Justus Winter, le Mon 06 Jan 2014 00:34:53 +0100, a écrit :
> The pattern is !x&y. An expression of this form is almost always
> meaningless, because it combines a boolean operator with a bit
> operator. In particular, if the rightmost bit of y is 0, the result
> will always be 0.
>
> Fixed using
The pattern is !x&y. An expression of this form is almost always
meaningless, because it combines a boolean operator with a bit
operator. In particular, if the rightmost bit of y is 0, the result
will always be 0.
Fixed using coccinelle.
// !x&y combines boolean negation with bitwise and
//
// Co
