size of struct in C

Question

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

Consider the following C code:

#include <stdio.h>    

struct employee
{
  int id;
  char name[30];  
};

int main()
{
  struct employee e1;      
  printf("%d %d %d", sizeof(e1.id), sizeof(e1.name), sizeof(e1));
  return(0);
}

The output is:

4 30 36

Why is the size of the structure not equal to the sum of the sizes of its individual component variables?

You can use the attribute packed in gcc.. This will drop padding and keep the structure as small as possible. struct test_t { int c; } __attribute__((__packed__)); — eaanon01, Dec 03 '09 at 18:57
Duplicate of (at least) http://stackoverflow.com/questions/119123/why-isnt-sizeof-for-a-struct-equal-to-the-sum-of-sizeof-of-each-member — dmckee --- ex-moderator kitten, Dec 03 '09 at 19:09
eaanon01. you shouldn't tell anybody about something as unportable as attribute packed unless there's a really really good reason and all the implications are understood. — Peeter Joot, Dec 03 '09 at 19:12
See also: http://stackoverflow.com/questions/833526/why-short-is-stored-as-4-bytes-in-a-struct-in-c and http://stackoverflow.com/questions/143025/how-do-i-find-the-size-of-a-struct — dmckee --- ex-moderator kitten, Dec 03 '09 at 19:16
See this C FAQ on memory alignment. http://c-faq.com/struct/align.esr.html — Richard Chambers, Apr 26 '13 at 12:37

Michael Burr · Accepted Answer · 2009-12-04T06:53:28.110

82

The compiler may add padding for alignment requirements. Note that this applies not only to padding between the fields of a struct, but also may apply to the end of the struct (so that arrays of the structure type will have each element properly aligned).

For example:

struct foo_t {
    int x;
    char c;
};

Even though the c field doesn't need padding, the struct will generally have a sizeof(struct foo_t) == 8 (on a 32-bit system - rather a system with a 32-bit int type) because there will need to be 3 bytes of padding after the c field.

Note that the padding might not be required by the system (like x86 or Cortex M3) but compilers might still add it for performance reasons.

edited Dec 04 '09 at 06:53

answered Dec 03 '09 at 18:23

Michael Burr

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1

+1, although aligning to 6 bytes sounds weird. Maybe I am a bit behind in low-level stuff, though. – Michael Krelin - hacker Dec 03 '09 at 18:25
2

Well, the name starts at offset 4 (plausible enough) and extends to 34. 34 is not a multiple of 4 so it's end-padded to 36, which is 9*4. Makes sense to me! – Carl Smotricz Dec 03 '09 at 18:27
3

Its aligning to 32bit boundaries (4,8,16,24,32,36,...) – Mordachai Dec 03 '09 at 18:27
9

It is not aligning to 6 bytes. int is taking 4 bytes and char[30] is taking 32 bytes, both multiples of 4 making the size of the structure also a multiple of 4 bytes. – Sinan Ünür Dec 03 '09 at 18:28
Ah, that makes perfect sense. – Michael Krelin - hacker Dec 03 '09 at 18:28
The 6-byte alignment is likely a coincidence - 36 is a multiple of 4. – Michael Burr Dec 03 '09 at 18:29
Also of note: the compiler can insert padding *between* members, too, not just at the end, e.g. `struct { char c; int x; }` will also have size 8 on most 32-bit systems. – Adam Rosenfield Dec 03 '09 at 18:36
+1 or mentioning arrays since that's the biggest reason for certain platforms – Jim Buck Dec 03 '09 at 18:40
While I know that it is faster to answer than to search up the duplicates, you had to expect that this has been asked and answered before. – dmckee --- ex-moderator kitten Dec 03 '09 at 19:08

score 3 · Answer 2 · answered Dec 03 '09 at 18:29

Aligning to 6 bytes is not weird, because it is aligning to addresses multiple to 4.

So basically you have 34 bytes in your structure and the next structure should be placed on the address, that is multiple to 4. The closest value after 34 is 36. And this padding area counts into the size of the structure.

score 3 · Answer 3 · answered Dec 03 '09 at 18:33

As mentioned, the C compiler will add padding for alignment requirements. These requirements often have to do with the memory subsystem. Some types of computers can only access memory lined up to some 'nice' value, like 4 bytes. This is often the same as the word length. Thus, the C compiler may align fields in your structure to this value to make them easier to access (e.g., 4 byte values should be 4 byte aligned) Further, it may pad the bottom of the structure to line up data which follows the structure. I believe there are other reasons as well. More info can be found at this wikipedia page.

score 3 · Answer 4 · answered Dec 03 '09 at 18:56

3

Your default alignment is probably 4 bytes. Either the 30 byte element got 32, or the structure as a whole was rounded up to the next 4 byte interval.

answered Dec 03 '09 at 18:56

Liz Albin

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size of struct in C

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