diff mbox

[04/11] Improve generic strchrnul

Message ID 20161217065729.28561-5-rth@twiddle.net
State New
Headers show

Commit Message

Richard Henderson Dec. 17, 2016, 6:57 a.m. UTC
* string/strchrnul.c: Use haszero.h, whichzero.h.
---
 string/strchrnul.c | 126 +++++++----------------------------------------------
 1 file changed, 15 insertions(+), 111 deletions(-)
diff mbox

Patch

diff --git a/string/strchrnul.c b/string/strchrnul.c
index 629db46..fb2d1a4 100644
--- a/string/strchrnul.c
+++ b/string/strchrnul.c
@@ -21,8 +21,10 @@ 
    <http://www.gnu.org/licenses/>.  */
 
 #include <string.h>
-#include <memcopy.h>
 #include <stdlib.h>
+#include <stdint.h>
+#include <haszero.h>
+#include <whichzero.h>
 
 #undef __strchrnul
 #undef strchrnul
@@ -37,130 +39,32 @@  STRCHRNUL (const char *s, int c_in)
 {
   const unsigned char *char_ptr;
   const unsigned long int *longword_ptr;
-  unsigned long int longword, magic_bits, charmask;
+  unsigned long int longword, repeated_c, found;
+  uintptr_t i, align;
   unsigned char c;
 
   c = (unsigned char) c_in;
+  char_ptr = (const unsigned char *) s;
 
   /* 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 = (const unsigned char *) s;
-       ((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 == c || *char_ptr == '\0')
-      return (void *) char_ptr;
-
-  /* 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 = -1;
-  magic_bits = magic_bits / 0xff * 0xfe << 1 >> 1 | 1;
+      return (char *) char_ptr;
 
   /* Set up a longword, each of whose bytes is C.  */
-  charmask = c | (c << 8);
-  charmask |= charmask << 16;
-  if (sizeof (longword) > 4)
-    /* Do the shift in two steps to avoid a warning if long has 32 bits.  */
-    charmask |= (charmask << 16) << 16;
-  if (sizeof (longword) > 8)
-    abort ();
+  repeated_c = (-1ul / 0xff) * c;
 
-  /* 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 (;;)
+  longword_ptr = (unsigned long int *) char_ptr;
+  do
     {
-      /* 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.
-
-	 3) But wait!  Aren't we looking for C as well as zero?
-	 Good point.  So what we do is XOR LONGWORD with a longword,
-	 each of whose bytes is C.  This turns each byte that is C
-	 into a zero.  */
-
       longword = *longword_ptr++;
-
-      /* Add MAGIC_BITS to LONGWORD.  */
-      if ((((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) != 0 ||
-
-	  /* That caught zeroes.  Now test for C.  */
-	  ((((longword ^ charmask) + magic_bits) ^ ~(longword ^ charmask))
-	   & ~magic_bits) != 0)
-	{
-	  /* Which of the bytes was C or zero?
-	     If none of them were, it was a misfire; continue the search.  */
-
-	  const unsigned char *cp = (const unsigned char *) (longword_ptr - 1);
-
-	  if (*cp == c || *cp == '\0')
-	    return (char *) cp;
-	  if (*++cp == c || *cp == '\0')
-	    return (char *) cp;
-	  if (*++cp == c || *cp == '\0')
-	    return (char *) cp;
-	  if (*++cp == c || *cp == '\0')
-	    return (char *) cp;
-	  if (sizeof (longword) > 4)
-	    {
-	      if (*++cp == c || *cp == '\0')
-		return (char *) cp;
-	      if (*++cp == c || *cp == '\0')
-		return (char *) cp;
-	      if (*++cp == c || *cp == '\0')
-		return (char *) cp;
-	      if (*++cp == c || *cp == '\0')
-		return (char *) cp;
-	    }
-	}
     }
+  while (!haszero2(longword, longword ^ repeated_c));
 
-  /* This should never happen.  */
-  return NULL;
+  found = whichzero2(longword, longword ^ repeated_c);
+  return (char *) (longword_ptr - 1) + found;
 }
 
 weak_alias (__strchrnul, strchrnul)