Well its not like that
first u have x=3;
now u have to do
y=++x + ++x + ++X;
before doing first addition
y=(++x + ++x) + ++x;
       ^          ^

above spotted x will increase , before their addition
and then
y= 5+5 + ++x;
now
y=10+ 6
y=16;

hope u now get it :)

On Sun, Jul 24, 2011 at 2:30 PM, Arun Vishwanathan
<aaron.nar...@gmail.com>wrote:

> *x= 3 initially
>
>
> On Sun, Jul 24, 2011 at 11:00 AM, Arun Vishwanathan <
> aaron.nar...@gmail.com> wrote:
>
>> if that is the case as u say, then wont it be 3+ 4+ 5 when x +3
>> initially?..and then x increments by one later due to the single post
>> increment
>>
>>
>> On Sun, Jul 24, 2011 at 10:15 AM, sagar pareek <sagarpar...@gmail.com>wrote:
>>
>>> nope compiler read it from left to right
>>>
>>>
>>> On Sun, Jul 24, 2011 at 12:05 AM, Arun Vishwanathan <
>>> aaron.nar...@gmail.com> wrote:
>>>
>>>> @sagar: if what u said previously holds as in when u say y=x++ + ++x is
>>>> evaluated as 4+4 since ++x results in 4 and 4 is used in x++ too (cos post
>>>> increment increments x later) then for y=x++ + ++x + ++x with x beginning 
>>>> as
>>>> 3 shud the expression not be evaluated as 5+5+4( from rhs ++x does a 3 to 4
>>>> and another ++x does 4 to 5 and 5 is used in x++) .later x becomes 6 ?
>>>>
>>>>
>>>>
>>>>
>>>> On Sat, Jul 23, 2011 at 2:39 PM, sagar pareek <sagarpar...@gmail.com>wrote:
>>>>
>>>>> sorry for above...typo mistake :-
>>>>>
>>>>> yup
>>>>> but what about this
>>>>> x=3;
>>>>>  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 6:09 PM, sagar pareek 
>>>>> <sagarpar...@gmail.com>wrote:
>>>>>
>>>>>> 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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>>>>>>
>>>>>> --
>>>>>> **Regards
>>>>>> SAGAR PAREEK
>>>>>> COMPUTER SCIENCE AND ENGINEERING
>>>>>> NIT ALLAHABAD
>>>>>>
>>>>>>
>>>>>
>>>>>
>>>>> --
>>>>> **Regards
>>>>> SAGAR PAREEK
>>>>> COMPUTER SCIENCE AND ENGINEERING
>>>>> NIT ALLAHABAD
>>>>>
>>>>>  --
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>>>>
>>>>
>>>>
>>>> --
>>>> Arun Vish
>>>> Graduate Student
>>>> Department of Computer Science
>>>> University of Southern California
>>>>
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>>>
>>>
>>>
>>> --
>>> **Regards
>>> SAGAR PAREEK
>>> COMPUTER SCIENCE AND ENGINEERING
>>> NIT ALLAHABAD
>>>
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>>
>>
>>
>> --
>> Arun Vish
>> Graduate Student
>> Department of Computer Science
>> University of Southern California
>>
>>
>
>
> --
> Arun Vish
> Graduate Student
> Department of Computer Science
> University of Southern California
>
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-- 
**Regards
SAGAR PAREEK
COMPUTER SCIENCE AND ENGINEERING
NIT ALLAHABAD

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