博客园-sundeepblue-最新评论http://www.cnblogs.com/sundeepblue/CommentsRSS.aspxComputer Graphics, CAGD, Demoscene, intro [crack each line of code, cram each bit of byte, create each idea of mind]zh-cnFri, 02 Nov 2007 20:15:09 GMTFri, 02 Nov 2007 20:15:09 GMTcnblogsre: Inline Assembly in VC++ and gcchttp://www.cnblogs.com/sundeepblue/archive/2008/05/26/846939.html#1209379lost_passengerlost_passengerMon, 26 May 2008 06:50:55 GMThttp://www.cnblogs.com/sundeepblue/archive/2008/05/26/846939.html#1209379
lost_passenger 2008-05-26 14:50 发表评论
]]>re: Airplane: Cool!!!http://www.cnblogs.com/sundeepblue/archive/2007/10/09/917333.html#918724越狱第三季越狱第三季Tue, 09 Oct 2007 14:53:00 GMThttp://www.cnblogs.com/sundeepblue/archive/2007/10/09/917333.html#918724
越狱第三季 2007-10-09 22:53 发表评论
]]>re: 我在博客园安家http://www.cnblogs.com/sundeepblue/archive/2007/08/25/837502.html#869505fskyfskySat, 25 Aug 2007 08:59:00 GMThttp://www.cnblogs.com/sundeepblue/archive/2007/08/25/837502.html#869505
fsky 2007-08-25 16:59 发表评论
]]>re: 快速求平方根的算法http://www.cnblogs.com/sundeepblue/archive/2007/07/31/837532.html#837556夜瞳の小漫夜瞳の小漫Tue, 31 Jul 2007 05:30:00 GMThttp://www.cnblogs.com/sundeepblue/archive/2007/07/31/837532.html#837556还没有深入的读过，就记下来而已。

夜瞳の小漫 2007-07-31 13:30 发表评论
]]>re: 快速求平方根的算法http://www.cnblogs.com/sundeepblue/archive/2007/07/31/837532.html#837553夜瞳の小漫夜瞳の小漫Tue, 31 Jul 2007 05:29:00 GMThttp://www.cnblogs.com/sundeepblue/archive/2007/07/31/837532.html#837553
float Q_rsqrt( float number )
{
long i;
float x2, y;
const float threehalfs = 1.5F;

x2 = number * 0.5F;
y = number;
i = * ( long * ) &y; // evil floating point bit level hacking
i = 0x5f3759df - ( i >> 1 ); // what the fuck?
y = * ( float * ) &i;
y = y * ( threehalfs - ( x2 * y * y ) ); // 1st iteration
// y = y * ( threehalfs - ( x2 * y * y ) ); // 2nd iteration, this can be removed

#ifndef Q3_VM
#ifdef __linux__
assert( !isnan(y) ); // bk010122 - FPE?
#endif
#endif
return y;
}
对于 double ,如果在单片机环境可以试试可以使用如下代码

double InvSqrt(double number)
{
__int64 i;
double x2, y;
const double threehalfs = 1.5F;

x2 = number * 0.5F;
y = number;
i = * ( __int64 * ) &y;
i = 0x5fe6ec85e7de30da - ( i >> 1 );
y = * ( double * ) &i;
y = y * ( threehalfs - ( x2 * y * y ) ); // 1st iteration
y = y * ( threehalfs - ( x2 * y * y ) ); // 2nd iteration, this can be removed
return y;
}

哈哈。网上找的。

夜瞳の小漫 2007-07-31 13:29 发表评论
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