On Thu, 12 Jul 2001 00:12:15 GMT, Jack Klein <jackklein@spamcop.net>
wrote:

>
>For floating point types and signed integer types the behavior on
>overflow is undefined.  No diagnostic is required at compiler time for
>undefined behavior.
>

Is the reason here that the implementation of signed types is
implementation defined? I always thought that signed means 2's
complement.


>There is no overflow possible with the unsigned integer types.  The
>result is the "true" result modulo TYPE_MAX + 1.

I can understand that - thanks.


------------------------------------------------
Martin Aupperle
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I have a question about overflow of arithmetic types.

Is the expression

 T t = maxT + 1;

well defined for arithmetic types T? maxT should be the literal with
the larges representable value for type t.

What is the compiler supposed to do in such a situation? Is a
warning/error message required?

Thanks - Martin


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Martin Aupperle
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On Wed, 11 Jul 2001 17:04:30 GMT, MikeAlpha@NoSpam_csi.com (Martin
Aupperle) wrote in comp.std.c++:

>
> I have a question about overflow of arithmetic types.
>
> Is the expression
>
>  T t = maxT + 1;
>
> well defined for arithmetic types T? maxT should be the literal with
> the larges representable value for type t.
>
> What is the compiler supposed to do in such a situation? Is a
> warning/error message required?
>
> Thanks - Martin

For floating point types and signed integer types the behavior on
overflow is undefined.  No diagnostic is required at compiler time for
undefined behavior.

There is no overflow possible with the unsigned integer types.  The
result is the "true" result modulo TYPE_MAX + 1.

--
Jack Klein
Home: http://JK-Technology.Com
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