#include <iostream>
class B
{
public:
virtual void f() {std::cout<<"HI";}
int x;
};
class A
{
public:
void f() {std::cout<<"HI";}
int x;
};
int main () {
A a;
B b;
std::cout<<sizeof(a)<<" "<<sizeof(b);
return 0;
}
The output is
4 16
I expected it to be 8 bytes longer - vptr pointer. But for what the rest 4 bytes is used?
I found quite many forum posts (all from some years ago) where people discussed that object from the class with vprt is 4 or 8 bytes longer. I checked also on online C++ shell - output there is the same.
It seems that the pointer to the table of pointers to virtual functions has a size of 8 bytes. So the class B is aligned to the boundary of 8 bytes that is it has 4 additional padding bytes.
Related
class A {
int a;
public:
virtual ~A() {}
};
class B : public A {
int b;
};
int main(int argc, char* argv[])
{
cout << sizeof(B) << endl;
return 0;
}
(Based on 64bits program), I notice the differences of the values under Visual Studio and gcc, respectively are 24 and 16.
By analyzing, the virtual table costs 8 bytes, int a and int b respectively is 4 bytes. So the alignment for VS is 8 but for gcc is 4.
What does cause the difference, how does the C++ standard say?
struct C {
char a;
int b;
char c;
};
int main(int argc, char* argv[])
{
cout << sizeof(C) << endl;
return 0;
}
Moreover, under VS, the size of this struct is 12 that implies the alignment is 4, different from the alignment for class B under VS.
Why?
You will find that alignof(A) == 8 && sizeof(A) == 16 on both compilers, laid out as 8 bytes for a pointer, 4 bytes for int A::a, and 4 bytes of padding.
Since A is a base class, it is a potentially overlapping subobject of any B objects. This means that the padding can be reused.
gcc/clang (More specifically, things that follow the Itanium ABI) have to allocate int B::b into the 4 bytes of padding in the base class, so sizeof(B) is also 16 (but alignof(B) == 8 still).
MSVC (and clang compiling for Windows) follow a different ABI, which happens to not reuse the tail padding, so the layout of B looks like 8 bytes for a pointer, 4 bytes for int A::a, 4 bytes of padding (inherited from A), 4 bytes for int B::b, and 4 more bytes of padding.
As for your last questions, typically vtables are implemented as a pointer stored at the front of each object. This pointer will usually have the size an alignment requirement of 8 on a 64 bit platform. C would be comparable if it looked like:
struct C {
void* p;
char a;
// 3 bytes padding
int b;
char c;
// 7 bytes padding to alignment of 8
};
static_assert(sizeof(C) == 24 && alignof(C) == 8);
With the other's help, I finally figure it out.
alignof(B) under VS and GCC are same which is 8.
1. Why sizeof(B) are different?
Because they store the virtual function pointer (vptr) at different places, for VS at beginning and for GCC at the end. Something like this,
//B for VS
class B{
class A{
void* vptr;
int a;
};
int b;};
//B for GCC
class B{
int a;
int b;
void* vptr;
};
For VS, sizeof(B) = sizeof(vptr) + (sizeof(int a) padding to 8) + (sizeof(int b) padding to 8) = 24.
For GCC, sizeof(B) = (sizeof(int a)+sizeof(int b) padding to 8) + sizeof(vptr) = 16.
2. Why alignof(C) is 4?
Because there is virtual function in struct C. If you add virutal ~C(){} for C, alignof(C) would be 8 either.
As explained in one of the solutions of this question (Size of virtual pointer-C++) that you can calculate virtual pointer size in following manner :
struct virtual_base {
int data;
virtual_base() {}
virtual ~virtual_base() {}
};
struct non_virtual_base {
int data;
non_virtual_base() {}
~non_virtual_base() {}
};
int main() {
std::cout << sizeof( virtual_base ) - sizeof( non_virtual_base ) << '\n';
return 0;
}
But when i try this on cpp.sh http://cpp.sh/7o5av, without data (member variable) i get the size as 7 and with data size is coming out to be 12, so i failed to understand this behavior and any insights will be helpful and i know that size of empty class is 1 and in second with data member i expect this should come as 11 and not 7
You get 7 with no data member becuase the empty class has size 1 so while the virtual one contains a pointer to the virtual table of size 8: 8-1=7.
When data members are involved the result you get depends on the actual type of the members. If you use int then the difference is 12 becuase the vptr has to be aligned to multiple of 8.
This means that the int data member occupies bytes frm 0 to 4 and the vptr cannot be stored at byte 4 but starting from at byte 8. So the total size is 8+8=16 for the virtual struct. Try using a double and you will see that the difference is 8 as in the following code.
#include <iostream>
using namespace std;
struct virtual_base {
double data;
virtual_base() {}
virtual ~virtual_base() {}
};
struct non_virtual_base {
double data;
non_virtual_base() {}
~non_virtual_base() {}
};
int main() {
std::cout << sizeof( virtual_base ) - sizeof( non_virtual_base ) << '\n';
return 0;
}
try here: https://www.ideone.com/Ycpg64
virtual_base only contains the vftable pointer which is apparently 8 bytes on your platform. virtual_base also has an int, and the vftable is aligned to 8 bytes. So it's something like:
4 bytes for int | 4 padding bytes | 8 bytes for vftable pointer |
| x | x | x | x | | | | | v | v | v | v | v | v | v | v |
Please have a look at this. It might be helpful.
With int data member:
sizeof(virtual_base) is 16 [4(int)+ 4(padding) +8(vftable)] bytes
sizeof(non_virtual_base) is 4 bytes i.e. size of int.
Without int data member:
sizeof(virtual_base) is 8 bytes i.e the size of vftable pointer. There is no padding.
sizeof(non_virtual_base) without any data member is 1 byte.
When you declare any function as virtual in C++, the class receives a hidden member vptr, which points to vtable. This is used to choose which function should actually be called when dynamic polymorphism is used.
Disclaimer: I'll be using results from online compiler you posted, with the options you set (C++14, O2 optimization)
We can easily see that empty class has sizeof equal to 1, and sizeof class with virtual functions is 8.
Then, with data member, class without virtual functions gets sizeof 4 and the one with virtual functions get sizeof 16.
We can also check that sizeof pointer type (for example int*) is equal to 8
So, what happens here: empty class gets assigned size 1 by default. But, a class with virtual functions must have vptr member, which itself is 8 bytes long. This gives you 8-1 = 7
When you add member to non-virtual class, it simply gets the size of all it's members, in this case it's int, so you have size 4.
For virtual class, you already have vptr of size 8 and add int of size 4 to it. Here the mechanism of structure alignment kicks in. Presumably, the system it's compiling allows for access only to bytes which are on offset of multiply of 8, so the compiler, to optimize access time, adds artifial padding bytes, which hold no data. Basically, object of your class would look like this:
[int(4B)|padding(4B)|vptr(8B)]
This results in size of class being 16. So we get 16 - 4 = 12.
This question already has an answer here:
Size of class with virtual function
(1 answer)
Closed 8 years ago.
Have a look at these 2 cases:
class A {
public:
int a;
A () { a = 10;}
void foo () {std::cout << "a = " << a << std::endl;}
};
Here sizeof(A) gives 4 bytes, which makes sense.
class A {
public:
int a;
A () { a = 10;}
virtual void foo () {std::cout << "a = " << a << std::endl;}
};
Here sizeof(A) gives 16 bytes as opposed to 12 bytes (4 + 8 for pointer).
Is there any explanation in terms of memory alignment for this ?
It's up to the compiler how virtual functions are implemented, but what's likely happening here is it wants/needs to align the 8-byte pointer to the virtual dispatch table on a multiple-of-8 memory address. Then there's either { 4 bytes a, 4 padding, 8 vdt pointer } or { 8 vdt pointer, 4 bytes a, 4 padding } - the latter's less obvious, but consider that arrays of A need to be contiguous and spaced per sizeof(A), so 12's rounded up to 16 given the 8-byte alignment.
FIRST in virtual dispatch table, visual functions need a pointer size , in 32 bits computer, the size of pointer is 4 bytes, int 64 bits it's 8 bytes. so I think your computer is 64 bits.
SECOND the sizeof(A) should consider the padding and memory alignment .
so 16 bytes is arranged like that: 4 bytes(int a) + 8 bytes(a virtual function pointer int 64 bits) + 4 bytes padding(because the max size of elements in A is the virtual function pointer which is 8 bytes, so the sizeof(A)should be the integral multiples of 8 bytes )
I have this chunk of code:
#include <stdio.h>
class CoolClass {
public:
virtual void set(int x){x_ = x;};
virtual int get(){return x_;};
private:
int x_;
};
class PlainOldClass {
public:
void set(int x) {x_ = x;};
int get(){return x_;}
private:
int x_;
};
int main(void) {
printf("CoolClass size: %ld\n", sizeof(CoolClass));
printf("PlainOldClass size: %ld\n", sizeof(PlainOldClass));
return 0;
}
I'm getting a little bit confused because it says that the size of CoolClass is 16? How? Why? Even with the pointer to the vtable, shouldn't the size be 8? Size of oldclass is 4 as expected.
edit: I'm running Linux Mint 64 bit with g++ 4.6.3.
You can't assume anything about the sizes of anything other than char or unsigned char. If you're building on a 64 bit platform, int is likely still 4 bytes, but the size of the virtual table pointer would likely be 8, and the extra 4 bytes are for padding (so that the pointer is aligned to 8 bytes).
64-bit
+----+----+----+----+
| vp | vp | x_ | p |
+----+----+----+----+
vp - virtual table pointer
x_ - member
p - padding byte
32-bit
+----+----+
| vp | x_ |
+----+----+
vp - virtual table pointer
x_ - member
p - padding byte
Padding not required because the pointer is already aligned
As a test, you can try
class PlainOldClass {
private:
int* x_;
};
and its size would be 8.
My best guess is that you're compiling for a platform with 64-bit pointers. Then you'll need 8 bytes for the virtual pointer, probably 4 bytes for the int (some 64-bit platforms will also give that 8 bytes - but you say that sizeof (PlainOldClass) is 4, so that doesn't apply here), and another 4 to give the class the 64-bit alignment required by the pointer - giving a total of 16 bytes.
I think the cost is:
platform with 64-bit pointers, so 8 bytes for the virtual pointer
4 bytes for the int (but might be also 8 bytes on some platforms)
4 to give the class the 64-bit alignment required by the pointer
sum=16 bytes.
It's because your system may be 64-bit system, due to which it becomes,
8 bytes for vptr.
4 bytes for int
and additional 4 bytes padding to give 64-bit alignment to the pointer.
hence sum becomes 16.
I encounter this problem when tackling with virtual inheritance. I remember that in a non-virtual inheritance hierarchy, object of sub-class hold an object of its direct super-class. What about virtual inheritance? In this situation, does object of sub-class hold an object of its super-class directly or just hold a pointer pointing to an object of its super-class?
By the way, why the output of the following code is:
sizeof(A): 8
sizeof(B): 20
sizeof(C): 20
sizeof(D): 36
Code:
#include <iostream>
using namespace std;
class A{
char k[ 3 ];
public:
virtual void a(){};
};
class B : public virtual A{
char j[ 3 ];
public:
virtual void b(){};
};
class C : public virtual A{
char i[ 3 ];
public:
virtual void c(){};
};
class D : public B, public C{
char h[ 3 ];
public:
virtual void d(){};
};
int main( int argc, char *argv[] ){
cout << "sizeof(A): " << sizeof( A ) << endl;
cout << "sizeof(B): " << sizeof( B ) << endl;
cout << "sizeof(C): " << sizeof( C ) << endl;
cout << "sizeof(D): " << sizeof( D ) << endl;
return 0;
}
Thanks in advance.
Kind regards.
The virtual base object is somewhere in the memory block that belongs to the object (the memory with size = sizeof(object)). Because several sub objects of different types can be combined in various ways but must share the same base object, a offset pointer is needed for each sub object to find out the virtual base object. Without virtual inheritance, the offset to find out the corresponding base object is fixed at compile time for each class type.
The sizeof values depend on your compiler and machine, but the following assumptions are very common:
assumption: pointer size is 4 bytes
assumption: class size is rounded up to multiple of 4 bytes
sizeof(A): 8 -> 1 pointer to vtable (virtual method)
+ 3 chars -> 4+3=7
-> round up to 8
sizeof(B): 20 -> 8 + 1 pointer to vtable (virtual method)
+ 1 offset pointer to virtual base
+ 3 chars -> 8 + 4 + 4 + 3 = 19
-> round up to 20
sizeof(C): 32 -> 20 + 1 pointer to vtable (virtual method)
+ 1 offset pointer to virtual base
+ 3 chars
-> 20 + 4 + 4 + 3 = 31 // this calculation refers to an older
-> round up to 32 // version of the question's example
// where C had B as base class
The calculations are guessed because the real calculation must exactly know how the compiler works.
Regards,
Oliver
More details why an extra offset pointer is needed:
Example:
class B : virtual public A {...};
class C : virtual public A {...};
class D1 : public B {...};
class D2 : public B, C {...};
possible memory layout for D1:
A
B
D1
possible memory layout for D2:
A
C
B
D2
in the second case sub object B needs another offset to find its base A
An object of type D2 consists of a memory block, where all the parent object parts are contained, i.e. the memory block for an object of type D2 has a section for the A member variables, the C member variables, the B member variables and the D2 member variables. The order of these sections is compiler dependent, but the example shows, that for multiple virtual inheritance a offset pointer is needed, that points within the object's total memory block to the virtual base object. This is needed because the methods of class B know only one this pointer to B and must somehow calculate where the A memory part is relative to the this pointer.
Calculation sizeof(D):
sizeof(D): 36 -> A:3 chars + A:vtable
+ B:3 chars + B:vtable + B:virtual base pointer
+ C:3 chars + C:vtable + C:virtual base pointer
+ D:3 chars + D:vtable
= 3 + 4
+ 3 + 4 + 4
+ 3 + 4 + 4
+ 3 + 4
= 36
The above calculation is probably wrong ;-) ...
I'm not sure whether the D part has its own vtable pointer or not (this is all highly compiler dependent).
I now think that it could be that the D part use the vtable pointer entry of its parent classes and that the 4 extra bytes are used for alignment each part (multiple of 8 bytes):
So this calculation is probably more correct:
sizeof(D): 36 -> A:3 chars + A:vtable + A:alignment
+ B:3 chars + B:vtable + B:virtual base pointer + B:alignment
+ C:3 chars + C:vtable + C:virtual base pointer + C:alignment
+ D:3 chars + D:alignment
= 3 + 4 + 1
+ 3 + 4 + 4 + 1
+ 3 + 4 + 4 + 1
+ 3 + 1
= 36
I see three point analysis for the above question
a. Virtual Inheritance
"Virtual inheritance is a mechanism whereby a class specifies that it is willing to share the state of its virtual base class. Under virtual inheritance, only one, shared base-class subobject is inherited for a given virtual base regardless of how many times the class occurs as a virtual base within the derivation hierarchy. The shared base-class subobject is called a virtual base class." ... From Lippman
Virtual inheritance only avoids duplicate sub-objects inherited from multiple inheritance. But this does not indicate in any way that the base class objects will not be sub-objects. On the contrary, the sub-object (atleast one copy would be present - I mean would be included in sizeof() operation) even during the virtual inheritance.
b. virtual function
Virtual function is for dynamic binding of member functions of objects involved in hierarchy. So even this does not have any significance towards sub-object arrangements.
c. Implementation of the sub-objects
This is totally compiler dependent, and for all reasons would be very difficult to determine - in its implementation. However, we can confirm that the sizeof() of the object would include the size of the base class (sub) objects also - and we can visualize them as having the base class object embedded in them.
Each object of the inherited function will definitely contain space for the sub-objects.
HTH
does object of sub-class hold an object of its super-class directly
Yes, that is how it works whether the inheritance is virtual or not. I would use the word "contain" vs. "hold" however.
If your hierarchy looked like this, with no virtual inheritances anywhere:
# W <--- base class
# / \
# X Y <--- subclasses of W
# \ /
# Z <--- most derived class
Then Z will have two copies of W. But if you make the X-->W and Y-->W inheritances virtual, then Z will only have one copy of W because Z's two superclasses share their common base class.
# W
# / \ <--- make these two virtual to eliminate duplicate W in Z.
# X Y
# \ /
# Z
In your example:
class A{...};
class B : public virtual A{...};
class C : public virtual B{...}; // Edit: OP's code had this typo when I answered
class D : public B, public C{...};
Having B inherit virtually from A isn't necessary. The only virtual inheritances you need are C-->B and D-->B, since that is where the diamond "merges" going up the inheritance hierarchy:
# What you have | What you want?
# A | A
# / | /
# /v | /
# / | /
# B | B
# / \ | / \
# /v \ | /v \v
# / \ | / \
# C ) | C )
# \ / | \ /
# \ / | \ /
# \ / | \ /
# D | D
Of course if you have other classes not shown that inherit from A as well as B, that changes things -- maybe the B-->A inheritance does need to be virtual if there is another diamond you didn't tell us about.
I remember that in a non-virtual inheritance hierarchy, object of sub-class hold an object of its direct super-class.
That is not correct. Several implementations are going to do it this way, but it's not defined that way by Standard C++. Standard C++ does not specify how any of these things are to be implemented.
Virtual Inheritance is used only for some cases of multiple inheritance where a derived class multiply inherits from two base classes which themselves inherit from a common base class. An example of this is the iostream library, where istream and ostream inherit from basic_ios, and iostream inherits from istream and ostream (and so one iostream would have two basic_ios without virtual inheritance).
Unless you are in this specific scenario, you should not use virtual inheritance.
What about virtual inheritance? In this situation, does object of sub-class hold an object of its super-class directly or just hold a pointer pointing to an object of its super-class?
That is implementation defined. You do not need to know nor should you ever make any assumptions about this. Suffice to say that there is a runtime penalty for virtual inheritance, which is why you should avoid it when it is not needed.
Compare:
struct A {
void *vptr; // offset 0 size 4 alignment 4
char k[3]; // offset 4 size 3 alignment 1
char unnamed_padding; // offset 7 size 1
// total size 8 alignment 4
};
// MS:
struct base_B {
void *vptr; // offset 0 size 4 alignment 4
char j[3]; // offset 4 size 3 alignment 1
char unnamed_padding; // offset 7 size 1
A &a_subobject; // offset 8 size 4 alignment 4
// total size 12 alignment 4
base_B (&a_subobject) :a_subobject(a_subobject) {}
};
struct B {
base_B b; // offset 0 size 12 alignment 4
A a_subobject; // offset 12 size 8 alignment 4
// total size 20 alignment 4
B () : b(a_subobject) {}
};
struct base_C {
void *vptr; // offset 0 size 4 alignment 4
char i[3]; // offset 4 size 3 alignment 1
char unnamed_padding; // offset 7 size 1
A &a_subobject; // offset 8 size 4 alignment 4
// total size 12 alignment 4
base_C (&a_subobject) : a_subobject(a_subobject) {}
};
struct C {
base_C c;
A a_subobject; // offset 12 size 8 alignment 4
// total size 20 alignment 4
C () : c(a_subobject) {}
};
struct D {
// no new vptr!
// base_B is used as primary base: b_subobject.vptr is used as vptr
base_B b_subobject; // offset 0 size 12 alignment 4
base_C c_subobject; // offset 12 size 12 alignment 4
char h[3]; // offset 24 size 3 alignment 1
char unnamed_padding; // offset 27 size 1
A a_subobject; // offset 28 size 8 alignment 4
// total size 36 alignment 4
D (): b_subobject(a_subobject), c_subobject(a_subobject) {}
};
// GCC:
struct base_B {
void *vptr; // offset 0 size 4 alignment 4
char j[3]; // offset 4 size 3 alignment 1
char unnamed_padding; // offset 7 size 1
// total size 8 alignment 4
};
struct B {
base_B b; // offset 0 size 12 alignment 4
A a_subobject; // offset 8 size 8 alignment 4
// total size 16 alignment 4
};
struct base_C {
void *vptr; // offset 0 size 4 alignment 4
char i[3]; // offset 4 size 3 alignment 1
char unnamed_padding; // offset 7 size 1
// total size 8 alignment 4
};
struct C {
base_C b; // offset 0 size 12 alignment 4
A a_subobject; // offset 8 size 8 alignment 4
// total size 16 alignment 4
};
struct D {
// no new vptr!
// base_B is used as primary base: b_subobject.vptr is used as vptr
base_B b_subobject; // offset 0 size 8 alignment 4
base_C c_subobject; // offset 8 size 8 alignment 4
char h[3]; // offset 16 size 3 alignment 1
char unnamed_padding; // offset 19 size 1
A a_subobject; // offset 20 size 8 alignment 4
// total size 24 alignment 4
};