Eric Blake asked:
> Also, is anyone interested in making gnulib's memchr and strchrnul more
> efficient by copying the optimizations learned in memchr2?
Checked in like this:
2008-04-26 Eric Blake <[EMAIL PROTECTED]>
Bruno Haible <[EMAIL PROTECTED]>
* lib/memchr.c: Include intprops.h.
(__memchr): Optimize parallel detection of matching bytes. Rename local
variables. Add explanatory comments.
*** lib/memchr.c.orig 2008-04-26 12:34:09.000000000 +0200
--- lib/memchr.c 2008-04-26 12:32:30.000000000 +0200
***************
*** 45,50 ****
--- 45,52 ----
# define BP_SYM(sym) sym
#endif
+ #include "intprops.h"
+
#undef __memchr
#ifdef _LIBC
# undef memchr
***************
*** 58,205 ****
void *
__memchr (void const *s, int c_in, size_t n)
{
const unsigned char *char_ptr;
! const unsigned long int *longword_ptr;
! unsigned long int longword, magic_bits, charmask;
unsigned reg_char c;
- int i;
c = (unsigned char) c_in;
! /* Handle the first few characters by reading one character at a time.
Do this until CHAR_PTR is aligned on a longword boundary. */
for (char_ptr = (const unsigned char *) s;
! n > 0 && (size_t) char_ptr % sizeof longword != 0;
--n, ++char_ptr)
if (*char_ptr == c)
return (void *) char_ptr;
/* All these elucidatory comments refer to 4-byte longwords,
but the theory applies equally well to any size longwords. */
! longword_ptr = (const unsigned long int *) char_ptr;
!
! /* Bits 31, 24, 16, and 8 of this number are zero. Call these bits
! the "holes." Note that there is a hole just to the left of
! each byte, with an extra at the end:
!
! bits: 01111110 11111110 11111110 11111111
! bytes: AAAAAAAA BBBBBBBB CCCCCCCC DDDDDDDD
!
! The 1-bits make sure that carries propagate to the next 0-bit.
! The 0-bits provide holes for carries to fall into. */
!
! /* Set MAGIC_BITS to be this pattern of 1 and 0 bits.
! Set CHARMASK to be a longword, each of whose bytes is C. */
!
! magic_bits = 0xfefefefe;
! charmask = c | (c << 8);
! charmask |= charmask << 16;
! #if 0xffffffffU < ULONG_MAX
! magic_bits |= magic_bits << 32;
! charmask |= charmask << 32;
! if (8 < sizeof longword)
! for (i = 64; i < sizeof longword * 8; i *= 2)
! {
! magic_bits |= magic_bits << i;
! charmask |= charmask << i;
! }
! #endif
! magic_bits = (ULONG_MAX >> 1) & (magic_bits | 1);
!
! /* Instead of the traditional loop which tests each character,
! we will test a longword at a time. The tricky part is testing
! if *any of the four* bytes in the longword in question are zero. */
! while (n >= sizeof longword)
{
! /* We tentatively exit the loop if adding MAGIC_BITS to
! LONGWORD fails to change any of the hole bits of LONGWORD.
!
! 1) Is this safe? Will it catch all the zero bytes?
! Suppose there is a byte with all zeros. Any carry bits
! propagating from its left will fall into the hole at its
! least significant bit and stop. Since there will be no
! carry from its most significant bit, the LSB of the
! byte to the left will be unchanged, and the zero will be
! detected.
!
! 2) Is this worthwhile? Will it ignore everything except
! zero bytes? Suppose every byte of LONGWORD has a bit set
! somewhere. There will be a carry into bit 8. If bit 8
! is set, this will carry into bit 16. If bit 8 is clear,
! one of bits 9-15 must be set, so there will be a carry
! into bit 16. Similarly, there will be a carry into bit
! 24. If one of bits 24-30 is set, there will be a carry
! into bit 31, so all of the hole bits will be changed.
!
! The one misfire occurs when bits 24-30 are clear and bit
! 31 is set; in this case, the hole at bit 31 is not
! changed. If we had access to the processor carry flag,
! we could close this loophole by putting the fourth hole
! at bit 32!
!
! So it ignores everything except 128's, when they're aligned
! properly.
!
! 3) But wait! Aren't we looking for C, not zero?
! Good point. So what we do is XOR LONGWORD with a longword,
! each of whose bytes is C. This turns each byte that is C
! into a zero. */
!
! longword = *longword_ptr++ ^ charmask;
!
! /* Add MAGIC_BITS to LONGWORD. */
! if ((((longword + magic_bits)
!
! /* Set those bits that were unchanged by the addition. */
! ^ ~longword)
!
! /* Look at only the hole bits. If any of the hole bits
! are unchanged, most likely one of the bytes was a
! zero. */
! & ~magic_bits) != 0)
{
! /* Which of the bytes was C? If none of them were, it was
! a misfire; continue the search. */
! const unsigned char *cp = (const unsigned char *) (longword_ptr - 1);
!
! if (cp[0] == c)
! return (void *) cp;
! if (cp[1] == c)
! return (void *) &cp[1];
! if (cp[2] == c)
! return (void *) &cp[2];
! if (cp[3] == c)
! return (void *) &cp[3];
! if (4 < sizeof longword && cp[4] == c)
! return (void *) &cp[4];
! if (5 < sizeof longword && cp[5] == c)
! return (void *) &cp[5];
! if (6 < sizeof longword && cp[6] == c)
! return (void *) &cp[6];
! if (7 < sizeof longword && cp[7] == c)
! return (void *) &cp[7];
! if (8 < sizeof longword)
! for (i = 8; i < sizeof longword; i++)
! if (cp[i] == c)
! return (void *) &cp[i];
}
! n -= sizeof longword;
}
char_ptr = (const unsigned char *) longword_ptr;
! while (n-- > 0)
{
if (*char_ptr == c)
return (void *) char_ptr;
- else
- ++char_ptr;
}
! return 0;
}
#ifdef weak_alias
weak_alias (__memchr, BP_SYM (memchr))
--- 60,173 ----
void *
__memchr (void const *s, int c_in, size_t n)
{
+ /* On 32-bit hardware, choosing longword to be a 32-bit unsigned
+ long instead of a 64-bit uintmax_t tends to give better
+ performance. On 64-bit hardware, unsigned long is generally 64
+ bits already. Change this typedef to experiment with
+ performance. */
+ typedef unsigned long longword;
+
const unsigned char *char_ptr;
! const longword *longword_ptr;
! longword repeated_one;
! longword repeated_c;
unsigned reg_char c;
c = (unsigned char) c_in;
! /* Handle the first few bytes by reading one byte at a time.
Do this until CHAR_PTR is aligned on a longword boundary. */
for (char_ptr = (const unsigned char *) s;
! n > 0 && (size_t) char_ptr % sizeof (longword) != 0;
--n, ++char_ptr)
if (*char_ptr == c)
return (void *) char_ptr;
+ longword_ptr = (const longword *) char_ptr;
+
/* All these elucidatory comments refer to 4-byte longwords,
but the theory applies equally well to any size longwords. */
! /* Compute auxiliary longword values:
! repeated_one is a value which has a 1 in every byte.
! repeated_c has c in every byte. */
! repeated_one = 0x01010101;
! repeated_c = c | (c << 8);
! repeated_c |= repeated_c << 16;
! if (0xffffffffU < TYPE_MAXIMUM (longword))
{
! repeated_one |= repeated_one << 31 << 1;
! repeated_c |= repeated_c << 31 << 1;
! if (8 < sizeof (longword))
{
! int i;
! for (i = 64; i < sizeof (longword) * 8; i *= 2)
! {
! repeated_one |= repeated_one << i;
! repeated_c |= repeated_c << i;
! }
}
+ }
+
+ /* Instead of the traditional loop which tests each byte, we will test a
+ longword at a time. The tricky part is testing if *any of the four*
+ bytes in the longword in question are equal to c. We first use an xor
+ with repeated_c. This reduces the task to testing whether *any of the
+ four* bytes in longword1 is zero.
+
+ We compute tmp =
+ ((longword1 - repeated_one) & ~longword1) & (repeated_one << 7).
+ That is, we perform the following operations:
+ 1. Subtract repeated_one.
+ 2. & ~longword1.
+ 3. & a mask consisting of 0x80 in every byte.
+ Consider what happens in each byte:
+ - If a byte of longword1 is zero, step 1 and 2 transform it into 0xff,
+ and step 3 transforms it into 0x80. A carry can also be propagated
+ to more significant bytes.
+ - If a byte of longword1 is nonzero, let its lowest 1 bit be at
+ position k (0 <= k <= 7); so the lowest k bits are 0. After step 1,
+ the byte ends in a single bit of value 0 and k bits of value 1.
+ After step 2, the result is just k bits of value 1: 2^k - 1. After
+ step 3, the result is 0. And no carry is produced.
+ So, if longword1 has only non-zero bytes, tmp is zero.
+ Whereas if longword1 has a zero byte, call j the position of the least
+ significant zero byte. Then the result has a zero at positions 0, ...,
+ j-1 and a 0x80 at position j. We cannot predict the result at the more
+ significant bytes (positions j+1..3), but it does not matter since we
+ already have a non-zero bit at position 8*j+7.
+
+ So, the test whether any byte in longword1 is zero is equivalent to
+ testing whether tmp is nonzero. */
! while (n >= sizeof (longword))
! {
! longword longword1 = *longword_ptr ^ repeated_c;
!
! if ((((longword1 - repeated_one) & ~longword1)
! & (repeated_one << 7)) != 0)
! break;
! longword_ptr++;
! n -= sizeof (longword);
}
char_ptr = (const unsigned char *) longword_ptr;
! /* At this point, we know that either n < sizeof (longword), or one of the
! sizeof (longword) bytes starting at char_ptr is == c. On little-endian
! machines, we could determine the first such byte without any further
! memory accesses, just by looking at the tmp result from the last loop
! iteration. But this does not work on big-endian machines. Choose code
! that works in both cases. */
!
! for (; n > 0; --n, ++char_ptr)
{
if (*char_ptr == c)
return (void *) char_ptr;
}
! return NULL;
}
#ifdef weak_alias
weak_alias (__memchr, BP_SYM (memchr))
*** modules/memchr.orig 2008-04-26 12:34:09.000000000 +0200
--- modules/memchr 2008-04-26 12:32:02.000000000 +0200
***************
*** 6,11 ****
--- 6,12 ----
m4/memchr.m4
Depends-on:
+ intprops
configure.ac:
gl_FUNC_MEMCHR
