[23] Inheritance — what your mother never told you  Updated! 
(Part of C++ FAQ Lite, Copyright © 1991-2006, Marshall Cline, cline@parashift.com)

FAQs in section [23]:

[23.1] Is it okay for a non-virtual function of the base class to call a virtual function?

Yes. It's sometimes (not always!) a great idea. For example, suppose all Shape objects have a common algorithm for printing, but this algorithm depends on their area and they all have a potentially different way to compute their area. In this case Shape's area() method would necessarily have to be virtual (probably pure virtual) but Shape::print() could, if we were guaranteed no derived class wanted a different algorithm for printing, be a non-virtual defined in the base class Shape.

 #include "Shape.h"
 
 void Shape::print() const
 {
     float a = this->area();  // area() is pure virtual
     ...
 }

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[23.2] That last FAQ confuses me. Is it a different strategy from the other ways to use virtual functions? What's going on?

Yes, it is a different strategy. Yes, there really are two different basic ways to use virtual functions:

  1. Suppose you have the situation described in the previous FAQ: you have a method whose overall structure is the same for each derived class, but has little pieces that are different in each derived class. So the algorithm is the same, but the primitives are different. In this case you'd write the overall algorithm in the base class as a public method (that's sometimes non-virtual), and you'd write the little pieces in the derived classes. The little pieces would be declared in the base class (they're often protected, they're often pure virtual, and they're certainly virtual), and they'd ultimately be defined in each derived class. The most critical question in this situation is whether or not the public method containing the overall algorithm should be virtual. The answer is to make it virtual if you think that some derived class might need to override it.
  2. Suppose you have the exact opposite situation from the previous FAQ, where you have a method whose overall structure is different in each derived class, yet it has little pieces that are the same in most (if not all) derived classes. In this case you'd put the overall algorithm in a public virtual that's ultimately defined in the derived classes, and the little pieces of common code can be written once (to avoid code duplication) and stashed somewhere (anywhere!). A common place to stash the little pieces is in the protected part of the base class, but that's not necessary and it might not even be best. Just find a place to stash them and you'll be fine. Note that if you do stash them in the base class, you should normally make them protected, since normally they do things that public users don't need/want to do. Assuming they're protected, they probably shouldn't be virtual: if the derived class doesn't like the behavior in one of them, it doesn't have to call that method.

For emphasis, the above list is a both/and situation, not an either/or situation. In other words, you don't have to choose between these two strategies on any given class. It's perfectly normal to have method f() correspond to strategy #1 while method g() corresponds to strategy #2. In other words, it's perfectly normal to have both strategies working in the same class.

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[23.3] Should I use protected virtuals instead of public virtuals?  New! 

[Recently created thanks to a question from Neil Morgenstern (in 10/05) and connected in the (previously named) "Public Overloaded Non-Virtuals Call Protected Non-Overloaded Virtuals" Idiom (in 3/06). Click here to go to the next FAQ in the "chain" of recent changes.]

Sometimes yes, sometimes no.

First, stay away from always/never rules, and instead use whichever approach is the best fit for the situation. There are at least two good reasons to use protected virtuals (see below), but just because you are sometimes better off with protected virtuals does not mean you should always use them. Consistency and symmetry are good up to a point, but at the end of the day the most important metrics are cost + schedule + risk, and unless an idea materially improves cost and/or schedule and/or risk, it's just symmetry for symmetry's sake (or consistency for consistency's sake, etc.).

The cheapest + fastest + lowest risk approach in my experience ends up resulting in most virtuals being public, with protected virtuals being used whenever you have either of these two cases: the situation discussed in FAQ [23.2], or the situation discussed in FAQ [23.9].

The latter deserves some additional commentary. Pretend you have a base class with a set of overloaded virtuals. To make the example easy, pretend there are just two: virtual void f(int) and virtual void f(double). The idea of the Public Overloaded Non-Virtuals Call Protected Non-Overloaded Virtuals idiom is to change the public overloaded methods to non-virtuals, and make those call protected non-overloaded virtuals.

Code using public overloaded virtuals:

 class Base {
 public:
   virtual void f(int x);    ← may or may not be pure virtual
   virtual void f(double x); ← may or may not be pure virtual
 };

Improving this via the Public Overloaded Non-Virtuals Call Protected Non-Overloaded Virtuals idiom:

 class Base {
 public:
   void f(int x)    { f_int(x); }    ← non-virtual
   void f(double x) { f_double(x); } ← non-virtual
 protected:
   virtual void f_int(int);
   virtual void f_dbl(double);
 };

Here's an overview of the original code:

Public? Inline? Virtual? Overloaded?
Yes No Yes Yes

Here's an overview of the improved code that uses the Public Overloaded Non-Virtuals Call Protected Non-Overloaded Virtuals idiom:

Public? Inline? Virtual? Overloaded?
Yes Yes No Yes
No No Yes No

The reason I and others use this idiom is to make life easier and less error-prone for the developers of the derived classes. Remember the goals stated above: schedule + cost + risk? Let's evaluate this Idiom in light of those goals. From a cost/schedule standpoint, the base class (singular) is slightly larger but the derived classes (plural) are slightly smaller, for a net (small) improvement in schedule and cost. The more signicant improvement is in risk: the idiom packs the complexity of properly managing the hiding rule into the base class (singular). This means the derived classes (plural) more-or-less automatically handle the hiding rule, so the various developers who produce those derived classes can remain almost completely focused on the details of the derived classes themselves — they need not concern themselves with the (subtle and often misunderstood) hiding rule. This greatly reduces the chance that the writers of the derived classes will screw up the hiding-rule.

With apologies to Spock, the good of the many (the derived classes (plural)) outweighs the good of the one (the base class (singular)).

(See FAQ [23.9] for why you need to be careful about overriding some-but-not-all of a set of overloaded methods, and therefore why the above makes life easier on derived classes.)

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[23.4] When should someone use private virtuals?  New! 

[Recently created thanks to a question from Neil Morgenstern (in 10/05). Click here to go to the next FAQ in the "chain" of recent changes.]

Almost never.

Protected virtuals are okay, but private virtuals are usually a net loss. Reason: private virtuals confuse new C++ programmers, and confusion increases cost, delays schedule, and degrades risk.

New C++ programmers get confused by private virtuals because they think a private virtual cannot be overridden. After all, a derived class cannot access members that are private in its base class so how, they ask, could it override a private virtual from its base class? There are explanations for the above, but that's academic. The real issue is that almost everyone gets confused the first time they run into private virtuals, and confusion is bad.

Unless there is a compelling reason to the contrary, avoid private virtuals.

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[23.5] When my base class's constructor calls a virtual function on its this object, why doesn't my derived class's override of that virtual function get invoked?

Because that would be very dangerous, and C++ is protecting you from that danger.

The rest of this FAQ gives a rationale for why C++ needs to protect you from that danger, but before we start that, be advised that you can get the effect as if dynamic binding worked on the this object even during a constructor via The Dynamic Binding During Initialization Idiom.

First, here is an example to explain exactly what C++ actually does:

 #include <iostream>
 #include <string>
 
 void println(const std::string& msg)
 { std::cout << msg << '\n'; }
 
 class Base {
 public:
   Base()              { println("Base::Base()");  virt(); }
   virtual void virt() { println("Base::virt()"); }
 };
 
 class Derived : public Base {
 public:
   Derived()           { println("Derived::Derived()");  virt(); }
   virtual void virt() { println("Derived::virt()"); }
 };
 
 int main()
 {
   Derived d;
   ...
 }

The output from the above program will be:

 Base::Base()
 Base::virt() // ← Not Derived::virt()
 Derived::Derived()
 Derived::virt()

The rest of this FAQ describes why C++ does the above. If you're happy merely knowing what C++ does without knowing why, feel free to skip this stuff.

The explanation for this behavior comes from combining two facts:

  1. When you create a Derived object, it first calls Base's constructor. That's why it prints Base::Base() before Derived::Derived().
  2. While executing Base::Base(), the this object is not yet of type Derived; its type is still merely Base. That's why the call to virtual function virt() within Base::Base() binds to Base::virt() even though an override exists in Derived.

Now some of you are still curious, saying to yourself, "Hmmmm, but I still wonder why the this object is merely of type Base during Base::Base()." If that's you, the answer is that C++ is protecting you from serious and subtle bugs. In particular, if the above rule were different, you could easily use objects before they were initialized, and that would cause no end of grief and havoc.

Here's how: imagine for the moment that calling this->virt() within Base::Base() ended up invoking the override Derived::virt(). Overrides can (and often do!) access non-static data members declared in the Derived class. But since the non-static data members declared in Derived are not initialized during the call to virt(), any use of them within Derived::virt() would be a "use before initialized" error. Bang, you're dead.

So fortunately the C++ language doesn't let this happen: it makes sure any call to this->virt() that occurs while control is flowing through Base's constructor will end up invoking Base::virt(), not the override Derived::virt().

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[23.6] Okay, but is there a way to simulate that behavior as if dynamic binding worked on the this object within my base class's constructor?

Yes: the Dynamic Binding During Initialization idiom (AKA Calling Virtuals During Initialization).

To clarify, we're talking about this situation:

 class Base {
 public:
   Base();
   ...
   virtual void foo(int n) const; // often pure virtual
   virtual double bar() const;    // often pure virtual
   // if you don't want outsiders calling these, make them protected
 };
 
 Base::Base()
 {
   ... foo(42) ... bar() ...
   // these will not use dynamic binding
   // goal: simulate dynamic binding in those calls
 }
 
 class Derived : public Base {
 public:
   ...
   virtual void foo(int n) const;
   virtual double bar() const;
 };

This FAQ shows some ways to simulate dynamic binding as if the calls made in Base's constructor dynamically bound to the this object's derived class. The ways we'll show have tradeoffs, so choose the one that best fits your needs, or make up another.

The first approach is a two-phase initialization. In Phase I, someone calls the actual constructor; in Phase II, someone calls an "init" method on the object. Dynamic binding on the this object works fine during Phase II, and Phase II is conceptually part of construction, so we simply move some code from the original Base::Base() into Base::init().

 class Base {
 public:
   void init();  // may or may not be virtual
   ...
   virtual void foo(int n) const; // often pure virtual
   virtual double bar() const;    // often pure virtual
 };
 
 void Base::init()
 {
   ... foo(42) ... bar() ...
   // most of this is copied from the original Base::Base()
 }
 
 class Derived : public Base {
 public:
   ...
   virtual void foo(int n) const;
   virtual double bar() const;
 };

The only remaining issues are determining where to call Phase I and where to call Phase II. There are many variations on where these calls can live; we will consider two.

The first variation is simplest initially, though the code that actually wants to create objects requires a tiny bit of programmer self-discipline, which in practice means you're doomed. Seriously, if there are only one or two places that actually create objects of this hierarchy, the programmer self-discipline is quite localized and shouldn't cause problems.

In this variation, the code that is creating the object explicitly executes both phases. When executing Phase I, the code creating the object either knows the object's exact class (e.g., new Derived() or perhaps a local Derived object), or doesn't know the object's exact class (e.g., the virtual constructor idiom or some other factory). The "doesn't know" case is strongly preferred when you want to make it easy to plug-in new derived classes.

Note: Phase I often, but not always, allocates the object from the heap. When it does, you should store the pointer in some sort of managed pointer, such as a std::auto_ptr, a reference counted pointer, or some other object whose destructor deletes the allocation. This is the best way to prevent memory leaks when Phase II might throw exceptions. The following example assumes Phase I allocates the object from the heap.

 #include <memory>
 
 void joe_user()
 {
   std::auto_ptr<Base> p(/*...somehow create a Derived object via new...*/);
   p->init();
   ...
 }

The second variation is to combine the first two lines of the joe_user function into some create function. That's almost always the right thing to do when there are lots of joe_user-like functions. For example, if you're using some kind of factory, such as a registry and the virtual constructor idiom, you could move those two lines into a static method called Base::create():

 #include <memory>
 
 class Base {
 public:
   ...
   typedef std::auto_ptr<Base> Ptr;  // typedefs simplify the code
   static Ptr create();
   ...
 };
 
 Base::Ptr Base::create()
 {
   Ptr p(/*...use a factory to create a Derived object via new...*/);
   p->init();
   return p;
 }

This simplifies all the joe_user-like functions (a little), but more importantly, it reduces the chance that any of them will create a Derived object without also calling init() on it.

 void joe_user()
 {
   Base::Ptr p = Base::create();
   ...
 }

If you're sufficiently clever and motivated, you can even eliminate the chance that someone could create a Derived object without also calling init() on it. An important step in achieving that goal is to make Derived's constructors, including its copy constructor, protected or private..

The next approach does not rely on a two-phase initialization, instead using a second hierarchy whose only job is to house methods foo() and bar(). This approach doesn't always work, and in particular it doesn't work in cases when foo() and bar() need to access the instance data declared in Derived, but it is conceptually quite simple and clean and is commonly used.

Let's call the base class of this second hierarchy Helper, and its derived classes Helper1, Helper2, etc. The first step is to move foo() and bar() into this second hierarchy:

 class Helper {
 public:
   virtual void foo(int n) const = 0;
   virtual double bar() const = 0;
 };
 
 class Helper1 : public Helper {
 public:
   virtual void foo(int n) const;
   virtual double bar() const;
 };
 
 class Helper2 : public Helper {
 public:
   virtual void foo(int n) const;
   virtual double bar() const;
 };

Next, remove init() from Base (since we're no longer using the two-phase approach), remove foo() and bar() from Base and Derived (foo() and bar() are now in the Helper hierarchy), and change the signature of Base's constructor so it takes a Helper by reference:

 class Base {
 public:
   Base(const Helper& h);
   ...   // remove init() since not using two-phase this time
   ...   // remove foo() and bar() since they're in Helper
 };
 
 class Derived : public Base {
 public:
   ...   // remove foo() and bar() since they're in Helper
 };

We then define Base::Base(const Helper&) so it calls h.foo(42) and h.bar() in exactly those places that init() used to call this->foo(42) and this->bar():

 Base::Base(const Helper& h)
 {
   ... h.foo(42) ... h.bar() ...
   // almost identical to the original Base::Base()
   // but with h. in calls to h.foo() and h.bar()
 }

Finally we change Derived's constructor to pass a (perhaps temporary) object of an appropriate Helper derived class to Base's constructor (using the init list syntax). For example, Derived would pass an instance of Helper2 if it happened to contain the behaviors that Derived wanted for methods foo() and bar():

 Derived::Derived()
   : Base(Helper2())   // ←the magic happens here
 {
   ...
 }

Note that Derived can pass values into the Helper derived class's constructor, but it must not pass any data members that actually live inside the this object. While we're at it, let's explicitly say that Helper::foo() and Helper::bar() must not access data members of the this object, particularly data members declared in Derived. (Think about when those data members are initialized and you'll see why.)

Of course the choice of which Helper derived class could be made out in the joe_user-like function, in which case it would be passed into the Derived ctor and then up to the Base ctor:

 Derived::Derived(const Helper& h)
   : Base(h)
 {
   ...
 }

If the Helper objects don't need to hold any data, that is, if each is merely a collection of its methods, then you can simply pass static member functions instead. This might be simpler since it entirely eliminates the Helper hierarchy.

 class Base {
 public:
   typedef void (*FooFn)(int);  // typedefs simplify
   typedef double (*BarFn)();   //    the rest of the code
   Base(FooFn foo, BarFn bar);
   ...
 };
 
 Base::Base(FooFn foo, BarFn bar)
 {
   ... foo(42) ... bar() ...
   // almost identical to the original Base::Base()
   // except calls are made via function pointers.
 }

The Derived class is also easy to implement:

 class Derived : public Base {
 public:
   Derived();
   static void foo(int n); // the static is important!
   static double bar();    // the static is important!
   ...
 };
 
 Derived::Derived()
   : Base(foo, bar)  // ←pass the function-ptrs into Base's ctor
 {
   ...
 }

As before, the functionality for foo() and/or bar() can be passed in from the joe_user-like functions. In that case, Derived's ctor just accepts them and passes them up into Base's ctor:

 Derived::Derived(FooFn foo, BarFn bar)
   : Base(foo, bar)
 {
   ...
 }

A final approach is to use templates to "pass" the functionality into the derived classes. This is similar to the case where the joe_user-like functions choose the initializer-function or the Helper derived class, but instead of using function pointers or dynamic binding, it wires the code into the classes via templates.

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[23.7] I'm getting the same mess with destructors: calling a virtual on my this object from my base class's destructor ends up ignoring the override in the derived class; what's going on?

C++ is protecting you from yourself. What you are trying to do is very dangerous, and if the compiler did what you wanted, you'd be in worse shape.

For rationale of why C++ needs to protect you from that danger, read FAQ [23.5]. The situation during a destructor is analogous to that during the constructor. In particular, within the {body} of Base::~Base(), an object that was originally of type Derived has already been demoted (devolved, if you will) to an object of type Base. If you call a virtual function that has been overridden in class Derived, the call will resolve to Base::virt(), not to the override Derived::virt(). Same goes for using typeid on the this object: the this object really has been demoted to type Base; it is no longer an object of type Derived.

Read FAQ [23.5] for more insight on this matter.

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[23.8] Should a derived class redefine ("override") a member function that is non-virtual in a base class?

It's legal, but it ain't moral.

Experienced C++ programmers will sometimes redefine a non-virtual function for efficiency (e.g., if the derived class implementation can make better use of the derived class's resources) or to get around the hiding rule. However the client-visible effects must be identical, since non-virtual functions are dispatched based on the static type of the pointer/reference rather than the dynamic type of the pointed-to/referenced object.

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[23.9] What's the meaning of, Warning: Derived::f(char) hides Base::f(double)?

It means you're going to die.

Here's the mess you're in: if Base declares a member function f(double x), and Derived declares a member function f(char c) (same name but different parameter types and/or constness), then the Base f(double x) is "hidden" rather than "overloaded" or "overridden" (even if the Base f(double x) is virtual).

 class Base {
 public:
   void f(double x);  ← doesn't matter whether or not this is virtual
 };
 
 class Derived : public Base {
 public:
   void f(char c);  ← doesn't matter whether or not this is virtual
 };
 
 int main()
 {
   Derived* d = new Derived();
   Base* b = d;
   b->f(65.3);  ← okay: passes 65.3 to f(double x)
   d->f(65.3);  ← bizarre: converts 65.3 to a char ('A' if ASCII) and passes it to f(char c); does NOT call f(double x)!!
   return 0;
 }

Here's how you get out of the mess: Derived must have a using declaration of the hidden member function. For example,

 class Base {
 public:
   void f(double x);
 };
 
 class Derived : public Base {
 public:
   using Base::f;  ← This un-hides Base::f(double x)
   void f(char c);
 };

If the using syntax isn't supported by your compiler, redefine the hidden Base member function(s), even if they are non-virtual. Normally this re-definition merely calls the hidden Base member function using the :: syntax. E.g.,

 class Derived : public Base {
 public:
   void f(double x) { Base::f(x); }  ← The redefinition merely calls Base::f(double x)
   void f(char c);
 };

Note: the hiding problem also occurs if class Base declares a method f(char).

Note: warnings are not part of the standard, so your compiler may or may not give the above warning.

Note: nothing gets hidden when you have a base-pointer. Think about it: what a derived class does or does not do is irrelevant when the compiler is dealing with a base-pointer. The compiler might not even know that the particular derived class exists. Even if it knows of the existence some particular derived class, it cannot assume that a specific base-pointer necessarily points at an object of that particular derived class. Hiding takes place when you have a derived pointer, not when you have a base pointer.

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[23.10] What does it mean that the "virtual table" is an unresolved external?

If you get a link error of the form "Error: Unresolved or undefined symbols detected: virtual table for class Fred," you probably have an undefined virtual member function in class Fred.

The compiler typically creates a magical data structure called the "virtual table" for classes that have virtual functions (this is how it handles dynamic binding). Normally you don't have to know about it at all. But if you forget to define a virtual function for class Fred, you will sometimes get this linker error.

Here's the nitty gritty: Many compilers put this magical "virtual table" in the compilation unit that defines the first non-inline virtual function in the class. Thus if the first non-inline virtual function in Fred is wilma(), the compiler will put Fred's virtual table in the same compilation unit where it sees Fred::wilma(). Unfortunately if you accidentally forget to define Fred::wilma(), rather than getting a Fred::wilma() is undefined, you may get a "Fred's virtual table is undefined". Sad but true.

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[23.11] How can I set up my class so it won't be inherited from?

This is known as making the class "final" or "a leaf." There are three ways to do it: an easy technical approach, an even easier non-technical approach, and a slightly trickier technical approach.

The (easy) technical approach is to make the class's constructors private and to use the Named Constructor Idiom to create the objects. No one can create objects of a derived class since the base class's constructor will be inaccessible. The "named constructors" themselves could return by pointer if you want your objects allocated by new or they could return by value if you want the objects created on the stack.

The (even easier) non-technical approach is to put a big fat ugly comment next to the class definition. The comment could say, for example, // We'll fire you if you inherit from this class or even just /*final*/ class Whatever {...};. Some programmers balk at this because it is enforced by people rather than by technology, but don't knock it on face value: it is quite effective in practice.

A slightly trickier technical approach is to exploit virtual inheritance. Since the most derived class's ctor needs to directly call the virtual base class's ctor, the following guarantees that no concrete class can inherit from class Fred:

 class Fred;
 
 class FredBase {
 private:
   friend class Fred;
   FredBase() { }
 };
 
 class Fred : private virtual FredBase {
 public:
   ...
 };

Class Fred can access FredBase's ctor, since Fred is a friend of FredBase, but no class derived from Fred can access FredBase's ctor, and therefore no one can create a concrete class derived from Fred.

If you are in extremely space-constrained environments (such as an embedded system or a handheld with limited memory, etc.), you should be aware that the above technique might add a word of memory to sizeof(Fred). That's because most compilers implement virtual inheritance by adding a pointer in objects of the derived class. This is compiler specific; your mileage may vary.

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[23.12] How can I set up my member function so it won't be overridden in a derived class?

This is known as making the method "final" or "a leaf." Here's an easy-to-use solution to this that gives you 90+% of what you want: simply add a comment next to the method and rely on code reviews or random maintenance activities to find violators. The comment could say, for example, // We'll fire you if you override this method or perhaps more likely, /*final*/ void theMethod();.

The advantages to this technique are (a) it is extremely easy/fast/inexpensive to use, and (b) it is quite effective in practice. In other words, you get 90+% of the benefit with almost no cost — lots of bang per buck.

(I'm not aware of a "100% solution" to this problem so this may be the best you can get. If you know of something better, please let me know, cline@parashift.com. But please do not email me objecting to this solution because it's low-tech or because it doesn't "prevent" people from doing the wrong thing. Who cares whether it's low-tech or high-tech as long as it's effective?!? And nothing in C++ "prevents" people from doing the wrong thing. Using pointer casts and pointer arithmetic, people can do just about anything they want. C++ makes it easy to do the right thing, but it doesn't prevent espionage. Besides, the original question (see above) asked for something so people won't do the wrong thing, not so they can't do the wrong thing.)

In any case, this solution should give you most of the potential benefit at almost no cost.

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Revised Mar 1, 2006