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    bitops.h: correctly handle rol32 with 0 byte shift · d7e35dfa
    Sasha Levin 提交于
    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>
    d7e35dfa
bitops.h 6.0 KB