Is there any reason why the following C++ code behaves as it does ?
int i;
i=1; cout << (++i)++; Output: 2
i=1; cout << ++(++i); Output: 3
i=1; cout << (i++)++; Output: Error. LValue required
i=1; cout << ++(i++); Output: Error. LValue required
13  1458 
bintom wrote:
Is there any reason why the following C++ code behaves as it does ?
int i;
i=1; cout << (++i)++; Output: 2
Because it's nonsensical undefined behaviour.
--
Ian Collins.
Ian Collins wrote:
bintom wrote:
>Is there any reason why the following C++ code behaves as it does ?
int i;
i=1; cout << (++i)++; Output: 2
Because it's nonsensical undefined behaviour.
Why is it non-sensical? Why is it undefined?
V
--
Please remove capital 'A's when replying by e-mail
I do not respond to top-posted replies, please don't ask
Victor Bazarov wrote:
Ian Collins wrote:
>bintom wrote:
>>Is there any reason why the following C++ code behaves as it does ?
int i;
i=1; cout << (++i)++; Output: 2
Because it's nonsensical undefined behaviour.
Why is it non-sensical? Why is it undefined?
You are right, it is neither. Sorry, I've seen way too many posts that
are and do...
--
Ian Collins.
On Thu, 8 May 2008 17:12:15 -0700 (PDT), bintom
<bi************ *@gmail.comwrot e in comp.lang.c++:
Is there any reason why the following C++ code behaves as it does ?
int i;
i=1; cout << (++i)++; Output: 2
i=1; cout << ++(++i); Output: 3
i=1; cout << (i++)++; Output: Error. LValue required
i=1; cout << ++(i++); Output: Error. LValue required
For build-in types, such as int, the behavior of the first two is
undefined. You are attempting to modify the value of 'i' more than
once without a sequence point in between. The output could be
anything, as far as the language is concerned.
The last two are errors because the post increment operator on a
built-in type does not yield an object, instead it yields the value
that the object had before the post increment. So it is the same as
if you had written:
cout << 1++;
cout << ++1;
You can't apply the increment operators to a value.
I'd suggest you consult your C++ book about sequence points and
undefined behavior.
--
Jack Klein
Home: http://JK-Technology.Com
FAQs for
comp.lang.c http://c-faq.com/
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alt.comp.lang.l earn.c-c++ http://www.club.cc.cmu.edu/~ajo/docs/FAQ-acllc.html
Hi Jack,
Thank for your inputs on undefined/invalid uses (rather than
unpredictable results) of ++ operator. It was a great help.
bintom
On May 9, 2:54 am, "Victor Bazarov" <v.Abaza...@com Acast.netwrote:
Ian Collins wrote:
bintom wrote:
Is there any reason why the following C++ code behaves as it does ?
int i;
i=1; cout << (++i)++; Output: 2
Because it's nonsensical undefined behaviour.
Why is it non-sensical? Why is it undefined?
Because the standard says so. You're modifying the same object
twice without an intervening sequence point.
--
James Kanze (GABI Software) email:ja******* **@gmail.com
Conseils en informatique orientée objet/
Beratung in objektorientier ter Datenverarbeitu ng
9 place Sémard, 78210 St.-Cyr-l'École, France, +33 (0)1 30 23 00 34
James Kanze wrote:
On May 9, 2:54 am, "Victor Bazarov" <v.Abaza...@com Acast.netwrote:
>Ian Collins wrote:
>>bintom wrote:
Is there any reason why the following C++ code behaves as it does ?
>>>int i;
>>>i=1; cout << (++i)++; Output: 2
>>Because it's nonsensical undefined behaviour.
>Why is it non-sensical? Why is it undefined?
Because the standard says so. You're modifying the same object
twice without an intervening sequence point.
Thanks. I bet you that was the explanation the OP would have been
looking for.
V
--
Please remove capital 'A's when replying by e-mail
I do not respond to top-posted replies, please don't ask
James Kanze dixit:
On May 9, 2:54 am, "Victor Bazarov" <v.Abaza...@com Acast.netwrote:
>Ian Collins wrote:
>>bintom wrote:
Is there any reason why the following C++ code behaves as it does ?
>>>int i;
>>>i=1; cout << (++i)++; Output: 2
>>Because it's nonsensical undefined behaviour.
>Why is it non-sensical? Why is it undefined?
Because the standard says so. You're modifying the same object
twice without an intervening sequence point.
Can you tell what is an intervening sequence point ?
Thanks
Ian Collins wrote:
Victor Bazarov wrote:
>Ian Collins wrote:
>>bintom wrote:
Is there any reason why the following C++ code behaves as it does ?
int i;
i=1; cout << (++i)++; Output: 2
Because it's nonsensical undefined behaviour.
Why is it non-sensical? Why is it undefined?
You are right, it is neither. Sorry, I've seen way too many posts that
are and do...
Um, undefined behaviour overload. I was right the first time...
--
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