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.

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 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.