In the previous article about How Does Virtual Base Class Works Internally? we have seen address resolution of virtual base class & why it’s needed. But I have not discussed Complications of Using the Virtual Base Class. Which we will see in this article. This is going to be a bit complex & clumsy then learning the internal working of the virtual base class. But if you are not tired, then read forward.
I am re-posting the previous example here to refresh some memory:
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| class Top { public: int t; };
class Left : virtual public Top { public: int l; };
class Right : virtual public Top { public: int r; };
class Bottom : public Left, public Right { public: int b; };
|
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| Top
/ \
Left Right
\ /
Bottom
|
Complications of Downcasting While Using the Virtual Base Class#
- As we have seen, casting of the object
Bottom to Right(in other words, upcasting) requires adding offset to a pointer. One might be tempted to think that downcasting can then simply be implemented by subtracting the same offset. - This process is not easy for the compiler as it seems. To understand this, let us go through an example.
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| class AnotherBottom : public Left, public Right
{
public:
int ab1;
int ab2;
};
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Bottom & AnotherBottom have the same inheritance hierarchy except for their own data members. Now consider the following code.
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| Bottom* bottom1 = new Bottom();
AnotherBottom* bottom2 = new AnotherBottom();
Top* top1 = bottom1;
Top* top2 = bottom2;
Left* left = static_cast<Left*>(top1);
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- Following is memory layout for
Bottom & AnotherBottom
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| | | | |
|------------------------|<---- Bottom |------------------------|<---- AnotherBottom
| Left::l | | Left::l |
|------------------------| |------------------------|
| Left::_vptr_Left | | Left::_vptr_Left |
|------------------------| |------------------------|
| Right::r | | Right::r |
|------------------------| |------------------------|
| Right::_vptr_Right | | Right::_vptr_Right |
|------------------------| |------------------------|
| Bottom::b | | AnotherBottom::ab1 |
top1-->|------------------------| |------------------------|
| Top::t | | AnotherBottom::ab2 |
|------------------------| top2----->|------------------------|
| | | Top::t |
|------------------------|
| |
|
- Now consider how to implement the
static_cast from top1 to left, while taking into account that we do not know whether top1 is pointing to an object of type Bottom or an object of type AnotherBottom. It can’t be done! The necessary offset depends on the runtime type of top1 (20 for Bottom and 24 for AnotherBottom). The compiler will complain:
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| error: cannot convert from a pointer to base class 'Top' to a pointer to derived class 'Left' because the base is virtual
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- Since we need runtime information, we need to use a dynamic_cast instead:
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| Left* left = dynamic_cast<Left*>(top1);
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- However, the compiler is still unhappy:
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| error: cannot dynamic_cast 'top1' (of type 'class Top*')to type 'class Left*' (source type is not polymorphic)
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- The problem is that a dynamic_cast (as well as the use of
typeid) needs runtime type information about the object pointed to by top1. The compiler did not include that because it did not think that was necessary. To force the compiler to include that, we can add a virtual destructor to Top:
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| class Top
{
public:
virtual ~Top() {} // This line creates magic for us
int t;
};
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- Thus, for a downcasting object having virtual base class we need to have at least one virtual function in the virtual base class.
Double Pointer Hack#
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| Bottom* b = new Bottom();
Right* r = b;
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- We already know that the value of
b gets adjusted by 8 bytes before it is assigned to r so that it points to the Right section of the Bottom object). Thus, we can legally assign a Bottom* to a Right*. What about Bottom** and Right**?
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| Bottom** bb = &b;
Right** rr = bb;
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- Should the compiler accept this? A quick test will show that the compiler will complain:
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| error: invalid conversion from `Bottom**' to `Right**'
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- Why? Suppose the compiler would accept the assignment of
bb to rr. We can visualise the result as:
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| |----------| --------> |---------| | |
| bb | | b | ------> |------------------------|<---- Bottom
|----------| /----> |---------| | Left::l |
/ |------------------------|
/ | Left::_vptr_Left |
|----------| / |---------| ------> |------------------------|
| rr | | r | | Right::r |
|----------| |---------| |------------------------|
| Right::_vptr_Right |
|------------------------|
| Bottom::b |
|------------------------|
| Top::t |
|------------------------|
| |
|
- So,
bb and rr both point to b, and b and r point to the appropriate sections of the Bottom object. Now consider what happens when we assign to *rr (note that the type of *rr is Right*, so this assignment is valid):
- This is essentially the same assignment as the assignment to
r above. Thus, the compiler will implement it the same way! In particular, it will adjust the value of b by 8 bytes before it assigns it to *rr. But *rr pointed to b! If we visualise the result again:
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| |----------| --------> |-----------| | |
| bb | | b | |------------------------|<---- Bottom
|----------| /----> |-----------|\ | Left::l |
/ \ |------------------------|
/ \ | Left::_vptr_Left |
|----------| / |-----------|---\-----> |------------------------|
| rr | | r | | Right::r |
|----------| |-----------| |------------------------|
| Right::_vptr_Right |
|------------------------|
| Bottom::b |
|------------------------|
| Top::t |
|------------------------|
| |
|
- This is correct as long as we access the
Bottom object through *rr, but as soon as we access it through b itself, all memory references will be off by 8 bytes — obviously a very undesirable situation. - So, in summary, even if
*a and *b are related by some subtyping relation, **a and **b are not.
Constructors of Virtual Bases#
- The compiler must guarantees that the constructor for all virtual bases of a class gets invoked, and get invoked only once. If you don’t explicitly call the constructors of your virtual base class (independent of how far up the tree they are), the compiler will automatically insert a call to their default constructors.
- This can lead to some unexpected results. Consider the same class hierarchy again we have been considering so far, extended with constructors:
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| class Top
{
public:
Top() { a = -1; }
Top(int _a) { a = _a; }
int a;
};
class Left : virtual public Top
{
public:
Left() { b = -2; }
Left(int _a, int _b) : Top(_a) { b = _b; }
int b;
};
class Right : virtual public Top
{
public:
Right() { c = -3; }
Right(int _a, int _c) : Top(_a) { c = _c; }
int c;
};
class Bottom : public Left, public Right
{
public:
Bottom() { d = -4; }
Bottom(int _a, int _b, int _c, int _d) : Left(_a, _b), Right(_a, _c)
{
d = _d;
}
int d;
};
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- What would you expect this to output:
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| Bottom bottom(1,2,3,4);
printf("%d %d %d %d %d\n", bottom.Left::a, bottom.Right::a, bottom.b, bottom.c, bottom.d);
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- I know you were expecting different. But if you trace the execution of the constructors, you will find
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| Top::Top()
Left::Left(1,2)
Right::Right(1,3)
Bottom::Bottom(1,2,3,4)
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- As explained above, the compiler has inserted a call to the default constructor in
Bottom, before the execution of the other constructors. Then when Left tries to call its base class constructor(Top), we find that Top has already been initialised and the constructor does not get invoked. - To avoid this situation, you should explicitly call the constructor of your virtual base(s):
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| Bottom(int _a, int _b, int _c, int _d): Top(_a), Left(_a,_b), Right(_a,_c)
{
d = _d;
}
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Pointer Equivalence
- Once again assuming the same (virtual) class hierarchy, would you expect this to print “Equal”?
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| Bottom* b = new Bottom();
Right* r = b;
if(r == b)
printf("Equal!\n");
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- Bear in mind that the two addresses are not actually equal (
r is off by 8 bytes). However, that should be completely transparent to the user; so, the compiler actually subtracts the 8 bytes from r before comparing it to b; thus, the two addresses are considered equal. - Although, this also stands true for the following code.
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| class base1{};
class base2{};
class derived : public base1, public base2{};
derived *d = new derived();
base2 *b2 = d;
if(b2 == d)
printf("Equal!\n");
|
Reference#
