strlen.c 5.3 KB

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  1. /* Find the length of STRING, but scan at most MAXLEN characters.
  2. Copyright (C) 1991,1993,1997,2000,2001,2005 Free Software Foundation, Inc.
  3. Contributed by Jakub Jelinek <jakub@redhat.com>.
  4. Based on strlen written by Torbjorn Granlund (tege@sics.se),
  5. with help from Dan Sahlin (dan@sics.se);
  6. commentary by Jim Blandy (jimb@ai.mit.edu).
  7. The GNU C Library is free software; you can redistribute it and/or
  8. modify it under the terms of the GNU Lesser General Public License as
  9. published by the Free Software Foundation; either version 2.1 of the
  10. License, or (at your option) any later version.
  11. The GNU C Library is distributed in the hope that it will be useful,
  12. but WITHOUT ANY WARRANTY; without even the implied warranty of
  13. MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU
  14. Lesser General Public License for more details.
  15. You should have received a copy of the GNU Lesser General Public
  16. License along with the GNU C Library; see the file COPYING.LIB. If not,
  17. write to the Free Software Foundation, Inc., 59 Temple Place - Suite 330,
  18. Boston, MA 02111-1307, USA. */
  19. #include "api.h"
  20. /* Find the length of S, but scan at most MAXLEN characters. If no
  21. '\0' terminator is found in that many characters, return MAXLEN. */
  22. size_t strnlen (const char *str, size_t maxlen)
  23. {
  24. const char *char_ptr, *end_ptr = str + maxlen;
  25. const unsigned long int *longword_ptr;
  26. unsigned long int longword, himagic, lomagic;
  27. if (maxlen == 0)
  28. return 0;
  29. if (__builtin_expect (end_ptr < str, 0))
  30. end_ptr = (const char *) ~0UL;
  31. /* Handle the first few characters by reading one character at a time.
  32. Do this until CHAR_PTR is aligned on a longword boundary. */
  33. for (char_ptr = str; ((unsigned long int) char_ptr
  34. & (sizeof (longword) - 1)) != 0;
  35. ++char_ptr)
  36. if (*char_ptr == '\0')
  37. {
  38. if (char_ptr > end_ptr)
  39. char_ptr = end_ptr;
  40. return char_ptr - str;
  41. }
  42. /* All these elucidatory comments refer to 4-byte longwords,
  43. but the theory applies equally well to 8-byte longwords. */
  44. longword_ptr = (unsigned long int *) char_ptr;
  45. /* Bits 31, 24, 16, and 8 of this number are zero. Call these bits
  46. the "holes." Note that there is a hole just to the left of
  47. each byte, with an extra at the end:
  48. bits: 01111110 11111110 11111110 11111111
  49. bytes: AAAAAAAA BBBBBBBB CCCCCCCC DDDDDDDD
  50. The 1-bits make sure that carries propagate to the next 0-bit.
  51. The 0-bits provide holes for carries to fall into. */
  52. himagic = 0x80808080L;
  53. lomagic = 0x01010101L;
  54. if (sizeof (longword) > 4)
  55. {
  56. /* 64-bit version of the magic. */
  57. /* Do the shift in two steps to avoid a warning if long has 32 bits. */
  58. himagic = ((himagic << 16) << 16) | himagic;
  59. lomagic = ((lomagic << 16) << 16) | lomagic;
  60. }
  61. if (sizeof (longword) > 8)
  62. /* There is no abort(). Instead I return 0 if the length is too long */
  63. return 0;
  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 (longword_ptr < (unsigned long int *) end_ptr)
  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. longword = *longword_ptr++;
  94. if ((longword - lomagic) & himagic)
  95. {
  96. /* Which of the bytes was the zero? If none of them were, it was
  97. a misfire; continue the search. */
  98. const char *cp = (const char *) (longword_ptr - 1);
  99. char_ptr = cp;
  100. if (cp[0] == 0)
  101. break;
  102. char_ptr = cp + 1;
  103. if (cp[1] == 0)
  104. break;
  105. char_ptr = cp + 2;
  106. if (cp[2] == 0)
  107. break;
  108. char_ptr = cp + 3;
  109. if (cp[3] == 0)
  110. break;
  111. if (sizeof (longword) > 4)
  112. {
  113. char_ptr = cp + 4;
  114. if (cp[4] == 0)
  115. break;
  116. char_ptr = cp + 5;
  117. if (cp[5] == 0)
  118. break;
  119. char_ptr = cp + 6;
  120. if (cp[6] == 0)
  121. break;
  122. char_ptr = cp + 7;
  123. if (cp[7] == 0)
  124. break;
  125. }
  126. }
  127. char_ptr = end_ptr;
  128. }
  129. if (char_ptr > end_ptr)
  130. char_ptr = end_ptr;
  131. return char_ptr - str;
  132. }
  133. size_t strlen (const char *str)
  134. {
  135. return strnlen(str, -1);
  136. }