/* Search no more than N bytes of S for C. */
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;
unsigned char c;
c = (unsigned char) c_in;
/* 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)
if (*char_ptr == c)
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. */
/* 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)
{
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_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)
{
if (*char_ptr == c)
return (void *) char_ptr;
}
return NULL;
}
