Re: [bash-bug] Can arithmetic evaluation trap more than just division by zero?

[Roman Rakus]

On 04/06/2010 05:32 PM, John Reiser wrote:
> > > > Program received signal SIGFPE, Arithmetic exception.
> > > > 0x0000000000462cd5 in exp2 () at expr.c:761
> > > > 761 val1 /= val2;
> > > > (gdb) print val1
> > > > $1 = -9223372036854775808
> > > > (gdb) print val2
> > > > $2 = -1
> > > >
> > > > which is strange.
>
> > > Not at all. Overflow invokes undefined behaviour.
>
> > But why there is no overflow on 32bit system?
>
> No *detected* overflow.  For instance, if computing (2 ** 63)
> gives 0 (the low-order 32 bits of the true result) and
> does not detect the overflow (or ignores the overflow,
> either by intention or by bug), then there is no problem
> with computing (0 / -1).
Thanks for clarifying my words.
So, on 32bit systems, evaluation of (2 ** 63 / -1) gives the wrong number.
RR