padding wud be between int & char..(for ur last case)
now u said if it starts at 4 , still the block will be 8 (size of double),
that is 7 bytes to be padded..
so double element of structure would be starting at 4 + char(1) + 7 byte
padding ==12
but 12 is not a multiple of 8..
so there is a problem??
On Sat, Aug 6, 2011 at 8:06 PM, siddharth srivastava <akssps...@gmail.com>wrote:
> but if the reference is not 0 .. if it wud have been 4 then for tht case
>> tell me do we really need that 7 bytes??
>>
>
> yes, because in any case the block would be of 8 bytes only
> so even if you start at 4, you would have only 7 bytes left in that
> location for the next variable(which in this case is a double of 8 bytes, so
> the allocation would start at the 8th byte from the first location)
>
> Look at the following stack: X represents occupied memory, - represents the
> memory to be padded (in bytes)
>
> X - - - - - - -
> XXXXXXXX
> XXXX - - - -
>
> So, first char occupies 1 byte, then there are only 7 bytes left at that
> index, which are not enough for a double to be stored (got my point)
> Moreover, if you had
> struct a{
> char c;
> int i;
> double d;
> }
>
> then the size would have been 16
>
> as after allocating a char, there was enough space for an int to be stored.
> (just not sure if padding would be after int or between int and char)
>
>
>
>
>>
>> and yeah it is dependent on compiler size..
>> if u compile this snippet
>>
>> struct demo
>> {
>> char c;
>> double d;
>> int s;
>>
>>> }
>>
>>
>> in turbo c , its giving 11,, that means compiler nt doing padding at all
>> in turbo c..
>>
>> On Sat, Aug 6, 2011 at 7:50 PM, siddharth srivastava <akssps...@gmail.com
>> > wrote:
>>
>>> Hi Sandeep
>>>
>>> On 6 August 2011 19:16, SANDEEP CHUGH <sandeep.aa...@gmail.com> wrote:
>>>
>>>> take this case
>>>>
>>>> struct demo
>>>> {
>>>> char c;
>>>> double d;
>>>> int s;
>>>>
>>>>> }
>>>>
>>>>
>>>> what wud be the size??
>>>>
>>>>
>>>> solution is 24 according to following:-->
>>>>
>>>> char (1) + 7 byte padding +double(8)+int(4)+ 4 byte padding
>>>>
>>>>
>>>> suppose address starts at 4..
>>>>
>>>> i just wanna ask .. why there is 7 byte padding.. because just after 3
>>>> bytes padding after char we are getting address that is multiple of 8(size
>>>> of largest)..
>>>>
>>>
>>>
>>> after 3 bytes of padding i.e. the 4th byte(if reference is 0), is not a
>>> multiple of 8.
>>> Moreover, try understanding with this example:
>>> visualize a stack of memory with each location having the size of the
>>> largest sized variable in the structure.
>>> So in your case, each stack element is ought to be 8 bytes (due to
>>> double)
>>> Now, when first character is allocated, we have only 7 bytes left at that
>>> location in the stack, hence double has to be allocated in the next
>>> location.i.e. 2nd stack element and 8th byte (again reference 0)
>>>
>>> So total size is ought to be occupied is 24 in this case
>>>
>>>
>>> say if you had this declaration:
>>> struct a
>>> {
>>> char a;
>>> double d;
>>> int c;
>>> int e;
>>> }
>>>
>>> then too size would have been 24 as after allocation of int c, we still
>>> have 4 bytes in the same location which are then occupied by e (and were
>>> padded in previous case)
>>>
>>> Let me know, if I am not clear.
>>>
>>> @All
>>> Is it really architecture dependent (32 bit or 64 bit) ?
>>>
>>>
>>>
>>>>
>>>> can u please tell me??
>>>>
>>>>
>>>>
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>>>
>>>
>>>
>>> --
>>> Regards
>>> Siddharth Srivastava
>>>
>>>
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>>
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>
>
>
> --
> Regards
> Siddharth Srivastava
>
>
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