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Monster Media 1996 #15
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POWI.C
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C/C++ Source or Header
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1995-12-12
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3KB
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163 lines
/* powi.c
*
* Real raised to integer power
*
*
*
* SYNOPSIS:
*
* double x, y, powi();
* int n;
*
* y = powi( x, n );
*
*
*
* DESCRIPTION:
*
* Returns argument x raised to the nth power.
* The routine efficiently decomposes n as a sum of powers of
* two. The desired power is a product of two-to-the-kth
* powers of x. Thus to compute the 32767 power of x requires
* 28 multiplications instead of 32767 multiplications.
*
*
*
* ACCURACY:
*
*
* Relative error:
* arithmetic x domain n domain # trials peak rms
* DEC .04,26 -26,26 100000 2.7e-16 4.3e-17
* IEEE .04,26 -26,26 50000 2.0e-15 3.8e-16
* IEEE 1,2 -1022,1023 50000 8.6e-14 1.6e-14
*
* Returns MAXNUM on overflow, zero on underflow.
*
*/
/* powi.c */
/*
Cephes Math Library Release 2.1: January, 1989
Copyright 1984, 1987, 1989 by Stephen L. Moshier
Direct inquiries to 30 Frost Street, Cambridge, MA 02140
*/
#include "mconf.h"
extern double MAXNUM, MAXLOG, MINLOG, LOGE2;
double powi( x, nn )
double x;
int nn;
{
int n, e, sign, asign, lx;
double w, y, s;
double log(), frexp();
if( x == 0.0 )
{
if( nn == 0 )
return( 1.0 );
else if( nn < 0 )
return( MAXNUM );
else
return( 0.0 );
}
if( nn == 0 )
return( 1.0 );
if( x < 0.0 )
{
asign = -1;
x = -x;
}
else
asign = 0;
if( nn < 0 )
{
sign = -1;
n = -nn;
}
else
{
sign = 1;
n = nn;
}
/* Overflow detection */
/* Calculate approximate logarithm of answer */
s = frexp( x, &lx );
e = (lx - 1)*n;
if( (e == 0) || (e > 64) || (e < -64) )
{
s = (s - 7.0710678118654752e-1) / (s + 7.0710678118654752e-1);
s = (2.9142135623730950 * s - 0.5 + lx) * nn * LOGE2;
}
else
{
s = LOGE2 * e;
}
if( s > MAXLOG )
{
mtherr( "powi", OVERFLOW );
y = MAXNUM;
goto done;
}
#if 0
if( s < MINLOG )
return(0.0);
/* Handle tiny denormal answer, but with less accuracy
* since roundoff error in 1.0/x will be amplified.
* The precise demarcation should be the gradual underflow threshold.
*/
if( (s < (-MAXLOG+2.0)) && (sign < 0) )
{
x = 1.0/x;
sign = -sign;
}
#else
/* do not produce denormal answer */
if( s < -MAXLOG )
return(0.0);
#endif
/* First bit of the power */
if( n & 1 )
y = x;
else
{
y = 1.0;
asign = 0;
}
w = x;
n >>= 1;
while( n )
{
w = w * w; /* arg to the 2-to-the-kth power */
if( n & 1 ) /* if that bit is set, then include in product */
y *= w;
n >>= 1;
}
done:
if( asign )
y = -y; /* odd power of negative number */
if( sign < 0 )
y = 1.0/y;
return(y);
}