/*
* e_expf.c - single-precision exp function
*
* Copyright (c) 2009-2018, Arm Limited.
* SPDX-License-Identifier: MIT
*/
/*
* Algorithm was once taken from Cody & Waite, but has been munged
* out of all recognition by SGT.
*/
#include <math.h>
#include <errno.h>
#include "math_private.h"
float
expf(float X)
{
int N; float XN, g, Rg, Result;
unsigned ix = fai(X), edgecaseflag = 0;
/*
* Handle infinities, NaNs and big numbers.
*/
if (__builtin_expect((ix << 1) - 0x67000000 > 0x85500000 - 0x67000000, 0)) {
if (!(0x7f800000 & ~ix)) {
if (ix == 0xff800000)
return 0.0f;
else
return FLOAT_INFNAN(X);/* do the right thing with both kinds of NaN and with +inf */
} else if ((ix << 1) < 0x67000000) {
return 1.0f; /* magnitude so small the answer can't be distinguished from 1 */
} else if ((ix << 1) > 0x85a00000) {
__set_errno(ERANGE);
if (ix & 0x80000000) {
return FLOAT_UNDERFLOW;
} else {
return FLOAT_OVERFLOW;
}
} else {
edgecaseflag = 1;
}
}
/*
* Split the input into an integer multiple of log(2)/4, and a
* fractional part.
*/
XN = X * (4.0f*1.4426950408889634074f);
#ifdef __TARGET_FPU_SOFTVFP
XN = _frnd(XN);
N = (int)XN;
#else
N = (int)(XN + (ix & 0x80000000 ? -0.5f : 0.5f));
XN = N;
#endif
g = (X - XN * 0x1.62ep-3F) - XN * 0x1.0bfbe8p-17F; /* prec-and-a-half representation of log(2)/4 */
/*
* Now we compute exp(X) in, conceptually, three parts:
* - a pure power of two which we get from N>>2
* - exp(g) for g in [-log(2)/8,+log(2)/8], which we compute
* using a Remez-generated polynomial approximation
* - exp(k*log(2)/4) (aka 2^(k/4)) for k in [0..3], which we
* get from a lookup table in precision-and-a-half and
* multiply by g.
*
* We gain a bit of extra precision by the fact that actually
* our polynomial approximation gives us exp(g)-1, and we add
* the 1 back on by tweaking the prec-and-a-half multiplication
* step.
*
* Coefficients generated by the command
./auxiliary/remez.jl --variable=g --suffix=f -- '-log(BigFloat(2))/8' '+log(BigFloat(2))/8' 3 0 '(expm1(x))/x'
*/
Rg = g * (
9.999999412829185331953781321128516523408059996430919985217971370689774264850229e-01f+g*(4.999999608551332693833317084753864837160947932961832943901913087652889900683833e-01f+g*(1.667292360203016574303631953046104769969440903672618034272397630620346717392378e-01f+g*(4.168230895653321517750133783431970715648192153539929404872173693978116154823859e-02f)))
);
/*
* Do the table lookup and combine it with Rg, to get our final
* answer apart from the exponent.
*/
{
static const float twotokover4top[4] = { 0x1p+0F, 0x1.306p+0F, 0x1.6ap+0F, 0x1.ae8p+0F };
static const float twotokover4bot[4] = { 0x0p+0F, 0x1.fc1464p-13F, 0x1.3cccfep-13F, 0x1.3f32b6p-13F };
static const float twotokover4all[4] = { 0x1p+0F, 0x1.306fep+0F, 0x1.6a09e6p+0F, 0x1.ae89fap+0F };
int index = (N & 3);
Rg = twotokover4top[index] + (twotokover4bot[index] + twotokover4all[index]*Rg);
N >>= 2;
}
/*
* Combine the output exponent and mantissa, and return.
*/
if (__builtin_expect(edgecaseflag, 0)) {
Result = fhex(((N/2) << 23) + 0x3f800000);
Result *= Rg;
Result *= fhex(((N-N/2) << 23) + 0x3f800000);
/*
* Step not mentioned in C&W: set errno reliably.
*/
if (fai(Result) == 0)
return MATHERR_EXPF_UFL(Result);
if (fai(Result) == 0x7f800000)
return MATHERR_EXPF_OFL(Result);
return FLOAT_CHECKDENORM(Result);
} else {
Result = fhex(N * 8388608.0f + (float)0x3f800000);
Result *= Rg;
}
return Result;
}