( p[4], p[0] ), p[1]) ,p[2]), p[3] ) ;}实际置换：int rep( unsigned x ) { x = SAG(x,p[0]); x = SAG(x,p[1]); x = SAG(x,p[2]); x = SAG(x,p[3]); x = SAG(x,p[4]); return x ;}二进制码到GRAY码的转换:unsigned B2G(unsigned B ){ return B ^ (B>>1) ;}GRAY码到二进制码:unsigned G2B(unsigned G){ unsigned B ; B = G ^ (G>>1) ; B = G ^ (G>>2) ; B = G ^ (G>>4) ; B = G ^ (G>>8) ; B = G ^ (G>>16) ; return B ;}找出最左0字节的位置:int zbytel( unsigned x ){ static cahr table[16] = { 4,3,2,2, 1,1,1,1, 0,0,0,0, 0,0,0,0 } ; unsigned y ; y = (x&0x7f7f7f7f) + 0x7f7f7f7f ; y = ~(y|x|0x7f7f7f7f) ; return table[y*0x00204081 >> 28] ;//乘法可用移位和加完成}
