Previous documentation of rs_modnn() states simple arithmetic modulo
return a wrong result for values >= (3 * rs->nn). However, that is not
true. The rs_modnn() does the exactly same job as (x % rs->nn). This can
be proved from following loop invariants:

  while (x >= rs->nn) {
    x -= rs->nn; // (1)
    x = (x >> rs->mm) + (x & rs->nn); // (2)
  }

Let x0 denote the value of x before assignment. At (1), it is obvious
that x % nn == x0 % nn. At (2), because nn == ((1 << mm) - 1), we have

  x0 % nn == x0 % nn
  x0 % nn == (((x0 >> mm) << mm) + (x0 & nn)) % nn
  x0 % nn == ((x0 >> mm) * (nn + 1) + (x0 & nn)) % nn
  x0 % nn == ((x0 >> mm) * ((nn + 1) % nn) + (x0 & nn)) % nn
  x0 % nn == ((x0 >> mm) * 1 + (x0 & nn)) % nn   // let's assume nn > 1
  x0 % nn == ((x0 >> mm) + (x0 & nn)) % nn
  x0 % nn == x % nn

When the loop exits, it is obvious that 0 <= x < nn, so the return value
must equal to (x % rs->nn).

Signed-off-by: Zhang Boyang <zhangboyang.id@gmail.com>
---
 include/linux/rslib.h | 3 +--
 1 file changed, 1 insertion(+), 2 deletions(-)

diff --git a/include/linux/rslib.h b/include/linux/rslib.h
index 238bb85243d3..507fa14c03b2 100644
--- a/include/linux/rslib.h
+++ b/include/linux/rslib.h
@@ -116,8 +116,7 @@ void free_rs(struct rs_control *rs);
  *  rs->mm = number of bits per symbol
  *  rs->nn = (2^rs->mm) - 1
  *
- *  Simple arithmetic modulo would return a wrong result for values
- *  >= 3 * rs->nn
+ *  Calculate (x % rs->nn), without using a div instruction
 */
 static inline int rs_modnn(struct rs_codec *rs, int x)
 {
-- 
2.30.2