@ puneet :
tell me the case if u take the address to be starting from 4 not 0..
On Sat, Aug 6, 2011 at 6:55 PM, SANDEEP CHUGH <sandeep.aa...@gmail.com>wrote:
> @ puneet : ryt !! gud explanation.
>
>
> On Sat, Aug 6, 2011 at 6:53 PM, Puneet Gautam <puneet.nsi...@gmail.com>wrote:
>
>> Order is important ... but in the main case here which is
>>
>> 1) struct list
>> {
>> int data;
>> list *next;
>> }
>> and
>> 2) struct list
>> {
>> list *next;
>> int data;
>> }
>> order is not affecting its size...!!
>>
>> On 8/6/11, Puneet Gautam <puneet.nsi...@gmail.com> wrote:
>> > See guys.. the order is important but the size of whole structure
>> > needs to be a multiple of its largest sized variable...
>> > eg:
>> >
>> > struct p
>> > {
>> > double data;
>> > char a;
>> > char b;
>> > char c;
>> > char d;
>> > }t;
>> >
>> >
>> > struct q
>> > { char c;
>> > char d;
>> > double data;
>> > char a;
>> > char b;
>> > }t1;
>> >
>> > sizeof(t)=16
>> > sizeof(t1)=24
>> >
>> > This can be explained:Lets say address starts at 0
>> >
>> > In q structure, c and d take one byte each so data starts at 3 but it
>> > cant start at 3 (not its size multiple)...
>> > so double data starts at 8, leaving all 3-7 positions padded to char c
>> and
>> > d
>> >
>> > double ends at 16 so char a and b occupy 17 and 18 addresses.
>> >
>> > But if next structure variable starts, it wud have to start at 19
>> > which is not 8's multiple..
>> >
>> > So , char a and b are padded till address 23 and hence next structure
>> > variable can start at 24..(8 * 3)
>> >
>> > Hence t1's size =24, neither 19 nor 12...
>> >
>> > Similarly, we can account for structure p's variable t..t=16
>> > bytes(char a,b,c,d occupy 4bytes, get padded upto 7 and double then
>> > starts at 8 upto 15, next variable starts at 16..)
>> >
>> > Am i clear...???
>> >
>> >
>> >
>> > On 8/6/11, Nitish Garg <nitishgarg1...@gmail.com> wrote:
>> >> I think that the order is important. Because when we consider an array
>> of
>> >> structures the order becomes extremely important just as shown in the
>> >> above
>> >> example.
>> >>
>> >> On Sat, Aug 6, 2011 at 6:18 PM, Prashant Gupta
>> >> <prashantatn...@gmail.com>wrote:
>> >>
>> >>> Interesting :
>> >>> #include<iostream>
>> >>> using namespace std;
>> >>> int main()
>> >>> {
>> >>> struct p{
>> >>> int i;
>> >>> char j;
>> >>> char k;
>> >>> };
>> >>> struct q{
>> >>> char j;
>> >>> int i;
>> >>> char k;
>> >>> };
>> >>> printf("p=%u q=%u",sizeof(p),sizeof(q));
>> >>> return 0;
>> >>> }
>> >>> o/p : p=8 q=12
>> >>>
>> >>> On Sat, Aug 6, 2011 at 2:55 PM, Tushar Bindal
>> >>> <tushicom...@gmail.com>wrote:
>> >>>
>> >>>> that means the order is immaterial.
>> >>>> the sizeof the struct always remains same irrespective of the order
>> and
>> >>>> just depends on the type of variables???
>> >>>> why char with double does not get size in multiples of 8??
>> >>>>
>> >>>>
>> >>>> On Sat, Aug 6, 2011 at 12:54 PM, Puneet Gautam
>> >>>> <puneet.nsi...@gmail.com>wrote:
>> >>>>
>> >>>>> Sorry guys, int is 4 bytes on 64 bit and 2 bytes on 32 bit system..
>> >>>>>
>> >>>>> But padding rule remains same for both structures as mentioned
>> >>>>> above...
>> >>>>>
>> >>>>>
>> >>>>>
>> >>>>>
>> >>>>> On 8/6/11, Puneet Gautam <puneet.nsi...@gmail.com> wrote:
>> >>>>> > There is no difference between the two...
>> >>>>> >
>> >>>>> > On 32 bit system, both structures need every address location
>> where
>> >>>>> > int and pointer are stored to be a multiple of 4(highest size is
>> >>>>> > 4)..
>> >>>>> >
>> >>>>> > On 64 bit,
>> >>>>> > even if pointer is 4bytes(say, in 64 bit system), and p1, p2 be
>> >>>>> > structure variables, then p2 should start at address which is
>> >>>>> > multiple
>> >>>>> > of 8 as int data is 8bytes. So, if p1 starts at 0, it should end
>> at
>> >>>>> > 16
>> >>>>> > not 12 so that p2 starts at 8's multiple.
>> >>>>> >
>> >>>>> > This is done by padding pointer by 4bytes in both I and II struct.
>> >>>>> > declarations.
>> >>>>> >
>> >>>>> >
>> >>>>> > Hope i made it clear...!
>> >>>>> >
>> >>>>> > Thanks.
>> >>>>> >
>> >>>>> >
>> >>>>> >
>> >>>>> >
>> >>>>> > On 8/6/11, Tushar Bindal <tushicom...@gmail.com> wrote:
>> >>>>> >>
>> >>>>>
>> http://www.serc.iisc.ernet.in/ComputingFacilities/systems/cluster/xlf/html/xlfug/ug35.htm
>> >>>>> >> this says int is always 4 bytes and pointer is 8 bytes on 64 bit
>> >>>>> >> compiler.
>> >>>>> >>
>> >>>>> >> so how does padding affect these structures because of the
>> >>>>> >> difference
>> >>>>> in
>> >>>>> >> size of int and pointer?
>> >>>>> >>
>> >>>>> >>
>> >>>>> >> I tried this program
>> >>>>> >> https://ideone.com/CRU6x#view_edit_box
>> >>>>> >> char always gets 4 bytes whenever it has int or double in the
>> same
>> >>>>> struct
>> >>>>> >> irrrespctive of the order of the declaration of variables.
>> >>>>> >> I thought char should get size 8 when there is a double in the
>> ame
>> >>>>> struct
>> >>>>> >> whereas it gets size 4 only.
>> >>>>> >> what is the problem here?
>> >>>>> >>
>> >>>>> >> On Sat, Aug 6, 2011 at 4:40 AM, Shashank Jain
>> >>>>> >> <shashan...@gmail.com>
>> >>>>> >> wrote:
>> >>>>> >>
>> >>>>> >>> i dont understand the diff btw dem, could u plz elaborate?
>> >>>>> >>>
>> >>>>> >>> Shashank Jain
>> >>>>> >>> IIIrd year
>> >>>>> >>> Computer Engineering
>> >>>>> >>> Delhi College of Engineering
>> >>>>> >>>
>> >>>>> >>>
>> >>>>> >>>
>> >>>>> >>> On Sat, Aug 6, 2011 at 12:32 AM, Kamakshii Aggarwal
>> >>>>> >>> <kamakshi...@gmail.com
>> >>>>> >>> > wrote:
>> >>>>> >>>
>> >>>>> >>>> in case of 64 bit,
>> >>>>> >>>> size of second structure will also be 16 not 8
>> >>>>> >>>>
>> >>>>> >>>>
>> >>>>> >>>> On Fri, Aug 5, 2011 at 11:40 PM, UTKARSH SRIVASTAV <
>> >>>>> >>>> usrivastav...@gmail.com> wrote:
>> >>>>> >>>>
>> >>>>> >>>>> I think voth are just same..................
>> >>>>> >>>>>
>> >>>>> >>>>>
>> >>>>> >>>>> On Fri, Aug 5, 2011 at 10:57 AM, priya v <pria....@gmail.com>
>> >>>>> wrote:
>> >>>>> >>>>>
>> >>>>> >>>>>> in case of 64 bit machine y doesn't padding happen in the 2nd
>> >>>>> >>>>>> structure?
>> >>>>> >>>>>>
>> >>>>> >>>>>>
>> >>>>> >>>>>> On Fri, Aug 5, 2011 at 11:21 PM, hary rathor
>> >>>>> >>>>>> <harry.rat...@gmail.com>wrote:
>> >>>>> >>>>>>
>> >>>>> >>>>>>> no ,if u r using 32 bit machine . that will use 4 byte
>> pointer
>> >>>>> size
>> >>>>> >>>>>>> ,
>> >>>>> >>>>>>> but in 64 machine that enforce to be size of 8 . where
>> >>>>> >>>>>>> padding
>> >>>>> >>>>>>> will
>> >>>>> >>>>>>> take int your given first structure
>> >>>>> >>>>>>>
>> >>>>> >>>>>>> so for 32 bit- size will 8 8 for both structure
>> >>>>> >>>>>>> for 64 bit - size will 16 and 12 respectively cause of 4 bit
>> >>>>> padding
>> >>>>> >>>>>>> in
>> >>>>> >>>>>>> one structure
>> >>>>> >>>>>>>
>> >>>>> >>>>>>> hence 2nd structure is good for use
>> >>>>> >>>>>>>
>> >>>>> >>>>>>> --
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>> >>>>> >>>>> --
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>> >>>>> >>>>> CSE-3
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>> >>>>> >>>> --
>> >>>>> >>>> Regards,
>> >>>>> >>>> Kamakshi
>> >>>>> >>>> kamakshi...@gmail.com
>> >>>>> >>>>
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>> >>>>> >>
>> >>>>> >>
>> >>>>> >>
>> >>>>> >> --
>> >>>>> >> Tushar Bindal
>> >>>>> >> Computer Engineering
>> >>>>> >> Delhi College of Engineering
>> >>>>> >> Mob: +919818442705
>> >>>>> >> E-Mail : tushicom...@gmail.com
>> >>>>> >> Website: www.jugadengg.com
>> >>>>> >>
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>> >>>>>
>> >>>>
>> >>>>
>> >>>> --
>> >>>> Tushar Bindal
>> >>>> Computer Engineering
>> >>>> Delhi College of Engineering
>> >>>> Mob: +919818442705
>> >>>> E-Mail : tushicom...@gmail.com
>> >>>> Website: www.jugadengg.com
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>> >>>>
>> >>>
>> >>>
>> >>>
>> >>> --
>> >>> Prashant Gupta
>> >>> B.Tech Final Year
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>> >>> NIT Trichy
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