overflow of unsigned int

[subramanian100in]

Suppose

unsigned int size = UINT_MAX;

Now consider

++size;

After increment operator, size value becomes zero in

1) VC++ 2005 Express Edition on Intel Pentium D and
2) Redhat Enterprise Linux gcc on Intel Pentium D

Will it be possible to tell whether this behaviour(ie becoming 0) be
the same on all machines and all standard C compliant compilers. Is
this a question about overflow ? I am reading K & R second edition.
Are there any internet links that I should read material to know
more ?

Thanks

[Richard Heathfield]

su************* *@yahoo.com, India said:

Suppose

unsigned int size = UINT_MAX;

Now consider

++size;

After increment operator, size value becomes zero in

....all conforming implementations .

>
1) VC++ 2005 Express Edition on Intel Pentium D and
2) Redhat Enterprise Linux gcc on Intel Pentium D

Will it be possible to tell whether this behaviour(ie becoming 0) be
the same on all machines and all standard C compliant compilers.

Yes.

Is this a question about overflow ?

No. Unsigned integer types cannot overflow. All unsigned integer
arithmetic is performed modulo one greater than the largest value that
the type can store.

I am reading K & R second edition.

Good.

Are there any internet links that I should read material to know
more ?

<http://www.open-std.org/JTC1/SC22/WG14/www/docs/n1124.pdfis a draft
of the latest C Standard plus revisions. Also, you may like to consider
checking out some other relevant links about learning C, as listed at
<http://www.cpax.org.uk/prg/portable/c/resources.php>.

--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999
http://www.cpax.org.uk
email: rjh at the above domain, - www.

[santosh]

su************* *@yahoo.com, India wrote:

Suppose

unsigned int size = UINT_MAX;

Now consider

++size;

After increment operator, size value becomes zero in

1) VC++ 2005 Express Edition on Intel Pentium D and
2) Redhat Enterprise Linux gcc on Intel Pentium D

Will it be possible to tell whether this behaviour(ie becoming 0) be
the same on all machines and all standard C compliant compilers.

Yes, it is consistent across all conformant C implementations .

Is this a question about overflow ?

No, the C standard ascribes modulo 2^n arithmetic for unsigned types,
which by definition, has no overflow. Rather it produces a ring with
wrap-around of the value. Overflow is defined only for signed integral
and floating point types.

[Eric Sosman]

Richard Heathfield wrote On 03/09/07 07:37,:

[...]
Unsigned integer types cannot overflow. [...]

This is literally true, but it seems to be a point of
confusion: the matter arises in c.l.c. again and again and
again. Perhaps we need a term to describe the non-overflow-
but-possibly-surprising thing that happens when unsigned
arithmetic operates on an expression whose "mathematic al"
value is outside the type's range. "Discontinu ity" doesn't
seem appropriate, but "wraparound " (in various forms) has
been used before and seems to convey the right meaning.
Richard's oft-repeated statement might be a little more
informative to the befuddled if it read

Unsigned integer types cannot overflow;
they wrap around.

My own preference would be to focus on the arithmetic
rather than on the type, and say

Unsigned arithmetic cannot overflow; it
wraps around.

.... but that's a minor stylistic point about which reasonable
people can disagree in amity.

(In the post I quoted, Richard goes on to describe
"wrap around" precisely, in terms of modular arithmetic.
That's still helpful, but perhaps "wrap around" would give
the confused reader an up-front handle for understanding.)

Thoughts?

--
Er*********@sun .com