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Cover image for Part 2: All About Virtual Keyword in C++: How Does Virtual Base Class Works Internally?

Part 2: All About Virtual Keyword in C++: How Does Virtual Base Class Works Internally?

visheshpatel profile image Vishal Chovatiya Originally published at vishalchovatiya.com ・7 min read

All About Virtual Keyword in C++ (3 Part Series)

1) Part 1: All About Virtual Keyword in C++: How Does Virtual Function Works Internally? 2) Part 2: All About Virtual Keyword in C++: How Does Virtual Base Class Works Internally? 3) Part 3: All About Virtual Keyword in C++: How Does Virtual Destructor Works?

In PART 1 of "All About Virtual Keyword in C++" series, we have discussed "How Does Virtual Function Works Internally?". So, in this article, I will discuss "How Does Virtual Base Class Works Internally?". I am iterating the same thing which I have mentioned in the earlier article as well that implementation of a virtual mechanism is purely compiler dependent. So, there is no C++ standard is defined for such dynamic dispatch. Hence, here I am discussing the general approach.

/!\: Originally published @ www.vishalchovatiya.com.

As usual, before learning anything new I usually start with “Why do we need it in the first place?”

Why Do We Need a Virtual Class/Inheritance?

  • When we use inheritance, we basically extending a derived class with base class functionality. In other words, the base class object would be treated as sub-object in the derived class.
  • As a result, this would create a problem in multiple inheritances if base class sharing the same mutual class as sub-object in the top-level hierarchy and you want to access its property.

Problem

  • I know above statement is a bit complex. So, let see an example.
class Top { public: int t; };
class Left : public Top { public: int l; };
class Right : public Top { public: int r; };
class Bottom : public Left, public Right { public: int b; };
  • The above class hierarchy/inheritance results in the "diamond" which looks like below:
    Top
   /   \
Left   Right
   \   /
   Bottom
  • An instance of Bottom will be made up of Left, which includes Top, and Right which also includes Top. So, we have two sub-object of Top. This will create ambiguity as follows:
Bottom *bot = new Bottom;
bot->t = 5; // is this Left's `t` or Right's `t` data member ??
  • This was by far the simplest reason for the need of the virtual base class. And consider the following scenarios as an example:
Top   *t_ptr1 = new Left;
Top   *t_ptr2 = new Right; 
  • These both will work fine as Left or Right object memory layout has Top subobject. You can see the memory layout of the Bottom object below for clear understanding.
|                      |
|----------------------|  <------ Bottom bot;   // Bottom object 
|    Left::Top::t      |
|----------------------|
|    Left::l           |
|----------------------|
|    Right::Top::t     |
|----------------------|
|    Right::r          |
|----------------------|
|    Bottom::b         |
|----------------------|
|                      |
  • Now, what happens when we upcast a Bottom pointer?
Left  *left = new Bottom;
  • This will work fine as Bottom object memory layout starts with Left subobject.
  • However, what happens when we upcast to Right?
Right  *right = new Bottom;
  • For this to work, we have to adjust the right pointer value to make it point to the corresponding section of the Bottom layout:
|                      |
|----------------------|
|    Left::Top::t      |
|----------------------|
|    Left::l           |
|----------------------|  <------ right;
|    Right::Top::t     |
|----------------------|
|    Right::r          |
|----------------------|
|    Bottom::b         |
|----------------------|
|                      |
|                      |
  • After this adjustment, we can access the Bottom through the right pointer as a normal Right object.
  • But, what would happen if we do
Top* Top = new Bottom;
  • This statement is ambiguous: the compiler will complain
error: `Top' is an ambiguous base of `Bottom'
Top* topL = (Left*) Bottom;
Top* topR = (Right*) Bottom;

Solution

  • Virtual inheritance uses to solve such kind of diamond problems. When you specify virtual while inheriting your classes, you're telling the compiler that you only want a single instance.
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; };
  • This means that there is only one "instance" of Top included in the hierarchy. Hence
Bottom *bot = new Bottom;
bot->t = 5; // no longer ambiguous
  • So, this may seem more obvious and simpler from a programmer's point of view. But from the compiler's point of view, this is vastly more complicated.
  • But an interesting question is how this bot->t will be addressed & handle by the compiler? Ok, this is the time to move on to next point.

How Does Virtual Base Class Addressing Works Internally?

  • A class containing one or more virtual base class as a subobject, such as Bottom, is divided into two regions:
    1. Invariant region
    2. Shared region
  • Data within the invariant region remains at a fixed offset(which decided in compilation step) from the start of the object regardless of subsequent derivations. So members within the invariant region can access directly. In our case, its Left & Right & Bottom.
  • Additionally, the shared region represents the virtual base class subobjects whose location within the shared region fluctuates with an order of derivation & subsequent derivation. So members within the shared region need to accessed indirectly.
  • As the name suggests, an invariant region placed at the start of objects memory layout and the shared region placed at the end.
  • The offset of the shared region updated in the virtual table. The code necessary for this augmented by the compiler at the time of object construction. See below image for reference.
|                        |          
|------------------------| <------ Bottom bot;   // Bottom object           
|    Left::l             |          
|------------------------|               |------------------| 
|    Left::_vptr_Left    |-------|       |  offset of Top   | // offset starts 
|------------------------|       |-------|------------------|       // from left subobject = 20
|    Right::r            |               |    ...           |
|------------------------|               |------------------|  
|    Right::_vptr_Right  |-------|       
|------------------------|       |       |------------------| 
|    Bottom::b           |       |       |  offset of Top   | // offset starts 
|------------------------|       |-------|------------------|       // from right subobject = 12                       
|    Top::t              |               |    ...           |                                    
|------------------------|               |------------------|                                            
|                        |       
  • Now coming back to our interesting question i.e. "How this bot->t will be addressed?"
Bottom *bot = new Bottom;
bot->t = 5;
  • Above code will probably be transformed into
Bottom *bot = new Bottom;
(bot + _vptr_Left[-1])->t = 5; // If you haven't got this, then consider the above memory map
  • So, as you can see addressing of virtual base class is done through the offset(of the base class) stored in the virtual table.

Handling of Virtual Functions in the Virtual Base Class

  • Handling of the virtual functions in the virtual base class is the same as we have discussed in our previous article with multiple inheritances. There is nothing special about it.

Summary

As I mentioned earlier this seems straight forward from the programmer's point of view. But from the compiler’s point of view, it is complicated. And, there are many other Complications of Using the Virtual Base Class which I have discussed in a separate article. Let's summaries few take away points:

  • There is no C++ standard on dynamic dispatch implementation & it only states behaviour
  • The virtual class needed to avoid the diamond problem
  • The object having virtual base class as subobject has memory layout divided in two regions:
    1. Invariant region: non-virtual classes
    2. Shared region: virtual classes
  • Shared region i.e. virtual base class subobject placed at the end of memory layout.
  • The addresses of the data members in the virtual base class resolved using offset(of virtual base class) store in the virtual table.

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All About Virtual Keyword in C++ (3 Part Series)

1) Part 1: All About Virtual Keyword in C++: How Does Virtual Function Works Internally? 2) Part 2: All About Virtual Keyword in C++: How Does Virtual Base Class Works Internally? 3) Part 3: All About Virtual Keyword in C++: How Does Virtual Destructor Works?

Posted on Nov 1 '19 by:

visheshpatel profile

Vishal Chovatiya

@visheshpatel

Software Developer⌨, Mentor🔦, Fitness Freak🏋, Geek🤓, Hipster🕴, Blogger👨‍💻, Productivity Hacker⌚, Technical Writer✍️, Always a Student👨‍🎓 & Learning Junkie📚.

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