thnx:)
On Sun, Aug 21, 2011 at 9:26 AM, Sanjay Rajpal <srn...@gmail.com> wrote:
> See you are considering one bit of bit1, and printing it as signed integer.
> So what i think is that '1' bit will be treated as 1111 1111 while treating
> it as Signed Integer, and this is the binary representation of
> -1. Hence the result.
>
> If you write it as unsigned int bit1:1, the result is 122. So my argument
> is correct.
>
> Try this and let me know.
>
> Correct me If m wrong.
> Sanju
> :)
>
>
>
> On Sat, Aug 20, 2011 at 8:44 PM, sukran dhawan <sukrandha...@gmail.com>wrote:
>
>> only one bit is reserved for it.so the binary representation is 1.since
>> only one bit is present,
>> that bit becomes sign nit and hence -1
>>
>> On Sun, Aug 21, 2011 at 8:07 AM, saurabh singh <saurab...@gmail.com>wrote:
>>
>>> Read bit field....
>>>
>>>
>>> On Sun, Aug 21, 2011 at 2:44 AM, Nitin <coolguyinat...@gmail.com> wrote:
>>>
>>>> #include<stdio.h>
>>>> main()
>>>> {
>>>> struct value
>>>> {
>>>> int bit1:1;
>>>> int bit2:4;
>>>> int bit3:4;
>>>> }bit={1,2,2};
>>>> printf("%d%d%d",bit.bit1,bit.bit2,bit.bit3);
>>>> }
>>>>
>>>> output is -1,2,2;
>>>> can anybody tell me the reason that y it is giving -1 ??
>>>>
>>>>
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>>>
>>>
>>>
>>> --
>>> Saurabh Singh
>>> B.Tech (Computer Science)
>>> MNNIT ALLAHABAD
>>>
>>>
>>>
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