double powi( x, nn ) double x; int nn;
{ int n, e, sign, asign, lx; double w, y, s;
/* See pow.c for these tests. */ if( x == 0.0 )
{ if( nn == 0 ) return( 1.0 ); elseif( nn < 0 ) return( INFINITY ); else
{ if( nn & 1 ) return( x ); else return( 0.0 );
}
}
if( nn == 0 ) return( 1.0 );
if( nn == -1 ) return( 1.0/x );
if( x < 0.0 )
{
asign = -1;
x = -x;
} else
asign = 0;
if( nn < 0 )
{
sign = -1;
n = -nn;
} else
{
sign = 1;
n = nn;
}
/* Even power will be positive. */ if( (n & 1) == 0 )
asign = 0;
/* 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 = INFINITY; goto done;
}
#if DENORMAL if( s < MINLOG )
{
y = 0.0; goto done;
}
/* Handle tiny denormal answer, but with less accuracy *sinceroundofferrorin1.0/xwillbeamplified. *Theprecisedemarcationshouldbethegradualunderflowthreshold.
*/ 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;
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;
}
if( sign < 0 )
y = 1.0/y;
done:
if( asign )
{ /* odd power of negative number */ if( y == 0.0 )
y = NEGZERO; else
y = -y;
} return(y);
}
Messung V0.5 in Prozent
¤ Dauer der Verarbeitung: 0.11 Sekunden
(vorverarbeitet am 2026-06-17)
¤
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