有时候, 我们需要用到一些数学函数, 但却不想用math库函数时, 怎么办呢? 移植fdlibm, 链接:
http://www.netlib.org/fdlibm/, 里边有一些数学库函数, 根据项目需要移植函数即可.
- #ifndef __LITTLE_ENDIAN
- #define __HI(x) *(1+(int*)&x)
- #define __LO(x) *(int*)&x
- #define __HIp(x) *(1+(int*)x)
- #define __LOp(x) *(int*)x
- #else
- #define __HI(x) *(int*)&x
- #define __LO(x) *(1+(int*)&x)
- #define __HIp(x) *(int*)x
- #define __LOp(x) *(1+(int*)x)
- #endif
- #ifdef __STDC__
- static const double
- #else
- static double
- #endif
- ln2_hi = 6.93147180369123816490e-01, /* 3fe62e42 fee00000 */
- ln2_lo = 1.90821492927058770002e-10, /* 3dea39ef 35793c76 */
- two54 = 1.80143985094819840000e+16, /* 43500000 00000000 */
- Lg1 = 6.666666666666735130e-01, /* 3FE55555 55555593 */
- Lg2 = 3.999999999940941908e-01, /* 3FD99999 9997FA04 */
- Lg3 = 2.857142874366239149e-01, /* 3FD24924 94229359 */
- Lg4 = 2.222219843214978396e-01, /* 3FCC71C5 1D8E78AF */
- Lg5 = 1.818357216161805012e-01, /* 3FC74664 96CB03DE */
- Lg6 = 1.531383769920937332e-01, /* 3FC39A09 D078C69F */
- Lg7 = 1.479819860511658591e-01; /* 3FC2F112 DF3E5244 */
- static double zero = 0.0;
- #ifndef __STDC__
- double __ieee754_log(double x)
- #else
- double __ieee754_log(x)
- double x;
- #endif
- {
- double hfsq,f,s,z,R,w,t1,t2,dk;
- int k,hx,i,j;
- unsigned lx;
- hx = __HI(x); /* high word of x */
- lx = __LO(x); /* low word of x */
- k=0;
- if (hx < 0x00100000) { /* x < 2**-1022 */
- if (((hx&0x7fffffff)|lx)==0)
- return -two54/zero; /* log(+-0)=-inf */
- if (hx<0) return (x-x)/zero; /* log(-#) = NaN */
- k -= 54; x *= two54; /* subnormal number, scale up x */
- hx = __HI(x); /* high word of x */
- }
- if (hx >= 0x7ff00000) return x+x;
- k += (hx>>20)-1023;
- hx &= 0x000fffff;
- i = (hx+0x95f64)&0x100000;
- __HI(x) = hx|(i^0x3ff00000); /* normalize x or x/2 */
- k += (i>>20);
- f = x-1.0;
- if((0x000fffff&(2+hx))<3) { /* |f| < 2**-20 */
- if(f==zero) if(k==0) return zero; else {dk=(double)k;
- return dk*ln2_hi+dk*ln2_lo;}
- R = f*f*(0.5-0.33333333333333333*f);
- if(k==0) return f-R; else {dk=(double)k;
- return dk*ln2_hi-((R-dk*ln2_lo)-f);}
- }
- s = f/(2.0+f);
- dk = (double)k;
- z = s*s;
- i = hx-0x6147a;
- w = z*z;
- j = 0x6b851-hx;
- t1= w*(Lg2+w*(Lg4+w*Lg6));
- t2= z*(Lg1+w*(Lg3+w*(Lg5+w*Lg7)));
- i |= j;
- R = t2+t1;
- if(i>0) {
- hfsq=0.5*f*f;
- if(k==0) return f-(hfsq-s*(hfsq+R)); else
- return dk*ln2_hi-((hfsq-(s*(hfsq+R)+dk*ln2_lo))-f);
- } else {
- if(k==0) return f-s*(f-R); else
- return dk*ln2_hi-((s*(f-R)-dk*ln2_lo)-f);
- }
- }
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同时搜到另外一种计算对数的方法, 原理写得比较详细, 移步:
http://www.openedv.com/posts/list/58801.htm, 感谢作者原创.
- const double LN10=2.3025850929940456840179914546844;
- const double LN02=0.69314718055994530941723212145818;
-
- /* 计算一个浮点数的幂函数 */
- float FunPower(float bottom_num,unsigned short exponent)
- {
- float power = 1;
- unsigned short cnt;
- for(cnt=0;cnt<exponent;cnt++)
- {
- power *= bottom_num;
- }
- return power;
- }
-
- /* 计算对数多项式的和,第一个参数是真数变换后的值
- * 第二个参数的展开的阶数N */
- float LogPoly(float transform,unsigned short step)
- {
- unsigned short n;
- float polysum = 0.0;
- for(n=0;n<step;n++)
- {
- polysum += (FunPower(transform,2*n)/(2*n+1));
- }
- return polysum*2*transform;
- }
-
- /* 计算一个浮点数的对数,输入参数是这个浮点数(真数),它应该比1.33大 */
- float Logarithm(float antilog)
- {
- float log_std = antilog; //规格化后的真数
- float log_tran; //对log_std变形后的真数(变形真数)
- float log_result; //对数的运算结果
- short log_exp_dec = 0; //对对数规格化的10进制阶码
- short log_exp_bin = 0; //对对数规格化的2进制阶码
- /* 规格化使真数在0.6667和1.3333之间,使他更接近1 */
- //第一步模10规,使真数在0.1333到1.3333之间
- while(1)
- {
- if(log_std<=1.3333)
- {
- break;
- }
- else
- {
- log_std /= 10; //尾数除10
- log_exp_dec++; //阶码加一
- }
- }
- //第二步模2规,使真数在0.6667到1.3333之间
- while(1)
- {
- if(log_std>=0.6667)
- {
- break;
- }
- else
- {
- log_std *=2; //尾数乘2
- log_exp_bin --; //阶码减一
- }
- if(log_std <= 0)
- {
- return 0;
- }
- }
- log_tran = (log_std-1)/(log_std+1); //得到的变形真数
-
- log_result = log_exp_dec*LN10 + log_exp_bin*LN02 +
- LogPoly(log_tran,10); //展开10阶
- return log_result;
- }
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BTW: 我只是代码搬运工, 感谢fdlibm和xianshasaman, 各位同学在项目中用到数学函数时是怎么做的, 分享一下心得吧.
代码有点甜~
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