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Re: [PATCH 07/11] Improve generic strnlen
On 17/12/2016 04:57, Richard Henderson wrote:
> * string/strnlen.c: Use haszero.h, whichzero.h.
As for strchr, since you already optimizing memchr why not base
strnlen on memchr instead:
--
size_t
__strnlen (const char *str, size_t maxlen)
{
const char *char_str = memchr (str, 0, maxlen);
return char_str ? char_str - s : maxlen;
}
--
> ---
> string/strnlen.c | 125 +++++++++----------------------------------------------
> 1 file changed, 19 insertions(+), 106 deletions(-)
>
> diff --git a/string/strnlen.c b/string/strnlen.c
> index b2b0664..9927b9d 100644
> --- a/string/strnlen.c
> +++ b/string/strnlen.c
> @@ -21,7 +21,9 @@
> not, see <http://www.gnu.org/licenses/>. */
>
> #include <string.h>
> -#include <stdlib.h>
> +#include <stdint.h>
> +#include <haszero.h>
> +#include <whichzero.h>
>
> /* Find the length of S, but scan at most MAXLEN characters. If no
> '\0' terminator is found in that many characters, return MAXLEN. */
> @@ -33,11 +35,12 @@
> size_t
> __strnlen (const char *str, size_t maxlen)
> {
> - const char *char_ptr, *end_ptr = str + maxlen;
> + const char *char_ptr = str, *end_ptr = str + maxlen;
> const unsigned long int *longword_ptr;
> - unsigned long int longword, himagic, lomagic;
> + unsigned long int longword;
> + size_t i, align;
>
> - if (maxlen == 0)
> + if (__glibc_unlikely (maxlen == 0))
> return 0;
>
> if (__glibc_unlikely (end_ptr < str))
> @@ -45,116 +48,26 @@ __strnlen (const char *str, size_t maxlen)
>
> /* 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)
> + align = -(uintptr_t)char_ptr % sizeof(longword);
> + for (i = 0; i < align; ++i, ++char_ptr)
> if (*char_ptr == '\0')
> - {
> - if (char_ptr > end_ptr)
> - char_ptr = end_ptr;
> - return char_ptr - str;
> - }
> + goto found;
>
> - /* All these elucidatory comments refer to 4-byte longwords,
> - but the theory applies equally well to 8-byte longwords. */
> + longword_ptr = (const unsigned long int *) char_ptr;
> + char_ptr = end_ptr;
>
> - 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. */
> - himagic = 0x80808080L;
> - lomagic = 0x01010101L;
> - if (sizeof (longword) > 4)
> + while (longword_ptr < (const unsigned long int *) end_ptr)
> {
> - /* 64-bit version of the magic. */
> - /* Do the shift in two steps to avoid a warning if long has 32 bits. */
> - 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. */
> - while (longword_ptr < (unsigned long int *) end_ptr)
> - {
> - /* 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 ((longword - lomagic) & himagic)
> + longword = *longword_ptr;
> + if (haszero (longword))
> {
> - /* 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);
> -
> - char_ptr = cp;
> - if (cp[0] == 0)
> - break;
> - char_ptr = cp + 1;
> - if (cp[1] == 0)
> - break;
> - char_ptr = cp + 2;
> - if (cp[2] == 0)
> - break;
> - char_ptr = cp + 3;
> - if (cp[3] == 0)
> - break;
> - if (sizeof (longword) > 4)
> - {
> - char_ptr = cp + 4;
> - if (cp[4] == 0)
> - break;
> - char_ptr = cp + 5;
> - if (cp[5] == 0)
> - break;
> - char_ptr = cp + 6;
> - if (cp[6] == 0)
> - break;
> - char_ptr = cp + 7;
> - if (cp[7] == 0)
> - break;
> - }
> + char_ptr = (const char *) longword_ptr + whichzero (longword);
> + break;
> }
> - char_ptr = end_ptr;
> + longword_ptr++;
> }
>
> + found:
> if (char_ptr > end_ptr)
> char_ptr = end_ptr;
> return char_ptr - str;
>