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📄 atanf.c

📁 sdcc是为51等小型嵌入式cpu设计的c语言编译器支持数种不同类型的cpu
💻 C
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/*  atanf.c: Computes arctan of a 32-bit float as outlined in [1]    Copyright (C) 2001, 2002  Jesus Calvino-Fraga, jesusc@ieee.org     This library is free software; you can redistribute it and/or    modify it under the terms of the GNU Lesser General Public    License as published by the Free Software Foundation; either    version 2.1 of the License, or (at your option) any later version.    This library is distributed in the hope that it will be useful,    but WITHOUT ANY WARRANTY; without even the implied warranty of    MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the GNU    Lesser General Public License for more details.    You should have received a copy of the GNU Lesser General Public    License along with this library; if not, write to the Free Software    Foundation, Inc., 59 Temple Place, Suite 330, Boston, MA  02111-1307 USA *//* [1] William James Cody and W.  M.  Waite.  _Software manual for the   elementary functions_, Englewood Cliffs, N.J.:Prentice-Hall, 1980. *//* Version 1.0 - Initial release *//*** $Id: atanf.c 3654 2005-01-24 10:38:53Z vrokas $*/#include <math.h>#include <errno.h>#define P0 -0.4708325141E+0#define P1 -0.5090958253E-1#define Q0  0.1412500740E+1#define Q1  0.1000000000E+1#define P(g,f) ((P1*g+P0)*g*f)#define Q(g) (Q1*g+Q0)#define K1  0.2679491924 /* 2-sqrt(3) */#define K2  0.7320508076 /* sqrt(3)-1 */#define K3  1.7320508076 /* sqrt(3)   */#ifdef SDCC_mcs51   #define myconst code#else   #define myconst const#endiffloat atanf(const float x) _MATH_REENTRANT{    float f, r, g;    int n=0;	static myconst float a[]={  0.0, 0.5235987756, 1.5707963268, 1.0471975512 };    f=fabsf(x);    if(f>1.0)    {        f=1.0/f;        n=2;    }    if(f>K1)    {        f=((K2*f-1.0)+f)/(K3+f);        // What it is actually wanted is this more accurate formula,        // but SDCC optimizes it and then it does not work:        // f=(((K2*f-0.5)-0.5)+f)/(K3+f);        n++;    }    if(fabsf(f)<EPS) r=f;    else    {        g=f*f;        r=f+P(g,f)/Q(g);    }    if(n>1) r=-r;    r+=a[n];    if(x<0.0) r=-r;    return r;}

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