What is the behaviour of the sizeof() operator with a class having one or more inner classes? Let's consider the following code:

#include <cstdint>
#include <iostream>

class A
{
	class B
	{
		uint64_t i64;
	};
	B vec[ 10 ];
};

int main()
{	
	std::cout << sizeof( A ) << std::endl;
	//...
	return( 0 );
}

So far so god, it will print "80".

But now consider the next example:

#include <cstdint>
#include <iostream>

class A
{
	class B
	{
		uint64_t i64;
	};
	B vec[ 10 ];
	uint8_t arr[ 10 ];
};

int main()
{	
	std::cout << sizeof( A ) << std::endl;
	//..
	return( 0 );
}

Instead of "90" (80 bytes for vec + 10 bytes for arr) it prints 96, it counts 6 "extra" bytes.

Please not that not having vec B array, sizeof returns 10.

Having a single uint8_t instead an array of ten it will print 88 instead of 81.

.... so.... why? Is all this due data alignment?

All this happens on MSVC12 (Visual studio 2013), compiling in both debug and release mode.

Yes, it is due to data alignment. I am sure Google can fill you in on the details.

Yes, it is due to data alignment. I am sure Google can fill you in on the details.

Thank you, i will verify it when as soon I am able to use my PC (SSD firmware failure -.-).

B contains a u64. which must be 8-byte aligned. A contains a B, so A must be 8-byte aligned.

There's a requirement that sizeof(x) must be a multiple of alignof(x).

e.g. if you created an array of A's, but A had size=90 and align=8, then the first item would be aligned correctly, but the second item would not!

To solve this, the size of A is 'padded' with a few wasted bytes.

A's members are only 90 bytes long, which is 11.25 * 8 bytes. An extra 0.75 * 8 bytes needs to be added to pad 'A' out, so that it's size is a multiple of it's alignment requirement.

Just to clarify, this doesn't have anything to do with the inner class.

Next time you make a comparison, only change one thing at a time and the result will be less confusing! ;)

Just as Rattenhrim said, this is not related to the inner class.

This code will have the exact same result as yours:

class A
{
	uint8_t arr[ 10 ];
	uint64_t i64[ 10 ];
};

Even though it has no inner class.

If you want, you can increase your arr array size up to 16 and A's size would still be 96.

And yet another note... the size of the objects is not necessarily portable. Often you can just sum the sizes of the contents and adjust for padding, but sometimes it is more complex than that.

On another compiler the size may be something else; the compiler may have different alignment rules, different packing rules, or different rules about the content of class data.

Compiler options, pragmas, and other commands can change alignment and packing requirements which in turn can modify the size of an object.

Various rules allow compilers to add and remove padding, adjust for alignment, and even add 'invisible' data for the compiler's own purposes.


In your case you know that your class is 96 bytes, or 88 bytes, for those objects. As long as you don't modify the objects or compiler options they will stay that size.

You probably aren't going to change from MSVC12 to clang or egcs, nor are you likely to move from x86 to arm9 or from a 32-bit x86 processor to a 128-bit processor or a 16-bit processor or make drastic changes to your data packing setting. But someday in your career you probably will, so recognize that these things can result in different object sizes. Something may be 96 bytes on one system and 128 bytes on another system, and both are valid.

It is good to know and understand why things are the size they are, but it is also good to know that many factors can cause the underlying assumptions and the final size to be different than you directly count.