12.5 — The virtual table

To implement virtual functions, C++ uses a special form of late binding known as the virtual table. The virtual table is a lookup table of functions used to resolve function calls in a dynamic/late binding manner. The virtual table sometimes goes by other names, such as “vtable”, “virtual function table”, “virtual method table”, or “dispatch table”.

Because knowing how the virtual table works is not necessary to use virtual functions, this section can be considered optional reading.

The virtual table is actually quite simple, though it’s a little complex to describe in words. First, every class that uses virtual functions (or is derived from a class that uses virtual functions) is given its own virtual table. This table is simply a static array that the compiler sets up at compile time. A virtual table contains one entry for each virtual function that can be called by objects of the class. Each entry in this table is simply a function pointer that points to the most-derived function accessible by that class.

Second, the compiler also adds a hidden pointer to the base class, which we will call *__vptr. *__vptr is set (automatically) when a class instance is created so that it points to the virtual table for that class. Unlike the *this pointer, which is actually a function parameter used by the compiler to resolve self-references, *__vptr is a real pointer. Consequently, it makes each class object allocated bigger by the size of one pointer. It also means that *__vptr is inherited by derived classes, which is important.

By now, you’re probably confused as to how these things all fit together, so let’s take a look at a simple example:

Because there are 3 classes here, the compiler will set up 3 virtual tables: one for Base, one for D1, and one for D2.

The compiler also adds a hidden pointer to the most base class that uses virtual functions. Although the compiler does this automatically, we’ll put it in the next example just to show where it’s added:

When a class object is created, *__vptr is set to point to the virtual table for that class. For example, when a object of type Base is created, *__vptr is set to point to the virtual table for Base. When objects of type D1 or D2 are constructed, *__vptr is set to point to the virtual table for D1 or D2 respectively.

Now, let’s talk about how these virtual tables are filled out. Because there are only two virtual functions here, each virtual table will have two entries (one for function1(), and one for function2()). Remember that when these virtual tables are filled out, each entry is filled out with the most-derived function an object of that class type can call.

The virtual table for Base objects is simple. An object of type Base can only access the members of Base. Base has no access to D1 or D2 functions. Consequently, the entry for function1 points to Base::function1(), and the entry for function2 points to Base::function2().

The virtual table for D1 is slightly more complex. An object of type D1 can access members of both D1 and Base. However, D1 has overridden function1(), making D1::function1() more derived than Base::function1(). Consequently, the entry for function1 points to D1::function1(). D1 hasn’t overridden function2(), so the entry for function2 will point to Base::function2().

The virtual table for D2 is similar to D1, except the entry for function1 points to Base::function1(), and the entry for function2 points to D2::function2().

Here’s a picture of this graphically:

Although this diagram is kind of crazy looking, it’s really quite simple: the *__vptr in each class points to the virtual table for that class. The entries in the virtual table point to the most-derived version of the function objects of that class are allowed to call.

So consider what happens when we create an object of type D1:

Because d1 is a D1 object, d1 has its *__vptr set to the D1 virtual table.

Now, let’s set a base pointer to D1:

Note that because dPtr is a base pointer, it only points to the Base portion of d1. However, also note that *__vptr is in the Base portion of the class, so dPtr has access to this pointer. Finally, note that dPtr->__vptr points to the D1 virtual table! Consequently, even though dPtr is of type Base, it still has access to D1’s virtual table (through __vptr).

So what happens when we try to call dPtr->function1()?

First, the program recognizes that function1() is a virtual function. Second, the program uses dPtr->__vptr to get to D1’s virtual table. Third, it looks up which version of function1() to call in D1’s virtual table. This has been set to D1::function1(). Therefore, dPtr->function1() resolves to D1::function1()!

Now, you might be saying, “But what if dPtr really pointed to a Base object instead of a D1 object. Would it still call D1::function1()?”. The answer is no.

In this case, when b is created, __vptr points to Base’s virtual table, not D1’s virtual table. Consequently, bPtr->__vptr will also be pointing to Base’s virtual table. Base’s virtual table entry for function1() points to Base::function1(). Thus, bPtr->function1() resolves to Base::function1(), which is the most-derived version of function1() that a Base object should be able to call.

By using these tables, the compiler and program are able to ensure function calls resolve to the appropriate virtual function, even if you’re only using a pointer or reference to a base class!

Calling a virtual function is slower than calling a non-virtual function for a couple of reasons: First, we have to use the *__vptr to get to the appropriate virtual table. Second, we have to index the virtual table to find the correct function to call. Only then can we call the function. As a result, we have to do 3 operations to find the function to call, as opposed to 2 operations for a normal indirect function call, or one operation for a direct function call. However, with modern computers, this added time is usually fairly insignificant.

Also as a reminder, any class that uses virtual functions has a __vptr, and thus each object of that class will be bigger by one pointer. Virtual functions are powerful, but they do have a performance cost.

12.6 -- Pure virtual functions, abstract base classes, and interface classes
12.4 -- Early binding and late binding

189 comments to 12.5 — The virtual table

  • Digendra Chand

    class Base
        virtual void function1() {}
        virtual void function2() {}

    class D1: public Base
        virtual void function1() {};

    As we can see in derived class D1 we have not overrided function2(). So as we know the V_table of class D1 will have only function1() in it.

    int main ()
      D1 d;
      D1*p =&d;
      return 0;

    here p->function2(); will call Base::function2().
    Can some body explain to me if Base::function2() is not present in vtable of D1 than how it is getting called?

  • sam

    I think "But what if Base really pointed to a Base object..." should be "But what if dPtr really pointed to a Base object...".

    Having great fun with these lessons by the way and I'm glad to see you're updating the site too.

  • Vijay

    Thank you for sharing valuable information guys. It is very helpful.

    Someone please clarify, if there is an overridden virtual function in derived class then does the derived class VTABLE will contain the addresses of non-overridden functions of base class as well?
    Please consider the folowing case:

    class base
        int x;
        virtual void f0() {}
        virtual void f1() {}

    class derived1 : public base
        int y;
        void f0() override {}
        void f1() override {}

    Please clarify: will the derived class VTABLE will also contain the addresses of base class's virtual functions:
    f0 (derived class overridden f0)
    f1 (derived class overridden f1)


    the derived class's VTABLE will only contain the addresses of overridden derived class's functions:
    f0 (derived class overridden f0)
    f1 (derived class overridden f1)

    As per the explanation provided in the below discussion, it seems like the derived class's VTABLE will only contain the addresses of overridden derived class's functions i.e. f0 and f1 (overridden in derived class). However, a clarification from someone would be very helpful to avoid ant confusion.

    Thank you in advance.


  • Mehul Shah

    Hi Alex,

    I have few  questions  regarding vptr

    1) As far as my understanding is there, vptr is a  intance variable not a class variable, is it corrent ?
    2) Base and Derived will have only 1 vptr or base has its own vptr and derived has its own vptr
    3) in the above code when i run in visual studio 2015 i see the address of obj1.vptr and obj2.vptr is same. how is this possible as vptr is a intance variable not class variable


    • nascardriver

      Hi Mehul!

      1) Correct.
      2) The classes don't have vtable pointers, they have vtables. @Base has it's own vtable and @Derived has it's own vtable.
      3) Each instance has it's own vtable pointer, this can point to whichever vtable you like. In your example you have two vtables, one for @Base and one for @Derived. You also have two vtable pointers, one in @obj1 and one in @obj2. Since both objects are of the same type (Base) they'll both point to the vtable of @Base.

  • Dimon

    moreover, the order of functions placed in the VFTs of each derived class is exact as they declared in the base object. that's all magic. let me show you:

    0:000> dps poi(bas) L4 (dump VFT of bas)
    0022aba0  00221154 tests!ILT+335(?f0baseUAEXXZ) f0 at index 0
    0022aba4  0022138e tests!ILT+905(?f1baseUAEXXZ) f1 at index 1
    0022aba8  00221230 tests!ILT+555(?f2baseUAEXXZ) f2 at index 2
    0022abac  002213c5 tests!ILT+960(?f3baseUAEXXZ) f3 at index 3

    0:000> dps poi(der1) L4 (dump VFT of der1, funcs was shuffled, do you remember? )
    0022abb8  002212d5 tests!ILT+720(?f0derived1UAEXXZ) f0 at index 0 as in base order
    0022abbc  002210aa tests!ILT+165(?f1derived1UAEXXZ) f1 at index 1 as in base order
    0022abc0  00221208 tests!ILT+515(?f2derived1UAEXXZ) f2 at index 2 as in base order
    0022abc4  00221096 tests!ILT+145(?f3derived1UAEXXZ) f3 at index 3 as in base order

    0:000> dps poi(der2) L6 (dump VFT of der2)
    0022abd0  00221154 tests!ILT+335(?f0baseUAEXXZ)     f0 at index 0 and not implemented by derived2 (points to base::f0)
    0022abd4  002211bd tests!ILT+440(?f1derived2UAEXXZ) f1 at index 1 as in base order
    0022abd8  00221316 tests!ILT+785(?f2derived2UAEXXZ) f2 at index 2 as in base order
    0022abdc  002213c5 tests!ILT+960(?f3baseUAEXXZ)     f3 at index 3 and not implemented by derived2 (points to base::f3)
    0022abe0  002212bc tests!ILT+695(?f6derived2UAEXXZ) f6 at index 4 and new VF starting from derived2 class
    0022abe4  002211f4 tests!ILT+495(?f5derived2UAEXXZ) f5 at index 5 and new VF starting from derived2 class

    thus the indexes of functions will always be kept, and with each new virtual function table will grow.
    and if the some virtual function is unimplemented in derived class, its place in class VF table will be replaced with pointer to base class VF.

    another words, if we will do like this:

    we will actually call base::f0. same for f3,
    hope this helps.

  • neil

    can D1 or D2 vtable holds function1 and 2 without virtual keywords?
    because it is accessible right? when we create base pointer to child object.

  • Mireska

    I'm so glad that the compiler is very smart and is able to do all this stuff, truly mindblowing

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