In C++11 (N3485) 10.1.4 [class.mi] it says:
For each distinct occurence of a non-virtual base class in the class lattice of the most derived class, the most derived object shall contain a corresponding distinct base class subobject of that type.
For each distinct base class that is specified virtual, the most derived class shall contain a single base class object of that type.
Consider the following C++11 code:
struct B {};
struct BV : virtual B {};
struct BN : B {};
struct C1 : BV, BN {};
struct C2 : BV, BN {};
struct D : C1, C2 {};
Firstly, for clarity, how many vertices does the class lattice of D have?
Secondly, how many distinct subobjects of type B does the standard require that a most derived object of type D have?
update:
Which of the following is the class lattice?
(1)
B B B B
^ ^ ^ ^
| | | |
BV BN BV BN
^ ^ ^ ^
| | | |
\ / \ /
C1 C2
\ /
\ /
- D -
(2)
B<---------
^ \
| |
| B | B
| ^ | ^
| | | |
BV BN BV BN
^ ^ ^ ^
| | | |
\ / \ /
C1 C2
\ /
\ /
- D -
(3)
B
/ \
/ \
BV BN
| \ / |
| \/ |
| / \ |
| / \|
C1 C2
\ /
\ /
D
If the intention is that it is (1) then isn't it impossible to have any DAG that isn't a tree? (ie a diamond is impossible) If so wouldn't it be better to call it the class tree?
If it is (2) wouldn't it be sufficient to say "for each occurence of a base class in the class lattice there is a corresponding base class subobject" ?. That is, if the construction of the lattice already depends on virtual and non-virtual base class relationships to select edges and vertices?
If it is (3) then isn't the language incorrect in the standard because there can only ever be one occurence of a class in the class lattice?
Which of the following is the class lattice?
2
Demonstration:
#include <iostream>
struct B {};
struct BV : virtual B {};
struct BN : B {};
struct C1 : BV, BN {};
struct C2 : BV, BN {};
struct D : C1, C2 {};
int
main()
{
D d;
C1* c1 = static_cast<C1*>(&d);
BV* bv1 = static_cast<BV*>(c1);
BN* bn1 = static_cast<BN*>(c1);
B* b1 = static_cast<B*>(bv1);
B* b2 = static_cast<B*>(bn1);
C2* c2 = static_cast<C2*>(&d);
BV* bv2 = static_cast<BV*>(c2);
BN* bn2 = static_cast<BN*>(c2);
B* b3 = static_cast<B*>(bv2);
B* b4 = static_cast<B*>(bn2);
std::cout << "d = " << &d << '\n';
std::cout << "c1 = " << c1 << '\n';
std::cout << "c2 = " << c2 << '\n';
std::cout << "bv1 = " << bv1 << '\n';
std::cout << "bv2 = " << bv2 << '\n';
std::cout << "bn1 = " << bn1 << '\n';
std::cout << "bn2 = " << bn2 << '\n';
std::cout << "b1 = " << b1 << '\n';
std::cout << "b2 = " << b2 << '\n';
std::cout << "b3 = " << b3 << '\n';
std::cout << "b4 = " << b4 << '\n';
}
My output:
d = 0x7fff5ca18998
c1 = 0x7fff5ca18998
c2 = 0x7fff5ca189a0
bv1 = 0x7fff5ca18998
bv2 = 0x7fff5ca189a0
bn1 = 0x7fff5ca18998
bn2 = 0x7fff5ca189a0
b1 = 0x7fff5ca189a8
b2 = 0x7fff5ca18998
b3 = 0x7fff5ca189a8
b4 = 0x7fff5ca189a0
If it is (2) wouldn't it be sufficient to say "for each occurence of a base class in the class lattice there is a corresponding base class subobject" ?. That is, if the construction of the lattice already depends on virtual and non-virtual base class relationships to select edges and vertices?
Merging your suggestion...
A base class specifier that contains the keyword virtual, specifies a virtual base class. For each
distinctoccurrence of anon-virtualbase class in the class latticeof the most derived class, the most derived object (1.8) shall containthere is a correspondingdistinctbase class subobjectof that type.For each distinct base class that is specified virtual, the most derived object shall contain a single base class subobject of that type.
I'm not an expert in the language half of the standard. However when I read your modified specification, I don't see how:
class V { /∗...∗/ };
class A : virtual public V { /∗ ... ∗/ };
class B : virtual public V { /∗ ... ∗/ };
class C : public A, public B { /∗...∗/ };
Results in Figure 4:
V
/ \
/ \
A B
\ /
\ /
C
I don't see another place in the standard that specifies that although V
appears twice in the class hierarchy below C
, only one subobject of type V
actually exists because of the use of the virtual
keyword.
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