+1 to Dipankar

Sanju
:)



On Sat, Aug 20, 2011 at 12:01 AM, Dipankar Patro <[email protected]>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 <[email protected]> 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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