bitops.h: correctly handle rol32 with 0 byte shift

ROL on a 32 bit integer with a shift of 32 or more is undefined and the
result is arch-dependent. Avoid this by handling the trivial case of
roling by 0 correctly.

The trivial solution of checking if shift is 0 breaks gcc's detection
of this code as a ROL instruction, which is unacceptable.

This bug was reported and fixed in GCC
(https://gcc.gnu.org/bugzilla/show_bug.cgi?id=57157):

	The standard rotate idiom,

	  (x << n) | (x >> (32 - n))

	is recognized by gcc (for concreteness, I discuss only the case that x
	is an uint32_t here).

	However, this is portable C only for n in the range 0 < n < 32. For n
	== 0, we get x >> 32 which gives undefined behaviour according to the
	C standard (6.5.7, Bitwise shift operators). To portably support n ==
	0, one has to write the rotate as something like

	  (x << n) | (x >> ((-n) & 31))

	And this is apparently not recognized by gcc.

Note that this is broken on older GCCs and will result in slower ROL.
Acked-by: NLinus Torvalds <torvalds@linux-foundation.org>
Signed-off-by: NSasha Levin <sasha.levin@oracle.com>
Signed-off-by: NLinus Torvalds <torvalds@linux-foundation.org>

include/linux/bitops.h

@@ -107,7 +107,7 @@ static inline __u64 ror64(__u64 word, unsigned int shift)
  */
 static inline __u32 rol32(__u32 word, unsigned int shift)
 {
-	return (word << shift) | (word >> (32 - shift));
+	return (word << shift) | (word >> ((-shift) & 31));
 }
 
 /**