query?

[Radde]

HI all,

#include<iostream.h>

class A
{
//char i;
};

class B
{
//char i;
};

class C: public A, public B
{
private:
int i;
};

void main()
{
C c;
cout<<"size is "<<sizeof(c)<<endl;
}

When i run above code, iam getting size as 8, actually there is only
one int variable, it should be 4, why is it 8??
Suppose if i move int variable from class C to clas A, then size is 4.
why is it 8 if we add variable to class C?

Cheers...

[Victor Bazarov]

Radde wrote:

#include<iostream.h>
#include <iostream>

class A
{
//char i;
};

class B
{
//char i;
};

class C: public A, public B
{
private:
int i;
};

void main()
int main()
{
C c;
cout<<"size is "<<sizeof(c)<<endl;
std::cout<<"size is "<<sizeof(c)<<std::endl;
}

When i run above code, iam getting size as 8, actually there is only
one int variable, it should be 4, why is it 8??
Probably because your default compiler options make alignment at 8-byte
boundary. No way to tell. Alignment and padding are compiler-specific.
Suppose if i move int variable from class C to clas A, then size is 4.
why is it 8 if we add variable to class C?

You need to ask in the newsgroup dedicated to your compiler. Those issues
are compiler-specific and are not defined in the language Standard.

V

[Greg]

Victor Bazarov wrote:

Radde wrote:

#include<iostream.h>

#include <iostream>

class A
{
//char i;
};

class B
{
//char i;
};

class C: public A, public B
{
private:
int i;
};

void main()

int main()

{
C c;
cout<<"size is "<<sizeof(c)<<endl;

std::cout<<"size is "<<sizeof(c)<<std::endl;

}

When i run above code, iam getting size as 8, actually there is only
one int variable, it should be 4, why is it 8??

Probably because your default compiler options make alignment at 8-byte
boundary. No way to tell. Alignment and padding are compiler-specific.

The question being asked is why an instance of class C is larger than
an instance of class A or class B, not why it is four bytes larger. The
question is a good one. There would seem to be no apparent reason why a
compiler would increase the size of C when it inherits from A and B,
even when the amount of data that class C stores remains unchanged. And
in fact the only reason that the C++ compiler makes C larger in this
situation is because the C++ Standard requires it.

The C++ Standard says that neither class A or class B in this situation
can be of size zero bytes. Therefore an object of class C will
necessarily have to be larger than either of its two base classes,
because C's size cannot be zero bytes either. The answer has nothing to
do with padding or alignment, but rather conformity with the Standards.
Granted, the Standard doesn't say how much larger C has to be, but it
does explain why C is larger than A or B. And that really is the
question being asked.

Note that the "empty base class optimization" would make classes A and
B add zero incremental bytes to objects of type C. To take advantage of
this optimization (if the compiler supports it) classes A and B must
each derive from an empty base class themselves.

Suppose if i move int variable from class C to clas A, then size is 4.
why is it 8 if we add variable to class C?

You need to ask in the newsgroup dedicated to your compiler. Those issues
are compiler-specific and are not defined in the language Standard.

The Standard does tell us that moving the declaration from a derived to
base class around will not change the relative sizes of the classes
involved. And knowing that fact is sufficient to answer this question.

Greg

[Maxim Yegorushkin]

Greg wrote:

[]

When i run above code, iam getting size as 8, actually there is only
one int variable, it should be 4, why is it 8??
Probably because your default compiler options make alignment at 8-byte
boundary. No way to tell. Alignment and padding are compiler-specific.

The question being asked is why an instance of class C is larger than
an instance of class A or class B, not why it is four bytes larger. The
question is a good one. There would seem to be no apparent reason why a
compiler would increase the size of C when it inherits from A and B,

Indeed, there is no such reason. The reason is sizeof(C) is always a
multiple of alignof(C), with the latter being compilers specific.
even when the amount of data that class C stores remains unchanged. And
in fact the only reason that the C++ compiler makes C larger in this
situation is because the C++ Standard requires it.
It does not. Refer to §10/5:

<q>
.... A base class subobject may be of zero size (clause 9); however, two
subobjects that have the same class type and that belong to the same
most derived object must not be allocated at the same address (5.10).
</q>
The C++ Standard says that neither class A or class B in this situation
can be of size zero bytes. Therefore an object of class C will
necessarily have to be larger than either of its two base classes,
because C's size cannot be zero bytes either.
This is wrong.
The answer has nothing to
do with padding or alignment, but rather conformity with the Standards.
This is also wrong. Alignment is the reason. Consider:

#include <stdio.h>

struct A {};
struct B {};
struct C {};
struct D {};
struct E {};
struct F {};
struct G {};

struct H : A, B, C, D, E, F, G
{
};

struct I : A, B, C, D, E, F, G
{
} __attribute__((aligned(8)));

int main()
{
printf("alignof(H) == %u\n", __alignof__(H));
printf("sizeof(H) == %u\n", sizeof(H));
printf("alignof(I) == %u\n", __alignof__(I));
printf("sizeof(I) == %u\n", sizeof(I));
}

[max@localhost exp]$ g++ -Wall exp.cpp -o exp

[max@localhost exp]$ ./exp
alignof(H) == 1
sizeof(H) == 1
alignof(I) == 8
sizeof(I) == 8
Granted, the Standard doesn't say how much larger C has to be, but it
does explain why C is larger than A or B. And that really is the
question being asked.

Note that the "empty base class optimization" would make classes A and
B add zero incremental bytes to objects of type C. To take advantage of
this optimization (if the compiler supports it) classes A and B must
each derive from an empty base class themselves.

Sorry, this is bullshit.