ya thts ryt, for big endian it will be 64 -90 102 102

On Wed, Jul 13, 2011 at 10:22 PM, sunny agrawal <[email protected]>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 <[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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>>>>>
>>>>>
>>>>>
>>>>> --
>>>>> *Regards,*
>>>>> *Piyush Kapoor,*
>>>>> *CSE-IT-BHU*
>>>>>
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>>>
>>>
>>>
>>> --
>>> *Regards,*
>>> *Piyush Kapoor,*
>>> *CSE-IT-BHU*
>>>
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>>
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>
>
>
> --
> Sunny Aggrawal
> B-Tech IV year,CSI
> Indian Institute Of Technology,Roorkee
>
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