*
* Relative error:
* arithmetic domain # trials peak rms
- * IEEE -45,+MAXLOG 200,000 1.2e-19 2.5e-20
- *
- * ERROR MESSAGES:
- *
- * message condition value returned
- * expm1l overflow x > MAXLOG MAXNUM
- *
+ * IEEE -45,+maxarg 200,000 1.2e-19 2.5e-20
*/
#include "libm.h"
return expm1(x);
}
#elif LDBL_MANT_DIG == 64 && LDBL_MAX_EXP == 16384
-static const long double MAXLOGL = 1.1356523406294143949492E4L;
/* exp(x) - 1 = x + 0.5 x^2 + x^3 P(x)/Q(x)
-.5 ln 2 < x < .5 ln 2
C2 = 1.428606820309417232121458176568075500134E-6L,
/* ln 2^-65 */
minarg = -4.5054566736396445112120088E1L,
-huge = 0x1p10000L;
+/* ln 2^16384 */
+maxarg = 1.1356523406294143949492E4L;
long double expm1l(long double x)
{
long double px, qx, xx;
int k;
- /* Overflow. */
- if (x > MAXLOGL)
- return huge*huge; /* overflow */
+ if (isnan(x))
+ return x;
+ if (x > maxarg)
+ return x*0x1p16383L; /* overflow, unless x==inf */
if (x == 0.0)
return x;
- /* Minimum value.*/
if (x < minarg)
- return -1.0L;
+ return -1.0;
xx = C1 + C2;
/* Express x = ln 2 (k + remainder), remainder not exceeding 1/2. */
- px = floorl (0.5 + x / xx);
+ px = floorl(0.5 + x / xx);
k = px;
/* remainder times ln 2 */
x -= px * C1;
/* exp(x) = exp(k ln 2) exp(remainder ln 2) = 2^k exp(remainder ln 2).
We have qx = exp(remainder ln 2) - 1, so
exp(x) - 1 = 2^k (qx + 1) - 1 = 2^k qx + 2^k - 1. */
- px = ldexpl(1.0L, k);
+ px = scalbnl(1.0, k);
x = px * qx + (px - 1.0);
return x;
}