#include<stdio.h>
main()
{
int i=1;
printf("\n%d ",i<<=1%2);
return 0;
}
Try only this It provides o/p as 2 so it must be side effect.
On Mon, Jul 25, 2011 at 11:04 PM, rajeev bharshetty <rajeevr...@gmail.com>wrote:
> @sameer I think that is right
>
>
> On Mon, Jul 25, 2011 at 11:02 PM, sameer.mut...@gmail.com <
> sameer.mut...@gmail.com> wrote:
>
>> its because of side effect where value of i is getting changed twice in a
>> single line.
>> correct me if i am wrong :)
>>
>> *Muthuraj R.
>> 4TH Year BE.**
>> Information Science Dept*
>> *PESIT, Bengaluru .
>> *
>>
>>
>>
>>
>> On Mon, Jul 25, 2011 at 11:01 PM, geek forgeek <geekhori...@gmail.com>wrote:
>>
>>> y not the output is 3 2 coz on right to left evaluation of printf i
>>> shud be left shifted by 1 bit wgich shud make it 2 ??
>>>
>>>
>>> On Mon, Jul 25, 2011 at 10:28 AM, sameer.mut...@gmail.com <
>>> sameer.mut...@gmail.com> wrote:
>>>
>>>> yeah output
>>>> 0
>>>> 1 1
>>>> *is dis because of side effect? *
>>>> * *
>>>> *
>>>> *
>>>> *Muthuraj R.
>>>> 4TH Year BE.**
>>>> Information Science Dept*
>>>> *PESIT, Bengaluru .
>>>> *
>>>>
>>>>
>>>>
>>>>
>>>>
>>>> On Mon, Jul 25, 2011 at 10:49 PM, Deoki Nandan <deok...@gmail.com>wrote:
>>>>
>>>>> run on gcc compiler it would be
>>>>> 0
>>>>> 1 1
>>>>>
>>>>>
>>>>> On Mon, Jul 25, 2011 at 10:35 PM, geek forgeek
>>>>> <geekhori...@gmail.com>wrote:
>>>>>
>>>>>> 1.
>>>>>> #include<stdio.h>
>>>>>> main()
>>>>>> {
>>>>>> int i=1;
>>>>>> printf("\n%d",i^=1%2);
>>>>>> printf("\n%d %d",i^=1%2,i<<=1%2);
>>>>>> return 0;
>>>>>> }
>>>>>>
>>>>>> output 3 3
>>>>>> hey shudnt the output be 3 2
>>>>>>
>>>>>>
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>>>>>
>>>>>
>>>>>
>>>>> --
>>>>> **With Regards
>>>>> Deoki Nandan Vishwakarma
>>>>>
>>>>> *
>>>>> *
>>>>>
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>
>
>
> --
> Regards
> Rajeev N B <http://www.opensourcemania.co.cc>
>
>
--
Regards
Rajeev N B <http://www.opensourcemania.co.cc>
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