yup
but what about this
x=4;
y= x++ + ++x + ++x; // it is executed as:-

during first addition, increase the value of x, now first addition will be
4+4 + ++x;
now for second addition it will be like
8+5
hence final value of y=13;
do it by urself

On Sat, Jul 23, 2011 at 2:54 PM, shady <sinv...@gmail.com> wrote:

> @sagar
> would it get evaluated like this ?
> supposing x = 3;
>
> y = x++ + ++x; becomes y = (x=x+1) + (x=x+1);
> then x=x+1;
>
> so x = 5, y = 8;
>
> On Sat, Jul 23, 2011 at 2:48 PM, sagar pareek <sagarpar...@gmail.com>wrote:
>
>> @Venga
>> if u are doing this
>> y= x++ + ++x; //x=3
>> then it would be
>> like that :-
>> ++x; //x=4
>> y=x+x;
>> x++;
>>
>> i thing this is sufficient  :)
>>
>> On Sat, Jul 23, 2011 at 1:20 PM, Interstellar Overdrive <
>> abhi123khat...@gmail.com> wrote:
>>
>>> The expression y = x++ + x++ + ++y;  is not a valid one. The result is
>>> compiler dependent
>>> Read this for reference :http://c-faq.com/expr/seqpoints.html
>>>
>>>
>>>
>>>
>>>
>>>
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>>
>>
>>
>> --
>> **Regards
>> SAGAR PAREEK
>> COMPUTER SCIENCE AND ENGINEERING
>> NIT ALLAHABAD
>>
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COMPUTER SCIENCE AND ENGINEERING
NIT ALLAHABAD

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