On Thu, May 24, 2007 at 12:29:43AM +0100, Derek M Jones wrote: > It was intended as a probabilit statement (ok, I did not make that > clear). An expression containing n+n is more likely to overflow > than one containing n-n. Gimme a break. a) s/int/unsigned and run that through gcc; no change in behaviour b) no fscking way in hell *either* is acceptable for bitfield width - definitely not with -std=c99 -pedantic. Violates 6.6p6 and 6.7.2.1p3. c) what's happening is pretty obvious - the difference is not in overflows, it's in expression tree structure (remember, + and - are left-to-right). gcc throws several cheap optimizations at the expression and checks if it has come up with a constant. Simple common factors are taken out (n*m - n*m is seen as 0), common subexpressions are not recognized ((n+m)-(n+m) is not seen as constant). d) (c) is an exercise in software proctology - gcc has an obvious bug in that area (mishandling recognition of integer constant expressions), period. - To unsubscribe from this list: send the line "unsubscribe linux-sparse" in the body of a message to majordomo@xxxxxxxxxxxxxxx More majordomo info at http://vger.kernel.org/majordomo-info.html
