log10使用什么算法实现？
在项目中要使用到log10计算，可所用系统的标准库里并没有移植该功能，需要自己实现。请问，采用什么算法实现该功能？
在网上找了一个例子，但不确定能否在嵌入式的一些平台使用。
double my_log10(double x)
{
register double ret;
__asm__(
"fldlg2\n\t"
"fxch\n\t"
"fyl2x"
:"=t"(ret)
:"0"(x)
);
return ret;
}
也查看了glibc的源码，可调用了其它的一些API，对库的依赖比较大，逐步移植的话比较麻烦。代码如下：
double __log10 (double x)
{
if (__builtin_expect (islessequal (x, 0.0), 0) && _LIB_VERSION != _IEEE_)
{
if (x == 0.0)
{
feraiseexcept (FE_DIVBYZERO);
return __kernel_standard (x, x, 18); /* log10(0) */
}
else
{
feraiseexcept (FE_INVALID);
return __kernel_standard (x, x, 19); /* log10(x<0) */
}
}
return __ieee754_log10 (x);
}
粉Se幻彩
10 years, 9 months ago
Answers
参考 这里 实现了一个c语言版本的log10快速算法，还不是很完善。
#include <math.h>
#include <string.h>
#include <stdio.h>
static double Sqrt(double x)
{
if (x < 0)
return -1;
if (x == 0)
return 0;
double y = (double)sqrt((double)x);
return (y + x / y) / 2;
}
static double NegativeLog(double q)
{
int p;
double pi2 = 6.283185307179586476925286766559;
double eps2 = 0.00000000000001; // 1e-14
double eps1; // 1e-28
double r = q, s = q, n = q, q2 = q * q, q1 = q2 * q;
eps1 = eps2 * eps2;
for (p = 1; (n *= q1) > eps1; s += n, q1 *= q2)
r += (p = !p) ? n : -n;
double u = 1 - 2 * r, v = 1 + 2 * s, t = u / v;
double a = 1, b = Sqrt(1 - t * t * t * t);
for (; a - b > eps2; b = Sqrt(a * b), a = t)
t = (a + b) / 2;
return pi2 / (a + b) / v / v;
}
static double Log(double x)
{
int k = 0;
double ln10 = 2.30258509299404568401799145468;
if (x <= 0)
return -1;
if (x == 1)
return 0;
for (; x > 0.1; k++)
x /= 10;
for (; x <= 0.01; k--)
x *= 10;
return k * ln10 - NegativeLog(x);
}
double Log10(double x)
{
double ln10 = 2.30258509299404568401799145468;
return Log(x) / ln10;
}
int main(void)
{
printf(" self Log10: %f\n math log10: %f\n\n", Log10(1000), log10(1000));
printf(" self Log10: %f\n math log10: %f\n\n", Log10(1), log10(1));
printf(" self Log10: %f\n math log10: %f\n\n", Log10(8192.1024), log10(8192.1024));
printf(" self Log10: %f\n math log10: %f\n\n", Log10(0.3), log10(0.3));
printf(" self Log10: %f\n math log10: %f\n\n", Log10(33.8), log10(33.8));
return 0;
}
倒霉的小三
answered 10 years, 9 months ago