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* Update the comment that explains the choice of values in the

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table and the requirement on trailing zero bits.

* Remove the __aligned() compiler directives as these were found
  to have a negative effect on the produced code.

Submitted by:	bde
Approved by:	das (mentor)
This commit is contained in:
kargl 2012-10-13 19:53:11 +00:00
parent f9c7822213
commit e323b9460e
1 changed files with 7 additions and 6 deletions
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13
lib/msun/ld80/s_expl.c
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@ -64,7 +64,7 @@ o_threshold = LD80C(0xb17217f7d1cf79ab, 13, 11356.5234062941439488L),
/* log(2**(-16381-64-1)) rounded towards zero: */
u_threshold = LD80C(0xb21dfe7f09e2baa9, 13, -11399.4985314888605581L);
static const double __aligned(64)
static const double
/*
* ln2/INTERVALS = L1+L2 (hi+lo decomposition for multiplication). L1 must
* have at least 22 (= log2(|LDBL_MIN_EXP-extras|) + log2(INTERVALS)) lowest
@ -86,17 +86,18 @@ P6 = 1.3888891738560272e-3; /* 0x16c16c651633ae.0p-62 */
/*
* 2^(i/INTERVALS) for i in [0,INTERVALS] is represented by two values where
* the first 53 bits of the significand is stored in hi and the next 53
* bits are in lo. Tang's paper states that the trailing 6 bits of hi should
* the first 53 bits of the significand are stored in hi and the next 53
* bits are in lo. Tang's paper states that the trailing 6 bits of hi must
* be zero for his algorithm in both single and double precision, because
* the table is re-used in the implementation of expm1() where a floating
* point addition involving hi must be exact. The conversion of a 53-bit
* double into a 64-bit long double gives 11 trailing bit, which are zero.
* point addition involving hi must be exact. Here hi is double, so
* converting it to long double gives 11 trailing zero bits.
*/
static const struct {
double hi;
double lo;
} s[INTERVALS] __aligned(16) = {
/* XXX should rename 's'. */
} s[INTERVALS] = {
0x1p+0, 0x0p+0,
0x1.0163da9fb3335p+0, 0x1.b61299ab8cdb7p-54,
0x1.02c9a3e778060p+0, 0x1.dcdef95949ef4p-53,