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 <[email protected]> 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 <[email protected]>wrote:
>
>> why is the order of numbers reversed?
>>
>>
>> On Wed, Jul 13, 2011 at 8:07 PM, Anika Jain <[email protected]>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 <[email protected]>wrote:
>>>
>>>> Shouldn't the value of "11000000" be -64????
>>>>
>>>>
>>>> On Wed, Jul 13, 2011 at 4:53 PM, Anika Jain <[email protected]>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 <[email protected]>wrote:
>>>>>
>>>>>> why do we need a NthIntWithKBits() in this topic?
>>>>>>
>>>>>>
>>>>>> On Tue, Jul 12, 2011 at 7:58 AM, oppilas . <[email protected]
>>>>>> > wrote:
>>>>>>
>>>>>>>
>>>>>>>
>>>>>>> On Mon, Jul 11, 2011 at 5:54 PM, Piyush Kapoor 
>>>>>>> <[email protected]>wrote:
>>>>>>>
>>>>>>>> Can anybody give a full explanation
>>>>>>>>
>>>>>>>> http://ideone.com/K1QmV
>>>>>>>
>>>>>>>>  On Sat, Jul 9, 2011 at 10:49 PM, sunny agrawal <
>>>>>>>> [email protected]> wrote:
>>>>>>>>
>>>>>>>>> try to find out the binary representation of float value 5.2
>>>>>>>>>
>>>>>>>>>
>>>>>>>>> On Sat, Jul 9, 2011 at 10:46 PM, Sangeeta <[email protected]
>>>>>>>>> > 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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>>>>>
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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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