Virtual Member Functions

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Member functions can also be declared virtual. In this case, if called on a pointer or reference to an instance, they will not be accessed directly; rather, they will look up the function in the virtual function table (a list of pointers-to-member-functions for virtual functions, more commonly known as the vtable or vftable), and use that to call the version appropriate for the instance’s dynamic (actual) type. If the function is called directly, from a variable of a class, no lookup is performed.

struct Base {
    virtual void func() { std::cout << "In Base." << std::endl; }

struct Derived : Base {
    void func() override { std::cout << "In Derived." << std::endl; }

void slicer(Base x) { x.func(); }

// ...

Base b;
Derived d;

Base *pb = &b, *pd = &d; // Pointers.
Base &rb = b, &rd = d;   // References.

b.func();   // Output:  In Base.
d.func();   // Output:  In Derived.

pb->func(); // Output:  In Base.
pd->func(); // Output:  In Derived.

rb.func();  // Output:  In Base.
rd.func();  // Output:  In Derived.

slicer(b);  // Output:  In Base.
slicer(d);  // Output:  In Base.

Note that while pd is Base*, and rd is a Base&, calling func() on either of the two calls Derived::func() instead of Base::func(); this is because the vtable for Derived updates the Base::func() entry to instead point to Derived::func(). Conversely, note how passing an instance to slicer() always results in Base::func() being called, even when the passed instance is a Derived; this is because of something known as data slicing, where passing a Derived instance into a Base parameter by value renders the portion of the Derived instance that isn’t a Base instance inaccessible.

When a member function is defined as virtual, all derived class member functions with the same signature override it, regardless of whether the overriding function is specified as virtual or not. This can make derived classes harder for programmers to parse, however, as there’s no indication as to which function(s) is/are virtual.

struct B {
    virtual void f() {}

struct D : B {
    void f() {} // Implicitly virtual, overrides B::f.
                //  You'd have to check B to know that, though.

Note, however, that a derived function only overrides a base function if their signatures match; even if a derived function is explicitly declared virtual, it will instead create a new virtual function if the signatures are mismatched.

struct BadB {
    virtual void f() {}

struct BadD : BadB {
    virtual void f(int i) {} // Does NOT override BadB::f.

As of C++11, intent to override can be made explicit with the context-sensitive keyword override. This tells the compiler that the programmer expects it to override a base class function, which causes the compiler to omit an error if it doesn’t override anything.

struct CPP11B {
    virtual void f() {}

struct CPP11D : CPP11B {
    void f() override {}
    void f(int i) override {} // Error: Doesn't actually override anything.

This also has the benefit of telling programmers that the function is both virtual, and also declared in at least one base class, which can make complex classes easier to parse.

When a function is declared virtual, and defined outside the class definition, the virtual specifier must be included in the function declaration, and not repeated in the definition.

This also holds true for override.

struct VB {
    virtual void f(); // "virtual" goes here.
    void g();
/* virtual */ void VB::f() {} // Not here.
virtual void VB::g() {} // Error.

If a base class overloads a virtual function, only overloads that are explicitly specified as virtual will be virtual.

struct BOverload {
    virtual void func() {}
    void func(int) {}

struct DOverload : BOverload {
    void func() override {}
    void func(int) {}

// ...

BOverload* bo = new DOverload;
bo->func(); // Calls DOverload::func().
bo->func(1); // Calls BOverload::func(int).

For more information, see the relevant topic.

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