Re: int8 & INT64_IS_BUSTED

"Florian G. Pflug"  writes:
> I still think int8mul is buggy. It calculates result as arg1 * arg2, and then
> checks for an overflow by dividing again, and seeing if the right answer
> comes out. Which sounds good. But it *skips* that check if both arguments
> fit into an int32 - check is
> (arg1 == (int64) ((int32) arg1) && arg2 == (int64) ((int32) arg2)).

Good point --- we should probably #ifdef out that part for
INT64_IS_BUSTED.
		regards, tom lane