C++ sizeof with bool

Question

It is a simple question. Code first.

struct A {
    int x; 
};
struct B {
    bool y;
};
struct C {
    int x;
    bool y;
};

In main function, I call

cout << " bool : " << sizeof(bool) <<
     "\n int : " << sizeof(int) <<
     "\n class A : " << sizeof(A) <<
     "\n class B : " << sizeof(B) <<
     "\n class C : " << sizeof(C) << "\n";

And the result is

bool : 1
int : 4
class A : 4
class B : 1
class C : 8

Why is the size of class C 8 instead of 5? Note that this is compiled with gcc in MINGW 4.7 / Windows 7 / 32 bit machine.

The extra space is *padding*, but the purpose is *alignment*: most architectures are better suited for aligned access to data, so a 32bit `int` aligned to 4 bytes is faster to access (sometimes even atomic) than unaligned memory. In some architectures it is even worse and you cannot use unaligned types directly. — David Rodríguez - dribeas, Oct 02 '12 at 01:28
Possible duplicate of [Why isn't sizeof for a struct equal to the sum of sizeof of each member?](http://stackoverflow.com/questions/119123/why-isnt-sizeof-for-a-struct-equal-to-the-sum-of-sizeof-of-each-member) — Frédéric Hamidi, Aug 25 '14 at 08:31

evpo · Accepted Answer · 2014-01-25T01:33:20.007

12

The alignment of an aggregate is that of its strictest member (the member with the largest alignment requirement). In other words the size of the structure is a multiple of the alignment of its strictest (with the largest alignment requirement) member.

struct D
{
  bool a;
  // will be padded with char[7]
  double b; // the largest alignment requirement (8 bytes in my environment)
};

The size of the structure above will be 16 bytes because 16 is a multiple of 8. In your example the strictest type is int aligning to 4 bytes. That's why the structure is padded to have 8 bytes. I'll give you another example:

struct E
{
  int a;
  // padded with char[4]
  double b;
};

The size of the structure above is 16. 16 is multiple of 8 (alignment of double in my environment).

I wrote a blog post about memory alignment for more detailed explanation http://evpo.wordpress.com/2014/01/25/memory-alignment-of-structures-and-classes-in-c-2/

edited Jan 25 '14 at 01:33

answered Oct 02 '12 at 01:33

evpo

2,436
16
23

1

It's not the longest type; it's the alignment of the type whose alignment is "strictest" (i.e. longest). In many ABIs, double's are 8-byte long but only 4-byte aligned, for example. (Theoretically, it should be the LCM of all the alignments, but the standard requires that "Every alignment value shall be a non-negative integral power of two" (3.11, para 4), so you can just take the longest one.) – rici Oct 02 '12 at 04:03
what if there is a struct in struct. – M.kazem Akhgary Aug 06 '19 at 04:08
@M.kazemAkhgary The alignment requirement of the struct will be considered in the same way as with the plain data types such as double. – evpo Aug 08 '19 at 00:51

score 2 · Answer 2 · answered Oct 02 '12 at 01:06

2

Aligning structures to the size of a word, which is 4 bytes here.

answered Oct 02 '12 at 01:06

djechlin

59,258
35
162
290

Can I ask further? What is the reason of the aligning structure? – Sungmin Oct 02 '12 at 01:24
This doesn't explain the alignment strategy or may be the answer is not complete. If it aligned to 4, then struct B would be 4 bytes too. However, its size is 1 byte. Also in my answer I demonstrated that the compiler aligned to 8 bytes which is not 4 again. My point is that it aligns the size of the structure to a multiple of the largest type it contains. That explains it better. – evpo Oct 02 '12 at 02:15

score -1 · Answer 3 · edited May 23 '17 at 10:30

-1

Looking at the definition of your struct, you have 1 byte value followed by 4 byte Integer. This integer needs to be allocated on 4 byte boundary, which will force compiler to insert a 3 byte padding after your 1 byte bool. Which makes the size of struct to 8 byte. To avoid this you can change order of elements in the struct.

Also for two sizeof calls returning different values, are you sure you do not have a typo here and you are not taking size of pointer or different type or some integer variable.

Answered by Rohit J on struct size is different from typedef version?

edited May 23 '17 at 10:30

Community

1
1

answered Oct 02 '12 at 01:10

zeroglobal

13
3

In my case, changing the order of elements in struct C did not influence the result. – Sungmin Oct 02 '12 at 01:22
@Sungmin: The order of the members does not usually affect the padding. Consider that if the `int` must be aligned to a 4 byte boundary, it can be ordered as `bool`, padding, `int`; or `int`, `bool`, padding. Note that in an array, each element is located `sizeof(type)` bytes beyond the previous one. – David Rodríguez - dribeas Oct 02 '12 at 01:29
@Sungmin: You're correct. In this case changing the order would not change the size of the `struct`. – David Hammen Oct 02 '12 at 01:30

C++ sizeof with bool

3 Answers3