How size of this structure comes out to be unexpected?

Question

Possible Duplicate:
Why isn't sizeof for a struct equal to the sum of sizeof of each member?

Hi all,

i wanted to know how size of below structure comes out to be 24.As per my calculation it should be 20.and one more thing is there any way this structure is taking size of its variable t into account??

Please ignore any syntax error and i am on 32 bit machine

 struct structc
{ 

char        c; 
double      d; 
int         s;
} t;

main()
{
printf("sizeof(structc_t) = %d\n", sizeof(t)); 
}

@ArtoAle t needs sizeof(char) + 7 byte padding + sizeof(double) + sizeof(int) = 1 + 7 + 8 + 4 = 20 bytes. — Amit Singh Tomar, Jun 15 '11 at 08:19
Yeah, right, but you didn't consider final 4 bytes padding :) — ArtoAle, Jun 15 '11 at 08:20
@AMIT: It is indeed a duplicate. Also, it's meaningless to pull numbers out of thin air and then wonder why they don't match reality. What is your compiler? Only its documentation can answer the "why 24?" question definitively. — Jon, Jun 15 '11 at 08:21
yaa @ArtoAle that is what i want to know final 4 bytes padding why its needed and for which variable?? — Amit Singh Tomar, Jun 15 '11 at 08:23

score 6 · Answer 1 · edited May 23 '17 at 12:26

6

The size of the struct includes padding bytes between is members due to packing alignment, which is compiler- and architecture-dependent (see here for an example)

Update: Not surprisingly, this question is a dupe. For a better answer, see Why isn't sizeof for a struct equal to the sum of sizeof of each member?.

edited May 23 '17 at 12:26

Community

1
1

answered Jun 15 '11 at 08:14

Jon

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How size of this structure comes out to be unexpected?

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