Defined bahavior or otherwise?

[Nick Landsberg]

If the following construct defined behaviour
in the standard?

v += v + x;

(Assuming all variables are of the
same type.)

I know what I would expect the construct
to do, but expected is not necessarily
defined.

--
"It is impossible to make anything foolproof
because fools are so ingenious"
- A. Bloch

[Leor Zolman]

On Thu, 08 Apr 2004 00:55:35 GMT, Nick Landsberg <hu*****@NOSPAM.att.net>
wrote:

If the following construct defined behaviour
in the standard?

v += v + x;
Since the result cannot vary due to permissible variations in evaluation
order between sequence points (of which there don't seem to be any here),
you should be all set.
(Assuming all variables are of the
same type.)
I'm not sure it would make any difference if they weren't.
-leor
I know what I would expect the construct
to do, but expected is not necessarily
defined.

--
Leor Zolman --- BD Software --- www.bdsoft.com
On-Site Training in C/C++, Java, Perl and Unix
C++ users: Download BD Software's free STL Error Message Decryptor at:
www.bdsoft.com/tools/stlfilt.html

[Nick Landsberg]

Leor Zolman wrote:

On Thu, 08 Apr 2004 00:55:35 GMT, Nick Landsberg <hu*****@NOSPAM.att.net>
wrote:

If the following construct defined behaviour
in the standard?

v += v + x;

Since the result cannot vary due to permissible variations in evaluation
order between sequence points (of which there don't seem to be any here),
you should be all set.

(Assuming all variables are of the
same type.)

I'm not sure it would make any difference if they weren't.
-leor

I know what I would expect the construct
to do, but expected is not necessarily
defined.

Thanks, Leor :)
--
"It is impossible to make anything foolproof
because fools are so ingenious"
- A. Bloch

[Christopher Benson-Manica]

Leor Zolman <le**@bdsoft.com> spoke thus:

I'm not sure it would make any difference if they weren't.

Well, if one of them is an unsigned short, and the other isn't, then
differences are likely to ensue :)

--
Christopher Benson-Manica | I *should* know what I'm talking about - if I
ataru(at)cyberspace.org | don't, I need to know. Flames welcome.

[Leor Zolman]

On Thu, 8 Apr 2004 02:29:42 +0000 (UTC), Christopher Benson-Manica
<at***@nospam.cyberspace.org> wrote:

Leor Zolman <le**@bdsoft.com> spoke thus:

I'm not sure it would make any difference if they weren't.

Well, if one of them is an unsigned short, and the other isn't, then
differences are likely to ensue :)

Well, okay...but as I read it, that would result in
implementation-/defined/ behavior, which would at least remain in the
"defined" category, and thus within the target range of the OP's question
;-)
-leor
--
Leor Zolman --- BD Software --- www.bdsoft.com
On-Site Training in C/C++, Java, Perl and Unix
C++ users: Download BD Software's free STL Error Message Decryptor at:
www.bdsoft.com/tools/stlfilt.html

[Jack Klein]

On Thu, 08 Apr 2004 00:55:35 GMT, Nick Landsberg
<hu*****@NOSPAM.att.net> wrote in comp.lang.c:

If the following construct defined behaviour
in the standard?

v += v + x;

(Assuming all variables are of the
same type.)

I know what I would expect the construct
to do, but expected is not necessarily
defined.

If all the variables are of signed or floating point type, there is
always the possibility of undefined behavior if the sum of v and x is
outside the range of the type.

But overflow/underflow aside, the expression if perfectly well
defined. v is modified only once, on the left hand side of the
assignment operator, and the other access to it, on the right hand
side, is used in calculating the new value.

--
Jack Klein
Home: http://JK-Technology.Com
FAQs for
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alt.comp.lang.learn.c-c++
http://www.contrib.andrew.cmu.edu/~a...FAQ-acllc.html

[Barry Schwarz]

On Thu, 08 Apr 2004 00:55:35 GMT, Nick Landsberg
<hu*****@NOSPAM.att.net> wrote:

If the following construct defined behaviour
in the standard?

v += v + x;

(Assuming all variables are of the
same type.)

I know what I would expect the construct
to do, but expected is not necessarily
defined.

Since this is syntactic sugar for
v = v+v+x;
the behavior of both must be equally defined. Other than pathological
cases of underflow or overflow, why would the question even be raised?
<<Remove the del for email>>

[Joona I Palaste]

Barry Schwarz <sc******@deloz.net> scribbled the following:

On Thu, 08 Apr 2004 00:55:35 GMT, Nick Landsberg
<hu*****@NOSPAM.att.net> wrote:

If the following construct defined behaviour
in the standard?

v += v + x;

(Assuming all variables are of the
same type.)

I know what I would expect the construct
to do, but expected is not necessarily
defined.

Since this is syntactic sugar for
v = v+v+x;
the behavior of both must be equally defined. Other than pathological
cases of underflow or overflow, why would the question even be raised?

If evaluating v has side effects, the two forms are not equivalent, and
so can hardly be called syntactic sugar.

--
/-- Joona Palaste (pa*****@cc.helsinki.fi) ------------- Finland --------\
\-- http://www.helsinki.fi/~palaste --------------------- rules! --------/
"A bee could, in effect, gather its junk. Llamas (no poor quadripeds) tune
and vow excitedly zooming."
- JIPsoft

[Christian Bau]

In article
<bq*********************@bgtnsc04-news.ops.worldnet.att.net>,
Nick Landsberg <hu*****@NOSPAM.att.net> wrote:

If the following construct defined behaviour
in the standard?

v += v + x;

Yes. The two exceptions that cause undefined behavior are:

1. Same object modified twice, but v is modified once only.
2. Object modified, and its value is used, but not to determine the
new value. This is not the case, because the old value of v _is_ used to
determine the new value.

[Nick Landsberg]

Jack Klein wrote:

On Thu, 08 Apr 2004 00:55:35 GMT, Nick Landsberg
<hu*****@NOSPAM.att.net> wrote in comp.lang.c:

If the following construct defined behaviour
in the standard?

v += v + x;

(Assuming all variables are of the
same type.)

I know what I would expect the construct
to do, but expected is not necessarily
defined.

If all the variables are of signed or floating point type, there is
always the possibility of undefined behavior if the sum of v and x is
outside the range of the type.

But overflow/underflow aside, the expression if perfectly well
defined. v is modified only once, on the left hand side of the
assignment operator, and the other access to it, on the right hand
side, is used in calculating the new value.

Thanks, again.

--
"It is impossible to make anything foolproof
because fools are so ingenious"
- A. Bloch

[Dan Pop]

IMHO, such behaviour is far more obnoxious than a TIA at the end of the
initial post.

Dan

In <4L*********************@bgtnsc04-news.ops.worldnet.att.net> Nick Landsberg <hu*****@NOSPAM.att.net> writes:

Jack Klein wrote:

On Thu, 08 Apr 2004 00:55:35 GMT, Nick Landsberg
<hu*****@NOSPAM.att.net> wrote in comp.lang.c:

If the following construct defined behaviour
in the standard?

v += v + x;

(Assuming all variables are of the
same type.)

I know what I would expect the construct
to do, but expected is not necessarily
defined.

If all the variables are of signed or floating point type, there is
always the possibility of undefined behavior if the sum of v and x is
outside the range of the type.

But overflow/underflow aside, the expression if perfectly well
defined. v is modified only once, on the left hand side of the
assignment operator, and the other access to it, on the right hand
side, is used in calculating the new value.

Thanks, again.

--
"It is impossible to make anything foolproof
because fools are so ingenious"
- A. Bloch

--
Dan Pop
DESY Zeuthen, RZ group
Email: Da*****@ifh.de

[Leor Zolman]

On 8 Apr 2004 16:42:28 GMT, Da*****@cern.ch (Dan Pop) wrote:

IMHO, such behaviour is far more obnoxious than a TIA at the end of the
initial post.

If what you're referring to as obnoxious are posts by OPs to say "Thank
you" to their responders, then I'd enthusiastically welcome two instances
of /that/ sort of obnoxiousness for every one of all the other types ;-)
-leor

--
Leor Zolman --- BD Software --- www.bdsoft.com
On-Site Training in C/C++, Java, Perl and Unix
C++ users: Download BD Software's free STL Error Message Decryptor at:
www.bdsoft.com/tools/stlfilt.html

[Nick Landsberg]

Leor Zolman wrote:

On 8 Apr 2004 16:42:28 GMT, Da*****@cern.ch (Dan Pop) wrote:

IMHO, such behaviour is far more obnoxious than a TIA at the end of the
initial post.

If what you're referring to as obnoxious are posts by OPs to say "Thank
you" to their responders, then I'd enthusiastically welcome two instances
of /that/ sort of obnoxiousness for every one of all the other types ;-)
-leor

Actually, I owe folks an apology about that.

I replied to the initial response with a thanks,
not realizing that there were other additional
reponses. (I should have scanned the headers.)
After the second one, I should have realized
that there may be others and just limited
it to a single "thanks to all who replied"
rather than "spamming" the NG.

Mea Culpa.

--
"It is impossible to make anything foolproof
because fools are so ingenious"
- A. Bloch

[Martijn]

Nick Landsberg wrote:

If the following construct defined behaviour
in the standard?

v += v + x;

This seems a lot trickier:

v = v += x;

;)

I reckon this causes undefined behaviour...

--
Martijn
http://www.sereneconcepts.nl

[Peter Pichler]

"Nick Landsberg" <hu*****@NOSPAM.att.net> wrote in message
news:bq*********************@bgtnsc04-news.ops.worldnet.att.net...

If the following construct defined behaviour
in the standard?

v += v + x;

(Assuming all variables are of the
same type.)

If all variablkes are of the same /pointer/ type, then the behaviour is
defined absolutely clearly :-)