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

[Richard Neill]

Bob Proulx wrote:
> Richard Neill wrote:
> >  b)Consistent with other cases, where bash does give warnings. For example:
> >
> > $ X=$((3+078))
> > bash: 3+078: value too great for base (error token is "078")
> > $ echo $?
> > 1
>
> That is not really a comparable case.  The problem there is that the
> leading zero specifies an octal constant and the 8 cannot be converted
> to octal.  The "3+" part is just a distraction.
>
>   echo $((08))
>   bash: 08: value too great for base (error token is "08")
>
> Bob

Are you sure this isn't comparable? After all, in both cases, the user 
has submitted something to which bash cannot give a sensible answer. In 
the integer-overflow case, bash simply returns the wrong answer, with no 
warning. But in the octal case, bash (quite correctly, and helpfully) 
prints a warning.

If bash were to be consistent, then it should display no error message 
in the case of $((3+078)); it should either "Do what I mean"  [evaluate 
3 + 078 as 0103, treating 78 as 7 *8 + 8], or "Do something daft" [eg 
evaluate as 073  (treating 8 as an integer-overflow in the units place, 
which is not checked for)]

There's obviously an advantage to the user in being warned when an 
integer overflow occurs; is there any possible downside (apart from a 
couple of extra lines of code?

Richard