output??

[Sandeep]

struct demo
{
char c;
double d;
int s;
};
what wud be the size of struct object?

[saurabh singh]

16 bytes....

suppose starting address is 2000

char c will occupy 2000th position

Now compiler will look for next address thats multiple of 4.This is done to improove access speed.

double starts from 2004-2012.(8 bytes)

s starts from 2012-2016.

Total size=2016-2000=16 bytes

--
Saurabh Singh
B.Tech (Computer Science)
MNNIT ALLAHABAD

[SANDEEP CHUGH]

OUTPUT IS 24 ACCORDING TO FOLLOWING

CHAR (1) + 7 BYTE PADDING + DOUBLE (8) +INT(4)+ 4 BYTE PADDING

AND COMPILER WILL LOOK FOR ADDRESS WHICH IS A MULTIPLE OF 8 , NOT 4

BECAUSE MAXIMUM SIZED ELEMENT IS DOUBLE AND IT TAKES 8 BYTES.

[saurabh singh]

answer is 16 according to my logic...check with a gcc compliant compiler.

[saurabh singh]

would be 8 on 64 bit m/c....4 on a 32  bit m/c

[SANDEEP CHUGH]

out put is 24 for 32 bit compiler

and if am using turbo c it wud be 11

[saurabh singh]

http://ideone.com/xubzQ

[Tushar]

@saurabh
i tried the same thing in the morning and got the answer as 16 bytes
but not able to understand why only 4 bytes are reserved for char when
double is also present.
why aren't char and int padded according to double?

shall we pad char to end in multiple of 4 only when any other data
type is present?

[saurabh singh]

If I were a compiler I would have tried to pad according to the m/c size rather than the type of members...I havent read much of struct padding but my logic says pad for multiples of 4 for 32 bit addresses and pad for 8 in 64 bit addresses,

[Tushar Bindal]

thanks for explaining that

--
Tushar Bindal
Computer Engineering
Delhi College of Engineering
Mob: +919818442705
E-Mail : tushi...@gmail.com
Website: www.jugadengg.com