Re: sum of 64 bits using 32 bits cpu

Tomás Ó hÉilidhe <toe@[EMAIL PROTECTED]
> writes:

> On May 14, 12:07 pm, Eligiusz Narutowicz<eligiuszdotn...@[EMAIL PROTECTED]
>
> wrote:
>
>> I often wonder about this nonsensical part of the standard.
>
>
> Are you referring to the overflow of signed integers having undefined
> behaviour? If so:
>
> The reason for this, I think, is that there are many processors that
> have a "STATUS" register which uses bits to store such info as:
>     * The result of the last arithmetic operation was 0.
>     * The last arithmetic operation overflowed.
>
> If you had:
>
>     int i = 7;
>
>     i -= 34;
>
>     /* Now the STATUS register says there was overflow */

This was probably not the xample you intended as there is no overflow.

>
>     if (i) DoSomething();
>
> When the compiler sees "if (i)", it will simply check the STATUS
> register to see if the last operation resulted in zero. If an unsigned
> type is involved, then it's possible to have overflow and zero at the
> same time. With signed however, it doesn't bother checking for
> overflow, it just checks for zero. This is unreliable for some reason.
> Something along those lines in anyway.
>
> Somebody posted a nice anecdote about how people were complaining to a
> compiler manufacturer about getting dodgy behaviour out of "if (i)"
> after an overflow occurred, but the compiler writer was in strict
> abidance of the C89 Standard.