- /* 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))