rawmemchr.c 5.5 KB

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  1. /* Copyright (C) 1991,93,96,97,99,2000,2002 Free Software Foundation, Inc.
  2. This file is part of the GNU C Library.
  3. Based on strlen implementation by Torbjorn Granlund (tege@sics.se),
  4. with help from Dan Sahlin (dan@sics.se) and
  5. commentary by Jim Blandy (jimb@ai.mit.edu);
  6. adaptation to memchr suggested by Dick Karpinski (dick@cca.ucsf.edu),
  7. and implemented by Roland McGrath (roland@ai.mit.edu).
  8. The GNU C Library is free software; you can redistribute it and/or
  9. modify it under the terms of the GNU Lesser General Public
  10. License as published by the Free Software Foundation; either
  11. version 2.1 of the License, or (at your option) any later version.
  12. The GNU C Library is distributed in the hope that it will be useful,
  13. but WITHOUT ANY WARRANTY; without even the implied warranty of
  14. MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU
  15. Lesser General Public License for more details.
  16. You should have received a copy of the GNU Lesser General Public
  17. License along with the GNU C Library; if not, write to the Free
  18. Software Foundation, Inc., 59 Temple Place, Suite 330, Boston, MA
  19. 02111-1307 USA. */
  20. #include <string.h>
  21. #include <stdlib.h>
  22. #include <limits.h>
  23. #include "memcopy.h"
  24. #define LONG_MAX_32_BITS 2147483647
  25. #undef rawmemchr
  26. /* Find the first occurrence of C in S. */
  27. void attribute_hidden *__rawmemchr (const void * s, int c_in)
  28. {
  29. const unsigned char *char_ptr;
  30. const unsigned long int *longword_ptr;
  31. unsigned long int longword, magic_bits, charmask;
  32. unsigned reg_char c;
  33. c = (unsigned char) c_in;
  34. /* Handle the first few characters by reading one character at a time.
  35. Do this until CHAR_PTR is aligned on a longword boundary. */
  36. for (char_ptr = (const unsigned char *) s;
  37. ((unsigned long int) char_ptr & (sizeof (longword) - 1)) != 0;
  38. ++char_ptr)
  39. if (*char_ptr == c)
  40. return (void *) char_ptr;
  41. /* All these elucidatory comments refer to 4-byte longwords,
  42. but the theory applies equally well to 8-byte longwords. */
  43. longword_ptr = (unsigned long int *) char_ptr;
  44. /* Bits 31, 24, 16, and 8 of this number are zero. Call these bits
  45. the "holes." Note that there is a hole just to the left of
  46. each byte, with an extra at the end:
  47. bits: 01111110 11111110 11111110 11111111
  48. bytes: AAAAAAAA BBBBBBBB CCCCCCCC DDDDDDDD
  49. The 1-bits make sure that carries propagate to the next 0-bit.
  50. The 0-bits provide holes for carries to fall into. */
  51. if (sizeof (longword) != 4 && sizeof (longword) != 8)
  52. abort ();
  53. #if LONG_MAX <= LONG_MAX_32_BITS
  54. magic_bits = 0x7efefeff;
  55. #else
  56. magic_bits = ((unsigned long int) 0x7efefefe << 32) | 0xfefefeff;
  57. #endif
  58. /* Set up a longword, each of whose bytes is C. */
  59. charmask = c | (c << 8);
  60. charmask |= charmask << 16;
  61. #if LONG_MAX > LONG_MAX_32_BITS
  62. charmask |= charmask << 32;
  63. #endif
  64. /* Instead of the traditional loop which tests each character,
  65. we will test a longword at a time. The tricky part is testing
  66. if *any of the four* bytes in the longword in question are zero. */
  67. while (1)
  68. {
  69. /* We tentatively exit the loop if adding MAGIC_BITS to
  70. LONGWORD fails to change any of the hole bits of LONGWORD.
  71. 1) Is this safe? Will it catch all the zero bytes?
  72. Suppose there is a byte with all zeros. Any carry bits
  73. propagating from its left will fall into the hole at its
  74. least significant bit and stop. Since there will be no
  75. carry from its most significant bit, the LSB of the
  76. byte to the left will be unchanged, and the zero will be
  77. detected.
  78. 2) Is this worthwhile? Will it ignore everything except
  79. zero bytes? Suppose every byte of LONGWORD has a bit set
  80. somewhere. There will be a carry into bit 8. If bit 8
  81. is set, this will carry into bit 16. If bit 8 is clear,
  82. one of bits 9-15 must be set, so there will be a carry
  83. into bit 16. Similarly, there will be a carry into bit
  84. 24. If one of bits 24-30 is set, there will be a carry
  85. into bit 31, so all of the hole bits will be changed.
  86. The one misfire occurs when bits 24-30 are clear and bit
  87. 31 is set; in this case, the hole at bit 31 is not
  88. changed. If we had access to the processor carry flag,
  89. we could close this loophole by putting the fourth hole
  90. at bit 32!
  91. So it ignores everything except 128's, when they're aligned
  92. properly.
  93. 3) But wait! Aren't we looking for C, not zero?
  94. Good point. So what we do is XOR LONGWORD with a longword,
  95. each of whose bytes is C. This turns each byte that is C
  96. into a zero. */
  97. longword = *longword_ptr++ ^ charmask;
  98. /* Add MAGIC_BITS to LONGWORD. */
  99. if ((((longword + magic_bits)
  100. /* Set those bits that were unchanged by the addition. */
  101. ^ ~longword)
  102. /* Look at only the hole bits. If any of the hole bits
  103. are unchanged, most likely one of the bytes was a
  104. zero. */
  105. & ~magic_bits) != 0)
  106. {
  107. /* Which of the bytes was C? If none of them were, it was
  108. a misfire; continue the search. */
  109. const unsigned char *cp = (const unsigned char *) (longword_ptr - 1);
  110. if (cp[0] == c)
  111. return (void *) cp;
  112. if (cp[1] == c)
  113. return (void *) &cp[1];
  114. if (cp[2] == c)
  115. return (void *) &cp[2];
  116. if (cp[3] == c)
  117. return (void *) &cp[3];
  118. #if LONG_MAX > 2147483647
  119. if (cp[4] == c)
  120. return (void *) &cp[4];
  121. if (cp[5] == c)
  122. return (void *) &cp[5];
  123. if (cp[6] == c)
  124. return (void *) &cp[6];
  125. if (cp[7] == c)
  126. return (void *) &cp[7];
  127. #endif
  128. }
  129. }
  130. }
  131. strong_alias(__rawmemchr, rawmemchr)