Please dont reply to thi post..wrng question....sry for that
On Fri, Oct 26, 2012 at 2:31 AM, rahul sharma <rahul23111...@gmail.com>wrote:
> address of each byte is printed...each byte is 32 bit long...so 32/4=8 hex
> for each row...but how these are separated with %.2x tin group of
> 2's....%.2x means min. 2 width ..if <2..then leading zeros...
> so it should be ("%.2x",start[i]);
> start[i] in 2 but it is more so ignored...print start[i]....still how
> these are able to grouped??
>
>
> On Fri, Oct 26, 2012 at 1:53 AM, rahul sharma <rahul23111...@gmail.com>wrote:
>
>> As per o/p below:
>> 00 00 80 3f
>> 01 00 00 00
>> 44 ff 28 00
>> 01 00 00 00
>>
>> first byte address is first row.
>> second byte address is second row.
>> third byte address third row
>> and so on
>>
>> but how first row containg 3 values...as 1byte=2hexdigits..si only two
>> digits must be there.
>> and u said 44 ff 28 00 is address of i..how???getting a lot confused..plz
>> clearify..thnx in advance
>>
>> On Sun, Oct 21, 2012 at 1:10 AM, Saurabh Kumar <srbh.ku...@gmail.com>wrote:
>>
>>> Sorry, about that.
>>> Read it as:
>>> Yes a hex digit is represented by 4 bits but 1 Byte is being read using
>>> a char pointer* and you're printing the values in those Bytes.
>>>
>>>
>>> On 21 October 2012 01:03, Saurabh Kumar <srbh.ku...@gmail.com> wrote:
>>>
>>>> Sorry, I don't understand your question. *%.2x *is only a precision
>>>> specifier still.
>>>> (%.2x was used for neat formatting only, because you are printing the
>>>> values only 1 Byte long and a Byte can occupy at max 2digits in hex)
>>>>
>>>> >>>>hex representated by 4 bits.
>>>> Yes hex is represented by 4 bits i.e. 1 Byte and that's what you are
>>>> reading with a char pointer*, 1 Byte each time and printing the values in
>>>> those Bytes.
>>>>
>>>> >>>>total we have to represent 32 bits and 8 bits in each....plz xplain
>>>> Each output represents 32bits only. 1 Byte each (in total 4Bytes)
>>>>
>>>> It's showing you the memory layout. You stored *i = 1; *and when
>>>> probed it using a char pointer. you found following four bytes written:
>>>> *01 00 00 00*
>>>> It shows that on your machine:
>>>> 1. int is 4bytes long. (4x1Byte)
>>>> 2. First byte stores the least significant value, hence you are working
>>>> on a Little endian machine.
>>>>
>>>> similarly, for pointer:
>>>> char pointer reads 1 Byte at a time. It read 4Bytes in total i.e. 32
>>>> bits. Hence, you are working on a 32 bit machine. (as pointer has
>>>> value: *44 ff 28 00, *address of i)*.*
>>>> *
>>>> *
>>>> *
>>>> *
>>>> PS: This is an algorithm group, please refrain from asking such
>>>> language specific questions.
>>>>
>>>> On 21 October 2012 00:19, rahul sharma <rahul23111...@gmail.com> wrote:
>>>>
>>>>> Actually i have taken form http://www.geeksforgeeks.org/archives/730
>>>>> Please explain me o/p...as hex representated by 4 bits....then how cum
>>>>> is following o/p
>>>>> 00 00 80 3f
>>>>> 01 00 00 00
>>>>> 44 ff 28 00
>>>>> 01 00 00 00
>>>>>
>>>>> total we have to represent 32 bits and 8 bits in each....plz xplain
>>>>>
>>>>> On Sun, Oct 21, 2012 at 12:05 AM, rahul sharma <
>>>>> rahul23111...@gmail.com> wrote:
>>>>>
>>>>>> void show_bytes(byte_pointer start, int len)
>>>>>> {
>>>>>> int i;
>>>>>> for (i = 0; i < len; i++)
>>>>>> printf(" %.2x", start[i]);
>>>>>> printf("\n");
>>>>>> }
>>>>>>
>>>>>>
>>>>>>
>>>>>> byte_pointr is unsigned char *...typedef unsigned char *
>>>>>> byte_pointer....
>>>>>> plz tell me use of %.2x i knowx is for hexa........does it mean
>>>>>> print 8 bites of address in 4 hexa of 2 bits???i cant get xactly plz
>>>>>> explain
>>>>>>
>>>>>
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>>>>
>>>>
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>>
>>
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