-/* Copyright (C) 1991, 1993, 1996, 1997, 1999, 2000, 2003, 2004, 2006,
- 2008 Free Software Foundation, Inc.
+/* Copyright (C) 1991, 1993, 1996-1997, 1999-2000, 2003-2004, 2006, 2008-2014
+ Free Software Foundation, Inc.
Based on strlen implementation by Torbjorn Granlund (tege@sics.se),
with help from Dan Sahlin (dan@sics.se) and
void *
memchr2 (void const *s, int c1_in, int c2_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 int longword;
+
const unsigned char *char_ptr;
- const uintmax_t *longword_ptr;
- uintmax_t longword1;
- uintmax_t longword2;
- uintmax_t magic_bits;
- uintmax_t charmask1;
- uintmax_t charmask2;
+ void const *void_ptr;
+ const longword *longword_ptr;
+ longword repeated_one;
+ longword repeated_c1;
+ longword repeated_c2;
unsigned char c1;
unsigned char c2;
- int i;
c1 = (unsigned char) c1_in;
c2 = (unsigned char) c2_in;
if (c1 == c2)
return memchr (s, c1, n);
- /* 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 longword1 != 0;
- --n, ++char_ptr)
- if (*char_ptr == c1 || *char_ptr == c2)
- return (void *) char_ptr;
+ /* Handle the first few bytes by reading one byte at a time.
+ Do this until VOID_PTR is aligned on a longword boundary. */
+ for (void_ptr = s;
+ n > 0 && (uintptr_t) void_ptr % sizeof (longword) != 0;
+ --n)
+ {
+ char_ptr = void_ptr;
+ if (*char_ptr == c1 || *char_ptr == c2)
+ return (void *) void_ptr;
+ void_ptr = char_ptr + 1;
+ }
+
+ longword_ptr = void_ptr;
/* All these elucidatory comments refer to 4-byte longwords,
but the theory applies equally well to any size longwords. */
- longword_ptr = (const uintmax_t *) 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;
- charmask1 = c1 | (c1 << 8);
- charmask2 = c2 | (c2 << 8);
- charmask1 |= charmask1 << 16;
- charmask2 |= charmask2 << 16;
-#if 0xffffffffU < UINTMAX_MAX
- magic_bits |= magic_bits << 32;
- charmask1 |= charmask1 << 32;
- charmask2 |= charmask2 << 32;
- if (8 < sizeof longword1)
- for (i = 64; i < sizeof longword1 * 8; i *= 2)
- {
- magic_bits |= magic_bits << i;
- charmask1 |= charmask1 << i;
- charmask2 |= charmask2 << i;
- }
-#endif
- magic_bits = (UINTMAX_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 longword1)
+ /* Compute auxiliary longword values:
+ repeated_one is a value which has a 1 in every byte.
+ repeated_c1 has c1 in every byte.
+ repeated_c2 has c2 in every byte. */
+ repeated_one = 0x01010101;
+ repeated_c1 = c1 | (c1 << 8);
+ repeated_c2 = c2 | (c2 << 8);
+ repeated_c1 |= repeated_c1 << 16;
+ repeated_c2 |= repeated_c2 << 16;
+ if (0xffffffffU < (longword) -1)
{
- /* 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. */
-
- longword1 = *longword_ptr ^ charmask1;
- longword2 = *longword_ptr++ ^ charmask2;
-
- /* Add MAGIC_BITS to LONGWORD. */
- if ((((longword1 + magic_bits)
-
- /* Set those bits that were unchanged by the addition. */
- ^ ~longword1)
-
- /* 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
- || (((longword2 + magic_bits) ^ ~longword2) & ~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] == c1 || cp[0] == c2)
- return (void *) cp;
- if (cp[1] == c1 || cp[1] == c2)
- return (void *) &cp[1];
- if (cp[2] == c1 || cp[2] == c2)
- return (void *) &cp[2];
- if (cp[3] == c1 || cp[3] == c2)
- return (void *) &cp[3];
- if (4 < sizeof longword1 && (cp[4] == c1 || cp[4] == c2))
- return (void *) &cp[4];
- if (5 < sizeof longword1 && (cp[5] == c1 || cp[5] == c2))
- return (void *) &cp[5];
- if (6 < sizeof longword1 && (cp[6] == c1 || cp[6] == c2))
- return (void *) &cp[6];
- if (7 < sizeof longword1 && (cp[7] == c1 || cp[7] == c2))
- return (void *) &cp[7];
- if (8 < sizeof longword1)
- for (i = 8; i < sizeof longword1; i++)
- if (cp[i] == c1 || cp[i] == c2)
- return (void *) &cp[i];
- }
-
- n -= sizeof longword1;
+ repeated_one |= repeated_one << 31 << 1;
+ repeated_c1 |= repeated_c1 << 31 << 1;
+ repeated_c2 |= repeated_c2 << 31 << 1;
+ if (8 < sizeof (longword))
+ {
+ size_t i;
+
+ for (i = 64; i < sizeof (longword) * 8; i *= 2)
+ {
+ repeated_one |= repeated_one << i;
+ repeated_c1 |= repeated_c1 << i;
+ repeated_c2 |= repeated_c2 << 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 c1 or c2. We first use
+ an xor with repeated_c1 and repeated_c2, respectively. This reduces
+ the task to testing whether *any of the four* bytes in longword1 or
+ longword2 is zero.
+
+ Let's consider longword1. We compute tmp1 =
+ ((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, tmp1 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.
+
+ Similarly, we compute tmp2 =
+ ((longword2 - repeated_one) & ~longword2) & (repeated_one << 7).
+
+ The test whether any byte in longword1 or longword2 is zero is equivalent
+ to testing whether tmp1 is nonzero or tmp2 is nonzero. We can combine
+ this into a single test, whether (tmp1 | tmp2) is nonzero. */
+
+ while (n >= sizeof (longword))
+ {
+ longword longword1 = *longword_ptr ^ repeated_c1;
+ longword longword2 = *longword_ptr ^ repeated_c2;
+
+ if (((((longword1 - repeated_one) & ~longword1)
+ | ((longword2 - repeated_one) & ~longword2))
+ & (repeated_one << 7)) != 0)
+ break;
+ longword_ptr++;
+ n -= sizeof (longword);
}
char_ptr = (const unsigned char *) longword_ptr;
- while (n-- > 0)
+ /* At this point, we know that either n < sizeof (longword), or one of the
+ sizeof (longword) bytes starting at char_ptr is == c1 or == c2. On
+ little-endian machines, we could determine the first such byte without
+ any further memory accesses, just by looking at the (tmp1 | tmp2) 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 == c1 || *char_ptr == c2)
- return (void *) char_ptr;
- ++char_ptr;
+ return (void *) char_ptr;
}
- return 0;
+ return NULL;
}
projects / gnulib.git / blobdiff