On Sun, Aug 21, 2011 at 11:11 AM, Rahul Tiwari <rahultiwari6...@gmail.com>wrote:
> @vijay ....
> u take normalised form of 5.375 wrong .....
> actual normalised form of 5.375 = 0100 0000 1010 1100 0000 0000 0000 0000
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> On Sun, Aug 21, 2011 at 10:12 AM, Vijay Khandar
> <vijaykhand...@gmail.com>wrote:
>
>> but why only o/p-00 00 AC 40 and not AC 40 00 00 or 00 AC 40 00 or 40 AC
>> 00 00etc , plz explain in detail how p[0] pts to 00 and p[1] pts to 00 and
>> p[2] pts AC or 1010 1100 and p[3] pts to 40 or 0100 0000 ONLY in this
>> way..............Vijay Khandar...........
>>
>>
>> On Sat, Aug 20, 2011 at 12:31 PM, Dipankar Patro <dip10c...@gmail.com>wrote:
>>
>>> float is 4 bytes.
>>> so a=3.75 will be stored in 4 bytes in memory.
>>>
>>> the moment you have a pointer referring to the same memory location but
>>> type cast to (char *), the pointer will refer to character i.e. 1 byte.
>>> ^^ this explains the p[0] , p[1], p[2], p[3] <- 4 bytes of the 3.75
>>>
>>> now finally the o/p
>>> 00 00 AC 40
>>> ^^ it is in little endian format. i.e the data bytes are stored in memory
>>> is reverse format.
>>>
>>> On 20 August 2011 11:21, Vijay Khandar <vijaykhand...@gmail.com> wrote:
>>>
>>>> If the binary equivalent of 5.375 in normalised form is - 0100 0000
>>>> 1010 0000 1100 0000 0000 0000
>>>>
>>>> what is the o/p of following code-
>>>> main()
>>>> {
>>>> float a=5.375;
>>>> char *p;
>>>> int i;
>>>> p=(char *)&a;
>>>> for(i=0;i<=3;i++)
>>>> printf("%02X",(unsigned char)p[i]);
>>>> }
>>>>
>>>> O/P= 00 00 AC 40
>>>> Plz, Plz anyone explain me in detail, how this o/p is coming?
>>>> Vijay..........
>>>>
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>>>
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> --
> Rahul Tiwari aka " DONE "
> B Tech Final Year
> Information Technology
> Motilal Nehru National Institute of Technology , Allahabad
> 9838339030
>
>
--
Rahul Tiwari aka " DONE "
B Tech Final Year
Information Technology
Motilal Nehru National Institute of Technology , Allahabad
9838339030
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