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

[Andreas Schwab]

Richard Neill <rn214@hermes.cam.ac.uk> writes:

> 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.

The answer is not really wrong, it's the same you get from the equivalent
expression when evaluated in C.

> But in the octal case, bash (quite correctly, and helpfully)
> prints a warning.

It's not a warning, it's an error (and the whole command containing the
expansion is aborted).

> If bash were to be consistent, then it should display no error message in
> the case of $((3+078));

This is a syntax error, so there is no sensible meaning attached to it.  A
syntax error is something quite different than an undefined behaviour.

Andreas.

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Andreas Schwab, SuSE Labs, schwab@suse.de
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