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1e2fea6)
fix special cases, use multiplication instead of scalbnl
/* Table size */
#define NXT 32
/* Table size */
#define NXT 32
-/* log2(Table size) */
-#define LNXT 5
/* log(1+x) = x - .5x^2 + x^3 * P(z)/Q(z)
* on the domain 2^(-1/32) - 1 <= x <= 2^(1/32) - 1
/* log(1+x) = x - .5x^2 + x^3 * P(z)/Q(z)
* on the domain 2^(-1/32) - 1 <= x <= 2^(1/32) - 1
volatile long double z=0;
long double w=0, W=0, Wa=0, Wb=0, ya=0, yb=0, u=0;
volatile long double z=0;
long double w=0, W=0, Wa=0, Wb=0, ya=0, yb=0, u=0;
- if (y == 0.0)
- return 1.0;
- if (isnan(x))
+ /* make sure no invalid exception is raised by nan comparision */
+ if (isnan(x)) {
+ if (!isnan(y) && y == 0.0)
+ return 1.0;
+ }
+ if (isnan(y)) {
+ if (x == 1.0)
+ return 1.0;
+ }
+ if (x == 1.0)
+ return 1.0; /* 1**y = 1, even if y is nan */
+ if (x == -1.0 && !isfinite(y))
+ return 1.0; /* -1**inf = 1 */
+ if (y == 0.0)
+ return 1.0; /* x**0 = 1, even if x is nan */
-
- // FIXME: this is wrong, see pow special cases in c99 F.9.4.4
- if (!isfinite(y) && (x == -1.0 || x == 1.0) )
- return y - y; /* +-1**inf is NaN */
- if (x == 1.0)
- return 1.0;
+ if (x > 1.0 || x < -1.0)
- if (x > 0.0 && x < 1.0)
- return 0.0;
- if (x < -1.0)
- return INFINITY;
- if (x > -1.0 && x < 0.0)
return 0.0;
}
if (y <= -LDBL_MAX) {
return 0.0;
}
if (y <= -LDBL_MAX) {
+ if (x > 1.0 || x < -1.0)
- if (x > 0.0 && x < 1.0)
- return INFINITY;
- if (x < -1.0)
- return 0.0;
- if (x > -1.0 && x < 0.0)
return INFINITY;
}
if (x >= LDBL_MAX) {
return INFINITY;
}
if (x >= LDBL_MAX) {
/* Set iyflg to 1 if y is an integer. */
iyflg = 0;
if (w == y)
/* Set iyflg to 1 if y is an integer. */
iyflg = 0;
if (w == y)
-
-
- nflg = 0; /* flag = 1 if x<0 raised to integer power */
+ nflg = 0; /* (x<0)**(odd int) */
if (x <= 0.0) {
if (x == 0.0) {
if (y < 0.0) {
if (signbit(x) && yoddint)
if (x <= 0.0) {
if (x == 0.0) {
if (y < 0.0) {
if (signbit(x) && yoddint)
- return -INFINITY;
- return INFINITY;
+ /* (-0.0)**(-odd int) = -inf, divbyzero */
+ return -1.0/0.0;
+ /* (+-0.0)**(negative) = inf, divbyzero */
+ return 1.0/0.0;
- if (y > 0.0) {
- if (signbit(x) && yoddint)
- return -0.0;
- return 0.0;
- }
- if (y == 0.0)
- return 1.0; /* 0**0 */
- return 0.0; /* 0**y */
+ if (signbit(x) && yoddint)
+ return -0.0;
+ return 0.0;
}
if (iyflg == 0)
return (x - x) / (x - x); /* (x<0)**(non-int) is NaN */
}
if (iyflg == 0)
return (x - x) / (x - x); /* (x<0)**(non-int) is NaN */
+ /* (x<0)**(integer) */
+ if (yoddint)
+ nflg = 1; /* negate result */
+ x = -x;
-
- /* Integer power of an integer. */
- if (iyflg) {
- i = w;
- w = floorl(x);
- if (w == x && fabsl(y) < 32768.0) {
- w = powil(x, (int)y);
- return w;
- }
+ /* (+integer)**(integer) */
+ if (iyflg && floorl(x) == x && fabsl(y) < 32768.0) {
+ w = powil(x, (int)y);
+ return nflg ? -w : w;
- if (nflg)
- x = fabsl(x);
-
/* separate significand from exponent */
x = frexpl(x, &i);
e = i;
/* separate significand from exponent */
x = frexpl(x, &i);
e = i;
z += x;
/* Compute exponent term of the base 2 logarithm. */
z += x;
/* Compute exponent term of the base 2 logarithm. */
- w = -i;
- // TODO: use w * 0x1p-5;
- w = scalbnl(w, -LNXT); /* divide by NXT */
w += e;
/* Now base 2 log of x is w + z. */
w += e;
/* Now base 2 log of x is w + z. */
H = Fb + Gb;
Ha = reducl(H);
H = Fb + Gb;
Ha = reducl(H);
- w = scalbnl( Ga+Ha, LNXT );
/* Test the power of 2 for overflow */
if (w > MEXP)
/* Test the power of 2 for overflow */
if (w > MEXP)
z = z + w;
z = scalbnl(z, i); /* multiply by integer power of 2 */
z = z + w;
z = scalbnl(z, i); /* multiply by integer power of 2 */
- if (nflg) {
- /* For negative x,
- * find out if the integer exponent
- * is odd or even.
- */
- w = 0.5*y;
- w = floorl(w);
- w = 2.0*w;
- if (w != y)
- z = -z; /* odd exponent */
- }
-
-/* powil.c
- *
- * Real raised to integer power, long double precision
+/*
+ * Positive real raised to integer power, long double precision
- * Returns argument x raised to the nth power.
+ * Returns argument x>0 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
* 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
{
long double ww, y;
long double s;
{
long double ww, y;
long double s;
- int n, e, sign, asign, lx;
-
- if (x == 0.0) {
- if (nn == 0)
- return 1.0;
- else if (nn < 0)
- return LDBL_MAX;
- return 0.0;
- }
- if (x < 0.0) {
- asign = -1;
- x = -x;
- } else
- asign = 0;
-
if (nn < 0) {
sign = -1;
n = -nn;
if (nn < 0) {
sign = -1;
n = -nn;
/* First bit of the power */
if (n & 1)
y = x;
/* First bit of the power */
if (n & 1)
y = x;
- if (asign)
- y = -y; /* odd power of negative number */
if (sign < 0)
y = 1.0/y;
return y;
if (sign < 0)
y = 1.0/y;
return y;