Re: Bash arithmetic doesn't give error message on wrap.

[Paul Jarc]

Andreas Schwab <schwab@suse.de> wrote:
> prj@po.cwru.edu (Paul Jarc) writes:
>> result=a*b;
>> if (result/a!=b) { report overflow; }
>
> That won't work, since (signed integer) overflow is undefined in C.  A
> compiler is allowed to optimize the condition to false.

Right, operations would have to be performed in unsigned arithmetic
and then adjusted for the proper sign, range chaecking, etc., to get
signed results.


paul