Will u guyz pls tell me frm where do u study terms like "endian" ,it is a
pity i had to google it  :( :(

On Wed, Jul 13, 2011 at 10:30 PM, Anika Jain <anika.jai...@gmail.com> wrote:

> ya thts ryt, for big endian it will be 64 -90 102 102
>
>
> On Wed, Jul 13, 2011 at 10:22 PM, sunny agrawal 
> <sunny816.i...@gmail.com>wrote:
>
>> I think it is machine dependent and current output is for a little endian
>> machine
>> and output should be reverse for a big endian machine
>>
>> On Wed, Jul 13, 2011 at 10:18 PM, Anika Jain <anika.jai...@gmail.com>wrote:
>>
>>> it is not unsigned its signed but positive thts y it is 0..
>>> the order isnt reversed.. look at the code, the code 1st prints lowest
>>> order byte then next and then next and then highest soo output is 102 102
>>> -90 64
>>>
>>>
>>> On Wed, Jul 13, 2011 at 8:08 PM, Piyush Kapoor <pkjee2...@gmail.com>wrote:
>>>
>>>> why is the order of numbers reversed?
>>>>
>>>>
>>>> On Wed, Jul 13, 2011 at 8:07 PM, Anika Jain <anika.jai...@gmail.com>wrote:
>>>>
>>>>> sorry its 01000000 not 11000000 coz 5.2 is a positive no. so sign bit
>>>>> is 0
>>>>>
>>>>>
>>>>> On Wed, Jul 13, 2011 at 7:58 PM, Piyush Kapoor <pkjee2...@gmail.com>wrote:
>>>>>
>>>>>> Shouldn't the value of "11000000" be -64????
>>>>>>
>>>>>>
>>>>>> On Wed, Jul 13, 2011 at 4:53 PM, Anika Jain 
>>>>>> <anika.jai...@gmail.com>wrote:
>>>>>>
>>>>>>> binary equivalent of 5.2 is
>>>>>>> 101.0011001100110011001100110011(nonterminating)..
>>>>>>>
>>>>>>> now it is actually stored in normalised frorm in 32 bits..
>>>>>>> like this
>>>>>>> <--1 bit for sign---><-----8 bits for exponent-----><--------23 bits
>>>>>>> for fraction------>
>>>>>>> this is from higher order byte to lower order for little endian..
>>>>>>>
>>>>>>> if no. is positive sign bit is 0 else it is 1
>>>>>>>
>>>>>>> The rule says change your floating point no. in such a form that
>>>>>>> after 1st digit that is 1 decimal point comes sooo
>>>>>>> here it becomes like
>>>>>>> 1.0100110011001100110011001100110011(nonterminating) * 2^2
>>>>>>>
>>>>>>> so i here get an exponent of 2 as 2 here.. now in exponent 8 bits
>>>>>>> this exponent is stored as 127+exponent so here it becomes 10000001..
>>>>>>>
>>>>>>> now fraction here is clearly value after the decimal point i.e.
>>>>>>> 01001100110011001100110011001100110011(non terminationg) but only 1st 23
>>>>>>> bits are saved rest are left
>>>>>>>
>>>>>>> so finally wht we get is:
>>>>>>>
>>>>>>> 11000000 10100110 01100110 01100110
>>>>>>> (64)          (-90)         (102)        (102)
>>>>>>>
>>>>>>>
>>>>>>>
>>>>>>> On Wed, Jul 13, 2011 at 2:49 PM, Piyush Kapoor 
>>>>>>> <pkjee2...@gmail.com>wrote:
>>>>>>>
>>>>>>>> why do we need a NthIntWithKBits() in this topic?
>>>>>>>>
>>>>>>>>
>>>>>>>> On Tue, Jul 12, 2011 at 7:58 AM, oppilas . <
>>>>>>>> jatka.oppimi...@gmail.com> wrote:
>>>>>>>>
>>>>>>>>>
>>>>>>>>>
>>>>>>>>> On Mon, Jul 11, 2011 at 5:54 PM, Piyush Kapoor <
>>>>>>>>> pkjee2...@gmail.com> wrote:
>>>>>>>>>
>>>>>>>>>> Can anybody give a full explanation
>>>>>>>>>>
>>>>>>>>>> http://ideone.com/K1QmV
>>>>>>>>>
>>>>>>>>>>  On Sat, Jul 9, 2011 at 10:49 PM, sunny agrawal <
>>>>>>>>>> sunny816.i...@gmail.com> wrote:
>>>>>>>>>>
>>>>>>>>>>> try to find out the binary representation of float value 5.2
>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> On Sat, Jul 9, 2011 at 10:46 PM, Sangeeta <
>>>>>>>>>>> sangeeta15...@gmail.com> wrote:
>>>>>>>>>>>
>>>>>>>>>>>> int main(){
>>>>>>>>>>>> int i;
>>>>>>>>>>>> float a=5.2;
>>>>>>>>>>>> char *ptr;
>>>>>>>>>>>> ptr=(char *)&a;
>>>>>>>>>>>> for(i=0;i<=3;i++)
>>>>>>>>>>>> printf("%d ",*ptr++);
>>>>>>>>>>>> }
>>>>>>>>>>>>
>>>>>>>>>>>> output:
>>>>>>>>>>>>  102 102 -90 64.explain?
>>>>>>>>>>>>
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>>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> --
>>>>>>>>>>> Sunny Aggrawal
>>>>>>>>>>> B-Tech IV year,CSI
>>>>>>>>>>> Indian Institute Of Technology,Roorkee
>>>>>>>>>>>
>>>>>>>>>>>
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>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> --
>>>>>>>>>> *Regards,*
>>>>>>>>>> *Piyush Kapoor,*
>>>>>>>>>> *CSE-IT-BHU*
>>>>>>>>>>
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>>>>>>>>
>>>>>>>>
>>>>>>>> --
>>>>>>>> *Regards,*
>>>>>>>> *Piyush Kapoor,*
>>>>>>>> *CSE-IT-BHU*
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>>>>>>
>>>>>>
>>>>>> --
>>>>>> *Regards,*
>>>>>> *Piyush Kapoor,*
>>>>>> *CSE-IT-BHU*
>>>>>>
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>>>>
>>>>
>>>>
>>>> --
>>>> *Regards,*
>>>> *Piyush Kapoor,*
>>>> *CSE-IT-BHU*
>>>>
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>>
>>
>>
>> --
>> Sunny Aggrawal
>> B-Tech IV year,CSI
>> Indian Institute Of Technology,Roorkee
>>
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-- 
*Regards,*
*Piyush Kapoor,*
*CSE-IT-BHU*

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