先搁这吧,晚上回去分析。
- #include <string.h>
- #include <stdlib.h>
- #undef strlen
- /*
- Return the length of the null-terminated string STR. Scan for
- the null terminator quickly by testing four bytes at a time.
- */
- size_t strlen (str)
- const char *str;
- {
- const char *char_ptr;
- const unsigned long int *longword_ptr;
- unsigned long int longword, magic_bits, himagic, lomagic;
- /*
- 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 = str; ((unsigned long int) char_ptr
- & (sizeof (longword) - 1)) != 0;++char_ptr)
- if (*char_ptr == '/0')
- return char_ptr - str;
- /*
- All these elucidatory comments refer to 4-byte longwords,
- but the theory applies equally well to 8-byte longwords.
- */
- longword_ptr = (unsigned long int *) 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.
- */
- magic_bits = 0x7efefeffL;
- himagic = 0x80808080L;
- lomagic = 0x01010101L;
- if (sizeof (longword) > 4)
- {
- /* 64-bit version of the magic. */
- /* Do the shift in two steps to avoid a warning if long has 32 bits. */
- magic_bits = ((0x7efefefeL << 16) << 16) | 0xfefefeffL;
- himagic = ((himagic << 16) << 16) | himagic;
- lomagic = ((lomagic << 16) << 16) | lomagic;
- }
- if (sizeof (longword) > 8)
- abort ();
- /*
- 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.
- */
- for (;;)
- {
- /* 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. */
- longword = *longword_ptr++;
- if (
- #if 0
- /* Add MAGIC_BITS to LONGWORD. */
- (((longword + magic_bits)
- /* Set those bits that were unchanged by the addition. */
- ^ ~longword)
- /* 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)
- #else
- ((longword - lomagic) & himagic)
- #endif
- != 0)
- {
- /*
- Which of the bytes was the zero? If none of them were, it was
- a misfire; continue the search.
- */
- const char *cp = (const char *) (longword_ptr - 1);
- if (cp[0] == 0)
- return cp - str;
- if (cp[1] == 0)
- return cp - str + 1;
- if (cp[2] == 0)
- return cp - str + 2;
- if (cp[3] == 0)
- return cp - str + 3;
- if (sizeof (longword) > 4)
- {
- if (cp[4] == 0)
- return cp - str + 4;
- if (cp[5] == 0)
- return cp - str + 5;
- if (cp[6] == 0)
- return cp - str + 6;
- if (cp[7] == 0)
- return cp - str + 7;
- }
- }
- }
- }
- libc_hidden_builtin_def (strlen)