On Tue, Aug 05, 2008 at 10:43:32PM +0200, Stefan wrote: > So it's up to the linker to decide whether the expression is constant or > not ? This is just one reason. The other is that text I cited from the standard says that the constructs you were using aren't allowed. > Accordingly the compiler inserts the code for the substraction and thus the > linker step comes > too late to see, that after the substraction the expression had still been > constant. > Anyway, without arithmetic the decision is made via a constant, correctly. Yes. Regards Uz -- Ullrich von Bassewitz uz@musoftware.de ---------------------------------------------------------------------- To unsubscribe from the list send mail to majordomo@musoftware.de with the string "unsubscribe cc65" in the body(!) of the mail.Received on Tue Aug 5 22:58:56 2008
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