C++: sizeof of struct with bit fields

Question

Why is gcc giving returning 13 as the sizeof of the following class ? It seems to me that we should get e (4 bytes) + d (4 bytes) + 1 byte (for a and b) = 9 bytes. If it was alignment, aren't most 32 bit systems aligned on 8 byte boundaries ?

class A {
  unsigned char a:1;
  unsigned char b:4;
  unsigned int d;
  A* e;
} __attribute__((__packed__));


int main( int argc, char *argv[] )
{
  cout << sizeof(A) << endl;
}

./a.out 13

I would guess 32 bit systems would actually be aligned on 4 byte (32 bits) boundaries. — K-ballo
– K-ballo, Commented Sep 15, 2011 at 21:05

Chetan Ahuja · Accepted Answer · 2011-09-15 21:07:59Z

12

You are very likely running on a 64 bit platform and the size of the pointer is not 4 but 8 bytes. Just do a sizeof on A * and print it out.

answered Sep 15, 2011 at 21:07

Chetan Ahuja

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K-ballo · Accepted Answer · 2011-09-15 21:04:52Z

6

The actual size of structs with bitfields is implementation dependent, so whatever size gcc decides it to be would be right.

answered Sep 15, 2011 at 21:04

K-ballo

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3 Comments

Daniel Over a year ago

This is correct, but I don't see any reason why it would allocate 5 bytes for 5 bits (unless its really stupid compiler without any internal bit-wise operations).

K-ballo Over a year ago

Yes, its a bit strange that it allocates an odd number of bytes. However, it can do it if it wants too, which is a strong hint that another method should be used in order to control the number of bytes allocated.

T.E.D. Over a year ago

+1 - Right answer. If you want to be able to specify exactly which bytes get which data in a struct and what its ultimate size is, you'd be better off using Ada.

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