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

From: Richard Neill
Subject: Re: Bash arithmetic doesn't give error message on wrap.
Date: Mon, 30 Apr 2007 21:06:02 +0100
User-agent: Thunderbird (X11/20070403)

Bob Proulx wrote:
Andreas Schwab wrote:
Richard Neill <address@hidden> 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
The answer is not really wrong, it's the same you get from the equivalent
expression when evaluated in C.

Let me phrase this in a different way.


[OK - agreed. Thanks for your explanation]

Avoiding overflow is actually a very difficult problem.  People have
been working with and around overflow issues for years and years.
There is no clear "right" answer.  On some cpus the result is done one
way and on others the result is done a different way.  It is in these
cases where typically POSIX would give up and declare it undefined
behavior.  This is why "4000000000*4000000000" appears as a completely
different problem than "08".

I thought testing for overflow was quite simple?
Isn't it just a case of looking at the carry-bit, and seeing whether it gets set? If so, then the warning message would be a two-line patch to the code.

That said, I don't know enough about CPU internals to know what the carry-bits do with multiplication. (Addition/Subtraction overflows just change the carry-flag; Integer Division never suffers from overflows).

About the only way for bash to avoid it would be to include a full
arbitrary precision math library to evaluate these expressions itself.

I wasn't suggesting that! We have bc anyway. I was only suggesting that, when the CPU detects an overflow, bash could pass on the warning.

But that would slow the shell down by a large amount and it would make
the shell much bigger than it is today.  Both of those things would
cause people problems.  The entire reason cpus have math processors is
because these operations can be quite slow when done in software.

To give some additional weight to this, note that perl also uses the
underlying cpu for numerical computations.  So this is the same issue
as would be true in Perl.  Or in C/C++ too.

  perl -e 'printf("%d\n",4000000000*4000000000);'

Yes... but that's actually the %d doing it. Perl would automatically convert to a float. Eg

$ perl -e 'print(4000000000*4000000000);'

Thanks very much for your explanation.

Best wishes,


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