@Sagar
Memory sizes of long double variables are compiler and system configuration
dependent. So obviously, in accordance with your compiler, the size of long
double is 8 bytes.
On Mon, Jul 18, 2011 at 1:22 AM, Nikhil Gupta <nikhilgupta2...@gmail.com>wrote:
> @Aditya
>
> Here is the padding effect :
>
> Address of char : starts anywhere
> Address of long double : starts at 11 address locations from char variable
> --> 1+11+12=24 bytes
>
>
> On Mon, Jul 18, 2011 at 1:10 AM, sagar pareek <sagarpar...@gmail.com>wrote:
>
>> @aditya
>> actually first see your post, you have written o/p=24 accordingly padding
>> done is perfect. But actually its printing 16.
>> So now question arises of padding
>>
>> and its pareek not prateek :)
>>
>>
>> On Mon, Jul 18, 2011 at 12:38 AM, aditya kumar <
>> aditya.kumar130...@gmail.com> wrote:
>>
>>> @prateek . can you explain me ?? i dint get padding logic in this example
>>> of mine.
>>>
>>> On Mon, Jul 18, 2011 at 12:30 AM, sagar pareek <sagarpar...@gmail.com>wrote:
>>>
>>>> sizeof long double is 12. So padding concept is perfectly working
>>>>
>>>>
>>>> On Mon, Jul 18, 2011 at 12:26 AM, aditya kumar <
>>>> aditya.kumar130...@gmail.com> wrote:
>>>>
>>>>> struct st
>>>>> {
>>>>> char ch1;
>>>>> long double ld;
>>>>> }s;
>>>>> printf("%d",sizeof(s));
>>>>> //output : 24 (for 32-bit compiler)
>>>>> ->as i have mentioned above the behaviour is undefined in case of
>>>>> sizeof (struct)
>>>>> can any one explain me why the padding concept does not work here ??
>>>>>
>>>>> On Mon, Jul 18, 2011 at 12:13 AM, Parthiban <jega...@gmail.com> wrote:
>>>>>
>>>>>> @Abhi:
>>>>>> Answers:
>>>>>> 1. whenever a 'const' qualifier is added previously to a variable
>>>>>> declaration it means that the value of the variable is automatically
>>>>>> initialized to '0'(because of the 'auto' type of the const variable) and
>>>>>> cannot be changed in any of the following assignment statements to the
>>>>>> const
>>>>>> variable.
>>>>>>
>>>>>> 2.
>>>>>> Here for the structure struct s1 since the entire structure is ending
>>>>>> within 8 bytes no padding is done which means
>>>>>> s1: [char a]
>>>>>> 1byte
>>>>>> but for the structure struct s2 consider the following:
>>>>>> s2: [ char a
>>>>>> 1byte
>>>>>> ------ int a-------
>>>>>> ------4bytes----]
>>>>>> so here the concept of padding comes to make all the variable aligned
>>>>>> in even boundaries and so the structure after aligning will look as:
>>>>>> s2: [ ------ char b----
>>>>>> ------4byte-----
>>>>>> ------ int a-------
>>>>>> ------4bytes----]
>>>>>>
>>>>>> so the size of strcut s2 will be 8bytes..........
>>>>>>
>>>>>> --
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>>>>>
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>>>>
>>>>
>>>>
>>>> --
>>>> **Regards
>>>> SAGAR PAREEK
>>>> COMPUTER SCIENCE AND ENGINEERING
>>>> NIT ALLAHABAD
>>>>
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>>>
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>>
>>
>>
>> --
>> **Regards
>> SAGAR PAREEK
>> COMPUTER SCIENCE AND ENGINEERING
>> NIT ALLAHABAD
>>
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>
>
>
> --
> Nikhil Gupta
> Senior Co-ordinator, Publicity
> CSI, NSIT Students' Branch
> NSIT, New Delhi, India
>
>
--
Nikhil Gupta
Senior Co-ordinator, Publicity
CSI, NSIT Students' Branch
NSIT, New Delhi, India
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