strlen.c 5.1 KB

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  1. /* Copyright (C) 1991, 1993, 1997, 2000, 2003 Free Software Foundation, Inc.
  2. This file is part of the GNU C Library.
  3. Written by Torbjorn Granlund (tege@sics.se),
  4. with help from Dan Sahlin (dan@sics.se);
  5. commentary by Jim Blandy (jimb@ai.mit.edu).
  6. The GNU C Library is free software; you can redistribute it and/or
  7. modify it under the terms of the GNU Lesser General Public
  8. License as published by the Free Software Foundation; either
  9. version 2.1 of the License, or (at your option) any later version.
  10. The GNU C Library is distributed in the hope that it will be useful,
  11. but WITHOUT ANY WARRANTY; without even the implied warranty of
  12. MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU
  13. Lesser General Public License for more details.
  14. You should have received a copy of the GNU Lesser General Public
  15. License along with the GNU C Library; if not, write to the Free
  16. Software Foundation, Inc., 59 Temple Place, Suite 330, Boston, MA
  17. 02111-1307 USA. */
  18. #include <string.h>
  19. #include <stdlib.h>
  20. libc_hidden_proto(strlen)
  21. libc_hidden_proto(abort)
  22. /* Return the length of the null-terminated string STR. Scan for
  23. the null terminator quickly by testing four bytes at a time. */
  24. size_t strlen (const char *str)
  25. {
  26. const char *char_ptr;
  27. const unsigned long int *longword_ptr;
  28. unsigned long int longword, magic_bits, himagic, lomagic;
  29. /* Handle the first few characters by reading one character at a time.
  30. Do this until CHAR_PTR is aligned on a longword boundary. */
  31. for (char_ptr = str; ((unsigned long int) char_ptr
  32. & (sizeof (longword) - 1)) != 0;
  33. ++char_ptr)
  34. if (*char_ptr == '\0')
  35. return char_ptr - str;
  36. /* All these elucidatory comments refer to 4-byte longwords,
  37. but the theory applies equally well to 8-byte longwords. */
  38. longword_ptr = (unsigned long int *) char_ptr;
  39. /* Bits 31, 24, 16, and 8 of this number are zero. Call these bits
  40. the "holes." Note that there is a hole just to the left of
  41. each byte, with an extra at the end:
  42. bits: 01111110 11111110 11111110 11111111
  43. bytes: AAAAAAAA BBBBBBBB CCCCCCCC DDDDDDDD
  44. The 1-bits make sure that carries propagate to the next 0-bit.
  45. The 0-bits provide holes for carries to fall into. */
  46. magic_bits = 0x7efefeffL;
  47. himagic = 0x80808080L;
  48. lomagic = 0x01010101L;
  49. if (sizeof (longword) > 4)
  50. {
  51. /* 64-bit version of the magic. */
  52. /* Do the shift in two steps to avoid a warning if long has 32 bits. */
  53. magic_bits = ((0x7efefefeL << 16) << 16) | 0xfefefeffL;
  54. himagic = ((himagic << 16) << 16) | himagic;
  55. lomagic = ((lomagic << 16) << 16) | lomagic;
  56. }
  57. if (sizeof (longword) > 8)
  58. abort ();
  59. /* Instead of the traditional loop which tests each character,
  60. we will test a longword at a time. The tricky part is testing
  61. if *any of the four* bytes in the longword in question are zero. */
  62. for (;;)
  63. {
  64. /* We tentatively exit the loop if adding MAGIC_BITS to
  65. LONGWORD fails to change any of the hole bits of LONGWORD.
  66. 1) Is this safe? Will it catch all the zero bytes?
  67. Suppose there is a byte with all zeros. Any carry bits
  68. propagating from its left will fall into the hole at its
  69. least significant bit and stop. Since there will be no
  70. carry from its most significant bit, the LSB of the
  71. byte to the left will be unchanged, and the zero will be
  72. detected.
  73. 2) Is this worthwhile? Will it ignore everything except
  74. zero bytes? Suppose every byte of LONGWORD has a bit set
  75. somewhere. There will be a carry into bit 8. If bit 8
  76. is set, this will carry into bit 16. If bit 8 is clear,
  77. one of bits 9-15 must be set, so there will be a carry
  78. into bit 16. Similarly, there will be a carry into bit
  79. 24. If one of bits 24-30 is set, there will be a carry
  80. into bit 31, so all of the hole bits will be changed.
  81. The one misfire occurs when bits 24-30 are clear and bit
  82. 31 is set; in this case, the hole at bit 31 is not
  83. changed. If we had access to the processor carry flag,
  84. we could close this loophole by putting the fourth hole
  85. at bit 32!
  86. So it ignores everything except 128's, when they're aligned
  87. properly. */
  88. longword = *longword_ptr++;
  89. if (
  90. #if 0
  91. /* Add MAGIC_BITS to LONGWORD. */
  92. (((longword + magic_bits)
  93. /* Set those bits that were unchanged by the addition. */
  94. ^ ~longword)
  95. /* Look at only the hole bits. If any of the hole bits
  96. are unchanged, most likely one of the bytes was a
  97. zero. */
  98. & ~magic_bits)
  99. #else
  100. ((longword - lomagic) & himagic)
  101. #endif
  102. != 0)
  103. {
  104. /* Which of the bytes was the zero? If none of them were, it was
  105. a misfire; continue the search. */
  106. const char *cp = (const char *) (longword_ptr - 1);
  107. if (cp[0] == 0)
  108. return cp - str;
  109. if (cp[1] == 0)
  110. return cp - str + 1;
  111. if (cp[2] == 0)
  112. return cp - str + 2;
  113. if (cp[3] == 0)
  114. return cp - str + 3;
  115. if (sizeof (longword) > 4)
  116. {
  117. if (cp[4] == 0)
  118. return cp - str + 4;
  119. if (cp[5] == 0)
  120. return cp - str + 5;
  121. if (cp[6] == 0)
  122. return cp - str + 6;
  123. if (cp[7] == 0)
  124. return cp - str + 7;
  125. }
  126. }
  127. }
  128. }
  129. libc_hidden_weak(strlen)