@@ -25,138 +25,74 @@
#endif
#include <string.h>
-
#include <stddef.h>
-
-#include <limits.h>
+#include <stdint.h>
+#include <string-fzb.h>
+#include <string-fzi.h>
+#include <string-opthr.h>
#undef __memchr
-#ifdef _LIBC
-# undef memchr
-#endif
+#undef memchr
-#ifndef weak_alias
-# define __memchr memchr
-#endif
-
-#ifndef MEMCHR
-# define MEMCHR __memchr
+#ifdef MEMCHR
+#define __memchr MEMCHR
#endif
/* Search no more than N bytes of S for C. */
void *
-MEMCHR (void const *s, int c_in, size_t n)
+__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 int longword;
-
const unsigned char *char_ptr;
- const longword *longword_ptr;
- longword repeated_one;
- longword repeated_c;
+ const op_t *word_ptr;
+ op_t word, repeated_c;
unsigned char c;
+ uintptr_t i, align;
c = (unsigned char) c_in;
+ char_ptr = (const unsigned char *) s;
/* 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)
+ Do this until CHAR_PTR is aligned on a word boundary, or
+ the entirety of small inputs. */
+ align = -(uintptr_t)char_ptr % sizeof (word);
+ if (n < OP_T_THRES || align > n)
+ align = n;
+ for (i = 0; i < align; ++i, ++char_ptr)
if (*char_ptr == c)
return (void *) char_ptr;
- longword_ptr = (const longword *) char_ptr;
+ /* Set up a word, each of whose bytes is C. */
+ repeated_c = ((op_t)-1 / 0xff) * c;
- /* All these elucidatory comments refer to 4-byte longwords,
- but the theory applies equally well to any size longwords. */
+ word_ptr = (const op_t *) char_ptr;
+ n -= align;
+ if (__glibc_unlikely (n == 0))
+ return NULL;
- /* 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 < (longword) -1)
+ /* Loop while we have more than one word remaining. */
+ while (n > sizeof (word))
{
- repeated_one |= repeated_one << 31 << 1;
- repeated_c |= repeated_c << 31 << 1;
- if (8 < sizeof (longword))
- {
- size_t i;
-
- for (i = 64; i < sizeof (longword) * 8; i *= 2)
- {
- repeated_one |= repeated_one << i;
- repeated_c |= repeated_c << i;
- }
- }
+ word = *word_ptr;
+ if (has_eq (word, repeated_c))
+ goto found;
+ word_ptr++;
+ n -= sizeof (word);
}
- /* 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))
+ /* Since our pointer is aligned, we can always read the last word. */
+ word = *word_ptr;
+ if (has_eq (word, repeated_c))
{
- 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;
+ found:
+ i = index_first_eq (word, repeated_c);
+ if (i < n)
+ return (char *) word_ptr + i;
}
return NULL;
}
-#ifdef weak_alias
+
+#ifndef MEMCHR
weak_alias (__memchr, memchr)
-#endif
libc_hidden_builtin_def (memchr)
+#endif