sry sry o/p ll be 0 1 0....in the 2nd printf the value ll be evaluated
from rite to left...
so in the 1st printf i's vlue ll be 0 and in the 2nd printf stmt rite
expression is evaluated first and then i value ll be modified in it..
and with hte help of it left one is evaluated...crct me if i'm
wrong......
On 7/26/11, kavitha nk <kavithan...@gmail.com> wrote:
> ans ll nt be 0 0 2 va???
>
> On 7/25/11, Puneet Gautam <puneet.nsi...@gmail.com> wrote:
>> @aditi: the "%2" in each of these printf statements is redundant coz
>> it is "1%2"...and "%" has higher precedence over left or right
>> shifts..thats y output of say :
>>
>> #include<stdio.h>
>> main()
>> {
>> int i=1;
>> printf("\n%d%d ",i<<=2,i<<=1%2);
>> return 0;
>> }
>>
>> is 8 8
>>
>> 1 gets left shifted twice in i<<=2, becomes 4
>> then i<<1%2 concludes to i<<1 as 1%2 is 1 only..
>> i<<1 ,makes i=8 and hence
>>
>> 8 8 is printed.....
>>
>> On 7/25/11, Puneet Gautam <puneet.nsi...@gmail.com> wrote:
>>> @rajeev: hey...does printf evaluate an expression right to left
>>> without considering any priority of operators..?
>>> Pls reply asap..
>>>
>>>
>>>
>>> On 7/25/11, Puneet Gautam <puneet.nsi...@gmail.com> wrote:
>>>> @aditi: y dont u try it out urself considering the priorities of
>>>> different operators used here...!!
>>>> that would help...
>>>>
>>>> On 7/25/11, aditi garg <aditi.garg.6...@gmail.com> wrote:
>>>>> Can u plz elaborate...im not able to understand...
>>>>>
>>>>> 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>
>>>>>>
>>>>>>
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>>>>>
>>>>>
>>>>>
>>>>> --
>>>>> Aditi Garg
>>>>> Undergraduate Student
>>>>> Electronics & Communication Divison
>>>>> NETAJI SUBHAS INSTITUTE OF TECHNOLOGY
>>>>> Sector 3, Dwarka
>>>>> New Delhi
>>>>>
>>>>> 9718388816
>>>>>
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>>>>>
>>>>
>>>
>>
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>>
>
>
> --
> //BE COOL// kavi
>
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//BE COOL// kavi
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