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float 请大虾们帮忙:求C语言的float 类型数据的开根号算法??谁有相关资料、程序啊?
问
http://sourceforge.net/projects/msp...0030207.tar.bz2
中有libm目录,其中有ef_sqrt.c:
(或者ICC430这套工具下也有)
/* ef_sqrtf.c -- float version of e_sqrt.c.
* Conversion to float by Ian Lance Taylor, Cygnus Support, iancygnus.com.
*/
/*
* ====================================================
* Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
*
* Developed at SunPro, a Sun Microsystems, Inc. business.
* Permission to use, copy, modify, and distribute this
* software is freely granted, provided that this notice
* is preserved.
* ====================================================
*/
#include "fdlibm.h"
#ifdef __STDC__
static const float one = 1.0, tiny=1.0e-30;
#else
static float one = 1.0, tiny=1.0e-30;
#endif
#ifdef __STDC__
float __ieee754_sqrtf(float x)
#else
float __ieee754_sqrtf(x)
float x;
#endif
{
float z;
__int32_t sign = (__int32_t)0x80000000;
__uint32_t r,hx;
__int32_t ix,s,q,m,t,i;
GET_FLOAT_WORD(ix,x);
hx = ix&0x7fffffff;
/* take care of Inf and NaN */
if(!FLT_UWORD_IS_FINITE(hx))
return x*x+x; /* sqrt(NaN)=NaN, sqrt(+inf)=+inf
sqrt(-inf)=sNaN */
/* take care of zero and -ves */
if(FLT_UWORD_IS_ZERO(hx)) return x;/* sqrt(+-0) = +-0 */
if(ix<0) return (x-x)/(x-x); /* sqrt(-ve) = sNaN */
/* normalize x */
m = (ix>>23);
if(FLT_UWORD_IS_SUBNORMAL(hx)) { /* subnormal x */
for(i=0;(ix&0x00800000L)==0;i++) ix<<=1;
m -= i-1;
}
m -= 127; /* unbias exponent */
ix = (ix&0x007fffffL)|0x00800000L;
if(m&1) /* odd m, double x to make it even */
ix += ix;
m >>= 1; /* m = [m/2] */
/* generate sqrt(x) bit by bit */
ix += ix;
q = s = 0; /* q = sqrt(x) */
r = 0x01000000L; /* r = moving bit from right to left */
while(r!=0) {
t = s+r;
if(t<=ix) {
s = t+r;
ix -= t;
q += r;
}
ix += ix;
r>>=1;
}
/* use floating add to find out rounding direction */
if(ix!=0) {
z = one-tiny; /* trigger inexact flag */
if (z>=one) {
z = one+tiny;
if (z>one)
q += 2;
else
q += (q&1);
}
}
ix = (q>>1)+0x3f000000L;
ix += (m <<23);
SET_FLOAT_WORD(z,ix);
return z;
}
#ifdef __STDC__
float sqrt(float x)
#else
float sqrt(x)
float x;
#endif
{
return __ieee754_sqrtf(x);
} 答 1: 直接调用浮点库不可以吗?只要对执行时间没有特别快的要求,你可以在IAR中仿真评估一下耗时,看项目能不能接受。
中有libm目录,其中有ef_sqrt.c:
(或者ICC430这套工具下也有)
/* ef_sqrtf.c -- float version of e_sqrt.c.
* Conversion to float by Ian Lance Taylor, Cygnus Support, iancygnus.com.
*/
/*
* ====================================================
* Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
*
* Developed at SunPro, a Sun Microsystems, Inc. business.
* Permission to use, copy, modify, and distribute this
* software is freely granted, provided that this notice
* is preserved.
* ====================================================
*/
#include "fdlibm.h"
#ifdef __STDC__
static const float one = 1.0, tiny=1.0e-30;
#else
static float one = 1.0, tiny=1.0e-30;
#endif
#ifdef __STDC__
float __ieee754_sqrtf(float x)
#else
float __ieee754_sqrtf(x)
float x;
#endif
{
float z;
__int32_t sign = (__int32_t)0x80000000;
__uint32_t r,hx;
__int32_t ix,s,q,m,t,i;
GET_FLOAT_WORD(ix,x);
hx = ix&0x7fffffff;
/* take care of Inf and NaN */
if(!FLT_UWORD_IS_FINITE(hx))
return x*x+x; /* sqrt(NaN)=NaN, sqrt(+inf)=+inf
sqrt(-inf)=sNaN */
/* take care of zero and -ves */
if(FLT_UWORD_IS_ZERO(hx)) return x;/* sqrt(+-0) = +-0 */
if(ix<0) return (x-x)/(x-x); /* sqrt(-ve) = sNaN */
/* normalize x */
m = (ix>>23);
if(FLT_UWORD_IS_SUBNORMAL(hx)) { /* subnormal x */
for(i=0;(ix&0x00800000L)==0;i++) ix<<=1;
m -= i-1;
}
m -= 127; /* unbias exponent */
ix = (ix&0x007fffffL)|0x00800000L;
if(m&1) /* odd m, double x to make it even */
ix += ix;
m >>= 1; /* m = [m/2] */
/* generate sqrt(x) bit by bit */
ix += ix;
q = s = 0; /* q = sqrt(x) */
r = 0x01000000L; /* r = moving bit from right to left */
while(r!=0) {
t = s+r;
if(t<=ix) {
s = t+r;
ix -= t;
q += r;
}
ix += ix;
r>>=1;
}
/* use floating add to find out rounding direction */
if(ix!=0) {
z = one-tiny; /* trigger inexact flag */
if (z>=one) {
z = one+tiny;
if (z>one)
q += 2;
else
q += (q&1);
}
}
ix = (q>>1)+0x3f000000L;
ix += (m <<23);
SET_FLOAT_WORD(z,ix);
return z;
}
#ifdef __STDC__
float sqrt(float x)
#else
float sqrt(x)
float x;
#endif
{
return __ieee754_sqrtf(x);
} 答 1: 直接调用浮点库不可以吗?只要对执行时间没有特别快的要求,你可以在IAR中仿真评估一下耗时,看项目能不能接受。
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