我抄过来改了下，只和sqrt()比较了下正确性
#ifdef USING_ASM
inline DWORD __fastcall Log2_Floor(DWORD x)
{
__asm
{
bsr eax,x
jnz RETURN
mov eax,0ffffffffh
RETURN:
}
}
#else
inline DWORD Log2_Floor(DWORD x)
{
DWORD l=0;
if(x&0xffff0000){l|=16;x>>=16;}
if(x&0x0000ff00){l|= 8;x>>= 8;}
if(x&0x000000f0){l|= 4;x>>= 4;}
if(x&0x0000000c){l|= 2;x>>= 2;}
if(x&0x00000002){l|= 1;/*x>>= 1;*/}
return l;
}
#endif
DWORD __fastcall Sqrt(DWORD n)
{
static WORD sqrtTab[]=
{
0,1,1,1,2,2,2,2,2,3,3,3,3,3,3,3,
4,4,4,4,4,4,4,4,4,5,5,5,5,5,5,5,
5,5,5,5,6,6,6,6,6,6,6,6,6,6,6,6,
6,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,
256,258,260,262,264,266,268,270,272,274,276,278,280,282,284,286,
288,288,290,292,294,296,298,300,300,302,304,306,308,310,310,312,
314,316,318,320,320,322,324,326,326,328,330,332,334,334,336,338,
340,340,342,344,346,346,348,350,352,352,354,356,356,358,360,362,
362,364,366,366,368,370,370,372,374,374,376,378,378,380,382,384,
384,386,386,388,390,390,392,394,394,396,398,398,400,402,402,404,
406,406,408,408,410,412,412,414,416,416,418,418,420,422,422,424,
424,426,428,428,430,430,432,434,434,436,436,438,438,440,442,442,
444,444,446,448,448,450,450,452,452,454,454,456,458,458,460,460,
462,462,464,464,466,468,468,470,470,472,472,474,474,476,476,478,
480,480,482,482,484,484,486,486,488,488,490,490,492,492,494,494,
496,496,498,498,500,500,502,502,504,504,506,506,508,508,510,512
};
static const DWORD Table2[]=
{
0<<10,1<<10,4<<10,9<<10,16<<10,25<<10,36<<10,49<<10,
};
if(n<64) return sqrtTab;
DWORD r;
if(n<65536)
{
r=sqrtTab;
n-=Table2;
r<<=10;
DWORD c=0x100|r;r>>=1;
if(n>=c){n-=c;r|=0x100;}
c=0x40|r;r>>=1;
if(n>=c){n-=c;r|=0x40;}
c=0x10|r;r>>=1;
if(n>=c){n-=c;r|=0x10;}
c=0x04|r;r>>=1;
if(n>=c){n-=c;r|=0x04;}
c=0x01|r;r>>=1;
if(n>=c) r|=0x01;
return r;
}
switch(Log2_Floor(n)&0xfe)
{
#define SQRT_ADJUST {r=(r+n/r)>>1;if(r*r>n) --r;break;}
case 16:r=sqrtTab ;SQRT_ADJUST;
case 18:r=sqrtTab<<1;SQRT_ADJUST;
case 20:r=sqrtTab<<2;SQRT_ADJUST;
case 22:r=sqrtTab<<3;SQRT_ADJUST;
case 24:r=sqrtTab<<4;SQRT_ADJUST;
case 26:r=sqrtTab<<5;SQRT_ADJUST;
case 28:r=sqrtTab<<6;SQRT_ADJUST;
case 30:r=sqrtTab<<7;SQRT_ADJUST;
#undef SQRT_ADJUST
}
return r;
}
多谢分享！
很多时间会跑来回顾下这里的经典算法，#137 应该是适合游戏中小浮点数的近似开方，然后也看到一段类似又有意思的话：
"人们很早就在Quake3源代码中发现了如下的C代码，它可以快速的求1/sqrt(x)，在3D图形向量计算方面应用很广。
float InvSqrt(float x){
float xhalf=0.5f*x;
long i=*(long*)&x;
i=0x5f3759df - (i>>1);
x=*(float *)&i;
x=x*(1.5f-xhalf*x*x);
return x;
}
Beyond3D.com的Ryszard Sommefeldt一直在想到底是哪个家伙写了这些神奇的代码？2003年Chris Lomont还写了一篇文章（PDF)对这些代码进行了分析。毫无疑问写出这些代码的人绝对是天才。"
是John Carmack？Michael Abrash？John Carmack在邮件回复中明确表示不是他，也不是Michael，可能是Terje Matheson。于是侦探Ryszard又向Terje Mathisen寻求答案。Terje说他写过类似的高效代码，但上面的不是。他猜测可能在MIT的旧文档中可以找到。Ryszard的求证之路显然无止境。