[03/11] Improve generic memchr

diff --git a/string/memchr.c b/string/memchr.c
index ca9fd99..858a494 100644
--- a/string/memchr.c
+++ b/string/memchr.c
@@ -25,10 +25,11 @@ 
 #endif
 
 #include <string.h>
-
 #include <stddef.h>
-
 #include <limits.h>
+#include <stdint.h>
+#include <haszero.h>
+#include <whichzero.h>
 
 #undef __memchr
 #ifdef _LIBC
@@ -47,111 +48,50 @@ 
 void *
 MEMCHR (void const *s, int c_in, size_t n)
 {
-  /* On 32-bit hardware, choosing longword to be a 32-bit unsigned
-     long instead of a 64-bit uintmax_t tends to give better
-     performance.  On 64-bit hardware, unsigned long is generally 64
-     bits already.  Change this typedef to experiment with
-     performance.  */
-  typedef unsigned long int longword;
-
   const unsigned char *char_ptr;
-  const longword *longword_ptr;
-  longword repeated_one;
-  longword repeated_c;
+  const unsigned long int *longword_ptr;
+  unsigned long int longword, repeated_c;
   unsigned char c;
+  uintptr_t i, align;
 
   c = (unsigned char) c_in;
+  char_ptr = (const unsigned char *) s;
 
   /* Handle the first few bytes by reading one byte at a time.
      Do this until CHAR_PTR is aligned on a longword boundary.  */
-  for (char_ptr = (const unsigned char *) s;
-       n > 0 && (size_t) char_ptr % sizeof (longword) != 0;
-       --n, ++char_ptr)
+  align = -(uintptr_t)char_ptr % sizeof(longword);
+  if (align > n)
+    align = n;
+  for (i = 0; i < align; ++i, ++char_ptr)
     if (*char_ptr == c)
       return (void *) char_ptr;
 
-  longword_ptr = (const longword *) char_ptr;
+  /* Set up a longword, each of whose bytes is C.  */
+  repeated_c = (-1ul / 0xff) * c;
 
-  /* All these elucidatory comments refer to 4-byte longwords,
-     but the theory applies equally well to any size longwords.  */
+  longword_ptr = (const unsigned long int *) char_ptr;
+  n -= align;
+  if (__glibc_unlikely(n == 0))
+    return NULL;
 
-  /* Compute auxiliary longword values:
-     repeated_one is a value which has a 1 in every byte.
-     repeated_c has c in every byte.  */
-  repeated_one = 0x01010101;
-  repeated_c = c | (c << 8);
-  repeated_c |= repeated_c << 16;
-  if (0xffffffffU < (longword) -1)
+  /* Loop while we have more than one longword remaining.  */
+  while (n > sizeof (longword))
     {
-      repeated_one |= repeated_one << 31 << 1;
-      repeated_c |= repeated_c << 31 << 1;
-      if (8 < sizeof (longword))
-	{
-	  size_t i;
-
-	  for (i = 64; i < sizeof (longword) * 8; i *= 2)
-	    {
-	      repeated_one |= repeated_one << i;
-	      repeated_c |= repeated_c << i;
-	    }
-	}
-    }
-
-  /* Instead of the traditional loop which tests each byte, 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 equal to c.  We first use an xor
-     with repeated_c.  This reduces the task to testing whether *any of the
-     four* bytes in longword1 is zero.
-
-     We compute tmp =
-       ((longword1 - repeated_one) & ~longword1) & (repeated_one << 7).
-     That is, we perform the following operations:
-       1. Subtract repeated_one.
-       2. & ~longword1.
-       3. & a mask consisting of 0x80 in every byte.
-     Consider what happens in each byte:
-       - If a byte of longword1 is zero, step 1 and 2 transform it into 0xff,
-	 and step 3 transforms it into 0x80.  A carry can also be propagated
-	 to more significant bytes.
-       - If a byte of longword1 is nonzero, let its lowest 1 bit be at
-	 position k (0 <= k <= 7); so the lowest k bits are 0.  After step 1,
-	 the byte ends in a single bit of value 0 and k bits of value 1.
-	 After step 2, the result is just k bits of value 1: 2^k - 1.  After
-	 step 3, the result is 0.  And no carry is produced.
-     So, if longword1 has only non-zero bytes, tmp is zero.
-     Whereas if longword1 has a zero byte, call j the position of the least
-     significant zero byte.  Then the result has a zero at positions 0, ...,
-     j-1 and a 0x80 at position j.  We cannot predict the result at the more
-     significant bytes (positions j+1..3), but it does not matter since we
-     already have a non-zero bit at position 8*j+7.
-
-     So, the test whether any byte in longword1 is zero is equivalent to
-     testing whether tmp is nonzero.  */
-
-  while (n >= sizeof (longword))
-    {
-      longword longword1 = *longword_ptr ^ repeated_c;
-
-      if ((((longword1 - repeated_one) & ~longword1)
-	   & (repeated_one << 7)) != 0)
-	break;
+      longword = *longword_ptr ^ repeated_c;
+      if (haszero(longword))
+	goto found;
       longword_ptr++;
       n -= sizeof (longword);
     }
 
-  char_ptr = (const unsigned char *) longword_ptr;
-
-  /* At this point, we know that either n < sizeof (longword), or one of the
-     sizeof (longword) bytes starting at char_ptr is == c.  On little-endian
-     machines, we could determine the first such byte without any further
-     memory accesses, just by looking at the tmp result from the last loop
-     iteration.  But this does not work on big-endian machines.  Choose code
-     that works in both cases.  */
-
-  for (; n > 0; --n, ++char_ptr)
+  /* Since our pointer is aligned, we can always read the last longword.  */
+  longword = *longword_ptr ^ repeated_c;
+  if (haszero(longword))
     {
-      if (*char_ptr == c)
-	return (void *) char_ptr;
+ found:
+      i = whichzero(longword);
+      if (i < n)
+	return (char *) longword_ptr + i;
     }
 
   return NULL;