const unsigned char *char_ptr;
const longword *longword_ptr;
- longword longword1;
- longword longword2;
- longword magic_bits;
- longword charmask1;
- longword charmask2;
+ longword repeated_one;
+ longword repeated_c1;
+ longword repeated_c2;
unsigned char c1;
unsigned char c2;
- int i;
c1 = (unsigned char) c1_in;
c2 = (unsigned char) c2_in;
if (c1 == c2)
return memchr (s, c1, n);
- /* Handle the first few characters by reading one character at a time.
+ /* 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 longword1 != 0;
+ n > 0 && (size_t) char_ptr % sizeof (longword) != 0;
--n, ++char_ptr)
if (*char_ptr == c1 || *char_ptr == c2)
return (void *) char_ptr;
+ longword_ptr = (const longword *) char_ptr;
+
/* All these elucidatory comments refer to 4-byte longwords,
but the theory applies equally well to any size longwords. */
- longword_ptr = (const longword *) char_ptr;
- magic_bits = 0x01010101;
- charmask1 = c1 | (c1 << 8);
- charmask2 = c2 | (c2 << 8);
- charmask1 |= charmask1 << 16;
- charmask2 |= charmask2 << 16;
+ /* Compute auxiliary longword values:
+ repeated_one is a value which has a 1 in every byte.
+ repeated_c1 has c1 in every byte.
+ repeated_c2 has c2 in every byte. */
+ repeated_one = 0x01010101;
+ repeated_c1 = c1 | (c1 << 8);
+ repeated_c2 = c2 | (c2 << 8);
+ repeated_c1 |= repeated_c1 << 16;
+ repeated_c2 |= repeated_c2 << 16;
if (0xffffffffU < TYPE_MAXIMUM (longword))
{
- magic_bits |= magic_bits << 31 << 1;
- charmask1 |= charmask1 << 31 << 1;
- charmask2 |= charmask2 << 31 << 1;
- if (8 < sizeof longword1)
- for (i = 64; i < sizeof longword1 * 8; i *= 2)
- {
- magic_bits |= magic_bits << i;
- charmask1 |= charmask1 << i;
- charmask2 |= charmask2 << i;
- }
+ repeated_one |= repeated_one << 31 << 1;
+ repeated_c1 |= repeated_c1 << 31 << 1;
+ repeated_c2 |= repeated_c2 << 31 << 1;
+ if (8 < sizeof (longword))
+ {
+ int i;
+
+ for (i = 64; i < sizeof (longword) * 8; i *= 2)
+ {
+ repeated_one |= repeated_one << i;
+ repeated_c1 |= repeated_c1 << i;
+ repeated_c2 |= repeated_c2 << i;
+ }
+ }
}
- /* 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.
-
- We first use an xor to convert target bytes into a NUL byte,
- since the test for a zero byte is more efficient. For all byte
- values except 0x00 and 0x80, subtracting 1 from the byte will
- leave the most significant bit unchanged. So detecting 0 is
- simply a matter of subtracting from all bytes in parallel, and
- checking for a most significant bit that changed to 1. */
-
- while (n >= sizeof longword1)
+ /* 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 c1 or c2. We first use
+ an xor with repeated_c1 and repeated_c2, respectively. This reduces
+ the task to testing whether *any of the four* bytes in longword1 or
+ longword2 is zero.
+
+ Let's consider longword1. We compute tmp1 =
+ ((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, tmp1 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.
+
+ Similary, we compute tmp2 =
+ ((longword2 - repeated_one) & ~longword2) & (repeated_one << 7).
+
+ The test whether any byte in longword1 or longword2 is zero is equivalent
+ to testing whether tmp1 is nonzero or tmp2 is nonzero. We can combine
+ this into a single test, whether (tmp1 | tmp2) is nonzero. */
+
+ while (n >= sizeof (longword))
{
- longword1 = *longword_ptr ^ charmask1;
- longword2 = *longword_ptr ^ charmask2;
+ longword longword1 = *longword_ptr ^ repeated_c1;
+ longword longword2 = *longword_ptr ^ repeated_c2;
- if (((((longword1 - magic_bits) & ~longword1)
- | ((longword2 - magic_bits) & ~longword2))
- & (magic_bits << 7)) != 0)
+ if (((((longword1 - repeated_one) & ~longword1)
+ | ((longword2 - repeated_one) & ~longword2))
+ & (repeated_one << 7)) != 0)
break;
longword_ptr++;
- n -= sizeof longword1;
+ n -= sizeof (longword);
}
char_ptr = (const unsigned char *) longword_ptr;
- while (n-- > 0)
+ /* At this point, we know that either n < sizeof (longword), or one of the
+ sizeof (longword) bytes starting at char_ptr is == c1 or == c2. On
+ little-endian machines, we could determine the first such byte without
+ any further memory accesses, just by looking at the (tmp1 | tmp2) 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)
{
if (*char_ptr == c1 || *char_ptr == c2)
return (void *) char_ptr;
- ++char_ptr;
}
return NULL;
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