Is there ever a good reason to not declare a virtual destructor for a class? When should you specifically avoid writing one?
There is no need to use a virtual destructor when any of the below is true:
No intention to derive classes from it
No instantiation on the heap
No intention to store with access via a pointer to a superclass
No specific reason to avoid it unless you are really so pressed for memory.
To answer the question explicitly, i.e. when should you not declare a virtual destructor.
C++ '98/'03
Adding a virtual destructor might change your class from being POD (plain old data)* or aggregate to non-POD. This can stop your project from compiling if your class type is aggregate initialized somewhere.
struct A {
// virtual ~A ();
int i;
int j;
};
void foo () {
A a = { 0, 1 }; // Will fail if virtual dtor declared
}
In an extreme case, such a change can also cause undefined behaviour where the class is being used in a way that requires a POD, e.g. passing it via an ellipsis parameter, or using it with memcpy.
void bar (...);
void foo (A & a) {
bar (a); // Undefined behavior if virtual dtor declared
}
[* A POD type is a type that has specific guarantees about its memory layout. The standard really only says that if you were to copy from an object with POD type into an array of chars (or unsigned chars) and back again, then the result will be the same as the original object.]
Modern C++
In recent versions of C++, the concept of POD was split between the class layout and its construction, copying and destruction.
For the ellipsis case, it is no longer undefined behavior it is now conditionally-supported with implementation-defined semantics (N3937 - ~C++ '14 - 5.2.2/7):
...Passing a potentially-evaluated argument of class type (Clause 9) having a non-trivial copy constructor, a non-trivial move constructor, or a on-trivial destructor, with no corresponding parameter, is conditionally-supported with implementation-defined semantics.
Declaring a destructor other than =default will mean it's not trivial (12.4/5)
... A destructor is trivial if it is not user-provided ...
Other changes to Modern C++ reduce the impact of the aggregate initialization problem as a constructor can be added:
struct A {
A(int i, int j);
virtual ~A ();
int i;
int j;
};
void foo () {
A a = { 0, 1 }; // OK
}
I declare a virtual destructor if and only if I have virtual methods. Once I have virtual methods, I don't trust myself to avoid instantiating it on the heap or storing a pointer to the base class. Both of these are extremely common operations and will often leak resources silently if the destructor is not declared virtual.
A virtual destructor is needed whenever there is any chance that delete might be called on a pointer to an object of a subclass with the type of your class. This makes sure the correct destructor gets called at run time without the compiler having to know the class of an object on the heap at compile time. For example, assume B is a subclass of A:
A *x = new B;
delete x; // ~B() called, even though x has type A*
If your code is not performance critical, it would be reasonable to add a virtual destructor to every base class you write, just for safety.
However, if you found yourself deleteing a lot of objects in a tight loop, the performance overhead of calling a virtual function (even one that's empty) might be noticeable. The compiler cannot usually inline these calls, and the processor might have a difficult time predicting where to go. It is unlikely this would have a significant impact on performance, but it's worth mentioning.
Virtual functions mean every allocated object increases in memory cost by a virtual function table pointer.
So if your program involves allocating a very large number of some object, it would be worth avoiding all virtual functions in order to save the additional 32 bits per object.
In all other cases, you will save yourself debug misery to make the dtor virtual.
Not all C++ classes are suitable for use as a base class with dynamic polymorphism.
If you want your class to be suitable for dynamic polymorphism, then its destructor must be virtual. In addition, any methods which a subclass could conceivably want to override (which might mean all public methods, plus potentially some protected ones used internally) must be virtual.
If your class is not suitable for dynamic polymorphism, then the destructor should not be marked virtual, because to do so is misleading. It just encourages people to use your class incorrectly.
Here's an example of a class which would not be suitable for dynamic polymorphism, even if its destructor were virtual:
class MutexLock {
mutex *mtx_;
public:
explicit MutexLock(mutex *mtx) : mtx_(mtx) { mtx_->lock(); }
~MutexLock() { mtx_->unlock(); }
private:
MutexLock(const MutexLock &rhs);
MutexLock &operator=(const MutexLock &rhs);
};
The whole point of this class is to sit on the stack for RAII. If you're passing around pointers to objects of this class, let alone subclasses of it, then you're Doing It Wrong.
A good reason for not declaring a destructor as virtual is when this saves your class from having a virtual function table added, and you should avoid that whenever possible.
I know that many people prefer to just always declare destructors as virtual, just to be on the safe side. But if your class does not have any other virtual functions then there is really, really no point in having a virtual destructor. Even if you give your class to other people who then derive other classes from it then they would have no reason to ever call delete on a pointer that was upcast to your class - and if they do then I would consider this a bug.
Okay, there is one single exception, namely if your class is (mis-)used to perform polymorphic deletion of derived objects, but then you - or the other guys - hopefully know that this requires a virtual destructor.
Put another way, if your class has a non-virtual destructor then this is a very clear statement: "Don't use me for deleting derived objects!"
If you have a very small class with a huge number of instances, the overhead of a vtable pointer can make a difference in your program's memory usage. As long as your class doesn't have any other virtual methods, making the destructor non-virtual will save that overhead.
I usually declare the destructor virtual, but if you have performance critical code that is used in an inner loop, you might want to avoid the virtual table lookup. That can be important in some cases, like collision checking. But be careful about how you destroy those objects if you use inheritance, or you will destroy only half of the object.
Note that the virtual table lookup happens for an object if any method on that object is virtual. So no point in removing the virtual specification on a destructor if you have other virtual methods in the class.
If you absolutely positively must ensure that your class does not have a vtable then you must not have a virtual destructor as well.
This is a rare case, but it does happen.
The most familiar example of a pattern that does this are the DirectX D3DVECTOR and D3DMATRIX classes. These are class methods instead of functions for the syntactic sugar, but the classes intentionally do not have a vtable in order to avoid the function overhead because these classes are specifically used in the inner loop of many high-performance applications.
On operation that will be performed on the base class, and that should behave virtually, should be virtual. If deletion can be performed polymorphically through the base class interface, then it must behave virtually and be virtual.
The destructor has no need to be virtual if you don't intend to derive from the class. And even if you do, a protected non-virtual destructor is just as good if deletion of base class pointers isn't required.
The performance answer is the only one I know of which stands a chance of being true. If you've measured and found that de-virtualizing your destructors really speeds things up, then you've probably got other things in that class that need speeding up too, but at this point there are more important considerations. Some day someone is going to discover that your code would provide a nice base class for them and save them a week's work. You'd better make sure they do that week's work, copying and pasting your code, instead of using your code as a base. You'd better make sure you make some of your important methods private so that no one can ever inherit from you.
Related
This question already has answers here:
When to use virtual destructors?
(20 answers)
Closed 3 years ago.
Based on what I found here and on other links on stackoverflow, we should always define a virtual destructor in the base class if we plan to use it polymorphically. I want to know if there is an exception to this rule.
I have seen production code that does not define virtual destructor for the pure abstract base classes and in one of cppcon 2014 video Accept no visitor, around 10:06 the BoolExp struct defined is a pure abstract class and has no virtual destructor.
So for a pure abstract class defined like this
class Base {
public:
virtual foo() = 0;
virtual bar() = 0;
}
My question is it absolutely must that we define a virtual destructor for "Base" class, even though it does have any data members? Are there any exceptions to the virtual destructor rule?
Thanks in advance.
Best,
RG
My question is it absolutely must that we define a virtual destructor for "Base" class, even though it does have any data members?
It depends. If you have a case like
base * foo = new child(stuff);
// doing stuff
delete foo;
then you absolutely must have a virtual destructor. Without it you'll never destroy the child part.
If you have a case like
child * foo = new child(stuff);
// doing stuff
delete foo;
Then you do not need a virtual destructor, as child's will be called.
So the rule is if you delete polymorphically, you need a polymorphic (virtual) destructor, if not, then you don't
The exception to the rule is if you never delete an object through a pointer to the base class. In that case the base class destructor does not need to be virtual.
But if you ever delete an object via a base class pointer, then the base class destructor must be virtual or your program has Undefined Behaviour.
My question is it absolutely must that we define a virtual destructor for "Base" class, even though it does have any data members?
Stricktly speaking, No.
However, whether the base class has any member variables is not relevant. If the destructor gets called using a pointer to the base class, your code has undefined behavior regardless of whether the base class has any member variables or not.
Are there any exceptions to the virtual destructor rule?
If you are able to manage lifetimes of derived classes in such a way that the call to delete the objects is done via derived class pointers, you don't invoke undefined behavior and your code will be well behaved, assuming everything else is in your code base is in good order.
we should always define a virtual destructor in the base class if we plan to use it polymorphically.
You should always define a virtual destructor in a base classs if we plan to delete it polymorphically through that base class.
Now, one problem is that "I don't intend to" isn't safe; you should make it impossible.
Make the destructor virtual (and empty), or make it protected (and empty). A protected destructor makes polymorphic deletion unlikely (it can be bypassed, but only through pretty insane means).
Barring that, you have to be careful. This is one of the reasons why inheriting from (say) std vector is a thing to be wary of.
is it absolutely must that we define a virtual destructor for "Base" class, even though it does have any data members? Are there any exceptions to the virtual destructor rule?
It is not a must. It's a good habit that can prevent bugs.
If you decide to create a base class that doesn't have a virtual destructor, it is the responsibility of you, the developer, to always ensure that derived objects are deleted as the correct type or as a base type that does have a virtual destructor.
Deleting a derived class object using a pointer to a base class that has a non-virtual destructor results in undefined behavior.
Otherwise you are ok to not to have virtual destructor.
I would like to make an important (at least, in my view) practical amendment to correct answers describing the deletion through base object.
In particular, the destructors are called non-virtually if object life-time is managed through std::shared_ptr<Base> allocated via
std::shared_ptr<Base> sptr = std::make_shared<Derived>(args);
We use a framework that relies on memcpy in certain functions. To my understanding I can give everything that is trivially copyable into these functions.
Now we want to use a simple class hierarchy. We are not sure whether we can have a class hierarchy that results in trivially copyable types because of the safe destruction. The example code looks like this.
class Timestamp; //...
class Header
{
public:
uint8_t Version() const;
const Timestamp& StartTime();
// ... more simple setters and getters with error checking
private:
uint8_t m_Version;
Timestamp m_StartTime;
};
class CanData : public Header
{
public:
uint8_t Channel();
// ... more setters and getters with error checking
private:
uint8_t m_Channel;
};
The base class is used in several similar subclasses. Here I omitted all constructors and destructors. Thus the classes are trivially copyable. I suppose though that the user can write a code that results in a memory leak like this:
void f()
{
Header* h = new CanData();
delete h;
}
Is it right that the class hierarchy without the virtual destructor is a problem even if all classes use the compiler's default destructor? Is it therefore right that I cannot have a safe class hierarchy that is trivially copyable?
This code
Header* h = new CanData();
delete h;
will trigger undefined behavior since §5.3.5/p3 states:
In the first alternative (delete object), if the static type of the object to be deleted is different from its
dynamic type, the static type shall be a base class of the dynamic type of the object to be deleted and the
static type shall have a virtual destructor or the behavior is undefined
and regardless of not having dynamically allocated objects contained in your derived class (really bad if you have), you shouldn't do it. Having a class hierarchy without the base class virtual destructor is not a problem per se, it becomes a problem when you try to mix static and dynamic types with delete.
Doing memcpy on a derived class object smells of bad design to me, I would rather address the need for a "virtual constructor" (i.e. a virtual clone() function in your base class) to duplicate your derived objects.
You can have your class hierarchy that is trivially copyable if you make sure that your object, its subobjects and base classes are trivially copyable. If you want to prevent users referring to your derived objects via base classes you could, as Mark first suggested, render the inheritance protected
class Header
{
public:
};
class CanData : protected Header
{ ^^^^^^^^^
public:
};
int main() {
Header *pt = new CanData(); // <- not allowed
delete pt;
}
Notice that you won't be able to use base pointers at all to refer to derived objects due to §4.10/p3 - pointer conversions.
If you delete a pointer to a derived type held as its base type and you don't have a virtual destructor, the derived types destructor won't be called, whether it's implicitly generated or not. And whether its implicitly generated or not, you want it to be called. If the derived type's destructor wouldn't actually do anything anyway though, it might not leak anything or cause a problem. If the derived type holds something like a std::string, std::vector, or anything with a dynamic allocation, you want the dtors to be called. As a matter of good practice, you always want a virtual destructor for base classes whether or not the derived classes destructors need to be called (since a base class shouldn't know about what derives from it, it shouldn't make an assumption like this).
If you copy a type like so:
Base* b1 = new Derived;
Base b2 = *b1;
You will only invoke Bases copy ctor. The parts of the object which are actually from Derived will not be involved. b2 will not secretly be a Derived, it will just be a Base.
My first instinct is "don't do that - find another way, a different framework, or fix the framework". But just for fun let's assume that for certain your class copy doesn't depend in any way on the copy constructor of the class or any of its comprised parts being called.
Then since you're clearly inheriting to implement rather than to substitute the solution is easy: Use protected inheritance and your problem is solved, because they can no longer polymorphically access or delete your object, preventing the undefined behavior.
It's almost safe. In particular, there is no memory leak in
Header* h = new CanData();
delete h;
delete h calls the destructor of Header and then frees the memory pointed to by h. The amount of memory freed is the same as was initially allocated at that memory address, not the sizeof(Header). Since Header and CanData are trivial, their destructors do nothing.
However, you must provide a virtual destructor to base even if it does nothing (by requirement of the standard to avoid undefined behaviour). A common guideline is that a destructor for a base class must be either public and virtual or protected and nonvirtual
Of course, you must beware slicing as usual.
Thanks all for posting various suggestions. I try a summarizing answer with an additional proposal for the solution.
The prerequisite of my question was to reach a class hierarchy that is trivially copyable. See http://en.cppreference.com/w/cpp/concept/TriviallyCopyable and especially the requirement of a trivial destructor (http://en.cppreference.com/w/cpp/language/destructor#Trivial_destructor). The class cannot need a destructor implemented. This restricts the allowed data members, but is fine for me. The example shows only C-compatible types without dynamic memory allocation.
Some pointed out that the problem of my code is undefined behaviour, not necessarily a memory leak. Marco quoted the standard regarding this. Thanks, really helpful.
From my understanding of the answers, possible solutions are the following. Please correct me if I am wrong. The solution's point is that the implementation of the base class must avoid that its destructor can be called.
Solution 1: The proposed solutions use protected inheritance.
class CanData : protected Header
{
...
};
It works but avoids that people can access the public interface of Header. This was the original intention to have a base class. CanData needs to forward these functions to Header. In the consequece, I would reconsider to use composition instead of inheritance here. But the solution should work.
Solution 2: Header's destructor must be protected, not the base class as a whole.
class Header
{
public:
uint8_t Version() const;
const Timestamp& StartTime();
// ... more simple setters and getters with error checking
protected:
~Header() = default;
private:
uint8_t m_Version;
Timestamp m_StartTime;
};
Then no user can delete Header. This is fine for me, because Header has no purpose on its own. With public derivation, the public interface remains available to the user.
My understanding is that CanData needs not implement a destructor to call the base class's desctructor. All can use the default destructor. I am not completely sure about this though.
All in all, the answers to my questions in the end of the origial positing are:
Is it right that the class hierarchy without the virtual destructor is a problem even if all classes use the compiler's default destructor?
It is only a problem if your destructor is public. You must avoid that people can access you desctrutor, except for derived classes. And you must ensure that derived classes call (implicitely) the base class's destructor.
Is it therefore right that I cannot have a safe class hierarchy that is trivially copyable?
You can make your base class safe with protected inheritance or a protected desctructor. Then you can have a hierarchy of trivially copyable classes.
In looking at the following simple code does it make sense to introduce a virtual destructor if I know that we are not deleting from a base pointer? It seems that we should try to avoid vtable look ups if possible for performance reasons. I understand about premature optimization etc. but this is just a question in general. I was wondering your thoughts on the following:
using a protected destructor if we are not deleting items through a base pointer
the overhead associated with introducing a single virtual method
Also, if my class only has the destructor as the virtual method would the lookup overhead only be for the destructor method and other methods would not incur a penalty or once you introduce a vptr everything suffers? I am assuming that each class would have an extra vptr inside of it but that it would only have to perform vptr lookups on the destructor.
class CardPlayer
{
public:
typedef std::vector<CardPlayer> CollectionType;
explicit CardPlayer()=default;
explicit CardPlayer(const Card::CollectionType& cards);
explicit CardPlayer(Card::CollectionType&& cards);
void receiveCard(const Card& card);
bool discardCard(Card&& card);
void foldCards();
inline const Card::CollectionType& getCards() { return cards_; }
// virtual ~CardPlayer() = default; // should we introduce vtable if not really needed?
protected:
~CardPlayer()=default;
Card::CollectionType cards_;
};
--------------------------------------------------------------------
#include "CardPlayer.h"
#include <functional>
class BlackJackPlayer : public CardPlayer
{
public:
typedef std::vector<BlackJackPlayer> CollectionType;
typedef std::function<bool(const Card::CollectionType&)> hitFnType;
BlackJackPlayer(hitFnType fn) : hitFn_(fn) {}
bool wantHit()
{
return hitFn_(getCards());
}
hitFnType hitFn_;
};
I'd avoid the virtual destructor and hence adding a vtbl to the class in your case. You can protect the class from being deleted through the base class pointer, so it seems that, without having any other virtual methods, it would be a premature pessimisation :)
Also, having one more pointer per instance (the vtbl) can add up in large projects. Performance is often depending on memory access and thus you should keep the object size as small as possible and also your memory access patterns as local as possible. The vtbl will be in a different memory location and in the worst case, you ask the processor to read another cache line just to delete an object.
To answer the other question you had: Only virtual methods are routed via the vtbl, all non-virtual calls are uneffected.
To look at it from the other direction, if you are assuming you will never call delete on a base class, there is no need to make the destructor virtual, and hiding the base class destructor is a good way to enforce your assumption.
Regarding simply making your destructor virtual, as soon as you declare any method virtual, that class will automatically be given a vtable, which can increase the memory footprint of the class. Only the virtual functions will incur a vtable lookup. However, both the extra memory ( a single pointer ) and the lookup have minimal performance costs, so you typically would not have to worry about it except in the most performance critical applications.
Making the destructor protected means that it can't be invoked via a base class pointer or reference, which means there's no strong need for it to be virtual.
In general, only public methods or methods that are called by other methods in the base class need to be virtual. Protected methods need only be virtual if they're called from other methods in the class.
Why doesn't C++ make destructors virtual by default for classes that have at least one other virtual function? In this case adding a virtual destructor costs me nothing, and not having one is (almost?) always a bug. Will C++0x address this?
You don't pay for what you don't need. If you never delete through base pointer, you may not want the overhead of the indirected destructor call.
Perhaps you were thinking that the mere existence of the vtable is the only overhead. But each individual function dispatch has to be considered, too, and if I want to make my destructor call dispatch directly, I should be allowed to do so.
It would be nice of your compiler to warn you if you do ever delete a base pointer and that class has virtual methods, I suppose.
Edit: Let me pull Simon's excellent comment in here: Check out this SO question on the code generated for destructors. As you can see, there's also code-bloat overhead to be considered.
Here's an example (not that I recommend writing such code):
struct base {
virtual void foo() const = 0;
virtual void bar () const = 0;
};
struct derived: base {
void foo() const {}
void bar() const {}
};
std::shared_ptr<base>
make_base()
{
return std::make_shared<derived>();
}
This is perfectly fine code that does not exhibit UB. This is possible because std::shared_ptr uses type-erasure; the final call to delete will delete a derived*, even if the last std::shared_ptr to trigger destruction is of type std::shared_ptr<void>.
Note that this behaviour of std::shared_ptr is not tailored to virtual destruction; it has a variety of other uses (e.g. std::shared_ptr<FILE> { std::fopen( ... ), std::fclose }). However since this technique already pays the cost of some indirection to work, some users may not be interested in having a virtual destructor for their base classes. That's what "pay only for what you need" means.
By the letter of the standard, a polymorphic class with a non-virtual destructor is not a bug. One specific action performed on such an object results in undefined behavior, but everything else is perfectly kosher. So given the otherwise lenient behavior of the standard in terms of what mistakes it allows programmers to make, why should the destructor be given special treatment?
And such a change would have costs, albeit mostly trivial ones: the virtual table will be one element larger, and the virtual dispatch associated with destructor calls.
To the best of my knowledge, no, there is no change in the behavior of destructors in this regard in C++11. I imagine it would say something in the section on special member functions, but it does not, and there is similarly nothing in the section of virtual functions in general.
Why does C++ not have a virtual constructor?
Hear it from the horse's mouth. :)
From Bjarne Stroustrup's C++ Style and Technique FAQ Why don't we have virtual constructors?
A virtual call is a mechanism to get work done given partial
information. In particular, "virtual" allows us to call a function
knowing only any interfaces and not the exact type of the object. To
create an object you need complete information. In particular, you
need to know the exact type of what you want to create. Consequently,
a "call to a constructor" cannot be virtual.
The FAQ entry goes on to give the code for a way to achieve this end without a virtual constructor.
Virtual functions basically provide polymorphic behavior. That is, when you work with an object whose dynamic type is different than the static (compile time) type with which it is referred to, it provides behavior that is appropriate for the actual type of object instead of the static type of the object.
Now try to apply that sort of behavior to a constructor. When you construct an object the static type is always the same as the actual object type since:
To construct an object, a constructor needs the exact type of the object it is to create [...] Furthermore [...]you cannot have a pointer to a constructor
(Bjarne Stroustup (P424 The C++ Programming Language SE))
Unlike object oriented languages such as Smalltalk or Python, where the constructor is a virtual method of the object representing the class (which means you don't need the GoF abstract factory pattern, as you can pass the object representing the class around instead of making your own), C++ is a class based language, and does not have objects representing any of the language's constructs. The class does not exist as an object at runtime, so you can't call a virtual method on it.
This fits with the 'you don't pay for what you don't use' philosophy, though every large C++ project I've seen has ended up implementing some form of abstract factory or reflection.
two reasons I can think of:
Technical reason
The object exists only after the constructor ends.In order for the constructor to be dispatched using the virtual table , there has to be an existing object with a pointer to the virtual table , but how can a pointer to the virtual table exist if the object still doesn't exist? :)
Logic reason
You use the virtual keyword when you want to declare a somewhat polymorphic behaviour. But there is nothing polymorphic with constructors , constructors job in C++ is to simply put an object data on the memory . Since virtual tables (and polymorphism in general) are all about polymorphic behaviour rather on polymorphic data , There is no sense with declaring a virtual constructor.
Summary: the C++ Standard could specify a notation and behaviour for "virtual constructor"s that's reasonably intuitive and not too hard for compilers to support, but why make a Standard change for this specifically when the functionality can already be cleanly implemented using create() / clone() (see below)? It's not nearly as useful as many other language proposal in the pipeline.
Discussion
Let's postulate a "virtual constructor" mechanism:
Base* p = new Derived(...);
Base* p2 = new p->Base(); // possible syntax???
In the above, the first line constructs a Derived object, so *p's virtual dispatch table can reasonably supply a "virtual constructor" for use in the second line. (Dozens of answers on this page stating "the object doesn't yet exist so virtual construction is impossible" are unnecessarily myopically focused on the to-be-constructed object.)
The second line postulates the notation new p->Base() to request dynamic allocation and default construction of another Derived object.
Notes:
the compiler must orchestrate memory allocation before calling the constructor - constructors normally support automatic (informally "stack") allocation, static (for global/namespace scope and class-/function-static objects), and dynamic (informally "heap") when new is used
the size of object to be constructed by p->Base() can't generally be known at compile-time, so dynamic allocation is the only approach that makes sense
it is possible to allocate runtime-specified amounts of memory on the stack - e.g. GCC's variable-length array extension, alloca() - but leads to significant inefficiencies and complexities (e.g. here and here respectively)
for dynamic allocation it must return a pointer so memory can be deleted later.
the postulated notation explicitly lists new to emphasise dynamic allocation and the pointer result type.
The compiler would need to:
find out how much memory Derived needed, either by calling an implicit virtual sizeof function or having such information available via RTTI
call operator new(size_t) to allocate memory
invoke Derived() with placement new.
OR
create an extra vtable entry for a function that combines dynamic allocation and construction
So - it doesn't seem insurmountable to specify and implement virtual constructors, but the million-dollar question is: how would it be better than what's possible using existing C++ language features...? Personally, I see no benefit over the solution below.
`clone()` and `create()`
The C++ FAQ documents a "virtual constructor" idiom, containing virtual create() and clone() methods to default-construct or copy-construct a new dynamically-allocated object:
class Shape {
public:
virtual ~Shape() { } // A virtual destructor
virtual void draw() = 0; // A pure virtual function
virtual void move() = 0;
// ...
virtual Shape* clone() const = 0; // Uses the copy constructor
virtual Shape* create() const = 0; // Uses the default constructor
};
class Circle : public Shape {
public:
Circle* clone() const; // Covariant Return Types; see below
Circle* create() const; // Covariant Return Types; see below
// ...
};
Circle* Circle::clone() const { return new Circle(*this); }
Circle* Circle::create() const { return new Circle(); }
It's also possible to change or overload create() to accept arguments, though to match the base class / interface's virtual function signature, arguments to overrides must exactly match one of the base class overloads. With these explicit user-provided facilities, it's easy to add logging, instrumentation, alter memory allocation etc..
We do, it's just not a constructor :-)
struct A {
virtual ~A() {}
virtual A * Clone() { return new A; }
};
struct B : public A {
virtual A * Clone() { return new B; }
};
int main() {
A * a1 = new B;
A * a2 = a1->Clone(); // virtual construction
delete a2;
delete a1;
}
Semantic reasons aside, there is no vtable until after the object is constructed, thus making a virtual designation useless.
Virtual functions in C++ are an implementation of run-time polymorphism, and they will do function overriding. Generally the virtual keyword is used in C++ when you need dynamic behavior. It will work only when object exists. Whereas constructors are used to create the objects. Constructors will be called at the time of object creation.
So if you create the constructor as virtual, as per the virtual keyword definition, it should have existing object to use, but constructor is used to to create the object, so this case will never exist. So you should not use the constructor as virtual.
So, if we try to declare virtual constructor compiler throw an Error:
Constructors cannot be declared virtual
You can find an example and the technical reason to why it is not allowed in #stefan 's answer. Now a logical answer to this question according to me is:
The major use of virtual keyword is to enable polymorphic behaviour when we don't know what type of the object the base class pointer will point to.
But think of this is more primitive way, for using virtual functionality you will require a pointer. And what does a pointer require? An object to point to! (considering case for correct execution of the program)
So, we basically require an object that already exists somewhere in the memory (we are not concerned with how the memory was allocated, it may be at compile time or either runtime) so that our pointer can correctly point to that object.
Now, think of the situation about the moment when the object of the class to be pointed is being assigned some memory -> Its constructor will be called automatically at that instance itself!
So we can see that we don't actually need to worry about the constructor being virtual, because in any of the cases you wish to use a polymorphic behaviour our constructor would have already been executed making our object ready for usage!
When people ask a question like this, I like to think to myself "what would happen if this were actually possible?" I don't really know what this would mean, but I guess it would have something to do with being able to override the constructor implementation based on the dynamic type of the object being created.
I see a number of potential problems with this. For one thing, the derived class will not be fully constructed at the time the virtual constructor is called, so there are potential issues with the implementation.
Secondly, what would happen in the case of multiple inheritance? Your virtual constructor would be called multiple times presumably, you would then need to have some way of know which one was being called.
Thirdly, generally speaking at the time of construction, the object does not have the virtual table fully constructed, this means it would require a large change to the language specification to allow for the fact that the dynamic type of the object would be known at construction time. This would then allow the base class constructor to maybe call other virtual functions at construction time, with a not fully constructed dynamic class type.
Finally, as someone else has pointed out you can implement a kind of virtual constructor using static "create" or "init" type functions that basically do the same thing as a virtual constructor would do.
Although the concept of virtual constructors does not fit in well since object type is pre-requisite for object creation, its not completly over-ruled.
GOF's 'factory method' design pattern makes use of the 'concept' of virtual constructor, which is handly in certain design situations.
Virtual functions are used in order to invoke functions based on the type of object pointed to by the pointer, and not the type of pointer itself. But a constructor is not "invoked". It is called only once when an object is declared. So, a constructor cannot be made virtual in C++.
Interview answer is : virtual ptr and table are related to objects but not the class.hence constructor builds the virtual table
hence we cant have virtual constructor as there is no Vtable before obj creation.
You shouldn't call virtual function within your constructor either. See : http://www.artima.com/cppsource/nevercall.html
In addition I'm not sure that you really need a virtual constructor. You can achieve polymorphic construction without it: you can write a function that will construct your object according to the needed parameters.
A virtual-table(vtable) is made for each Class having one or more 'virtual-functions'. Whenever an Object is created of such class, it contains a 'virtual-pointer' which points to the base of corresponding vtable. Whenever there is a virtual function call, the vtable is used to resolve to the function address.
Constructor can not be virtual, because when constructor of a class is executed there is no vtable in the memory, means no virtual pointer defined yet. Hence the constructor should always be non-virtual.
Cant we simply say it like.. We cannot inherit constructors. So there is no point declaring them virtual because the virtual provides polymorphism .
The virtual mechanism only works when you have a based class pointer to a derived class object. Construction has it's own rules for the calling of base class constructors, basically base class to derived. How could a virtual constructor be useful or called? I don't know what other languages do, but I can't see how a virtual constructor could be useful or even implemented. Construction needs to have taken place for the virtual mechanism to make any sense and construction also needs to have taken place for the vtable structures to have been created which provides the mechanics of the polymorphic behaviour.
C++ virtual constructor is not possible.For example you can not mark a constructor as virtual.Try this code
#include<iostream.h>
using namespace std;
class aClass
{
public:
virtual aClass()
{
}
};
int main()
{
aClass a;
}
It causes an error.This code is trying to declare a constructor as virtual.
Now let us try to understand why we use virtual keyword. Virtual keyword is used to provide run time polymorphism. For example try this code.
#include<iostream.h>
using namespace std;
class aClass
{
public:
aClass()
{
cout<<"aClass contructor\n";
}
~aClass()
{
cout<<"aClass destructor\n";
}
};
class anotherClass:public aClass
{
public:
anotherClass()
{
cout<<"anotherClass Constructor\n";
}
~anotherClass()
{
cout<<"anotherClass destructor\n";
}
};
int main()
{
aClass* a;
a=new anotherClass;
delete a;
getchar();
}
In main a=new anotherClass; allocates a memory for anotherClass in a pointer a declared as type of aClass.This causes both the constructor (In aClass and anotherClass) to call automatically.So we do not need to mark constructor as virtual.Because when an object is created it must follow the chain of creation (i.e first the base and then the derived classes).
But when we try to delete a delete a; it causes to call only the base destructor.So we have to handle the destructor using virtual keyword. So virtual constructor is not possible but virtual destructor is.Thanks
There's a very basic reason: Constructors are effectively static functions, and in C++ no static function can be virtual.
If you have much experience with C++, you know all about the difference between static & member functions. Static functions are associated with the CLASS, not the objects (instances), so they don't see a "this" pointer. Only member functions can be virtual, because the vtable- the hidden table of function pointers that makes 'virtual' work- is really a data member of each object.
Now, what is the constructor's job? It is in the name- a "T" constructor initializes T objects as they're allocated. This automatically precludes it being a member function! An object has to EXIST before it has a "this" pointer and thus a vtable. That means that even if the language treated constructors as ordinary functions (it doesn't, for related reasons I won't get into), they'd have to be static member functions.
A great way to see this is to look at the "Factory" pattern, especially factory functions. They do what you're after, and you'll notice that if class T has a factory method, it is ALWAYS STATIC. It has to be.
If you think logically about how constructors work and what the meaning/usage of a virtual function is in C++ then you will realise that a virtual constructor would be meaningless in C++. Declaring something virtual in C++ means that it can be overridden by a sub-class of the current class, however the constructor is called when the objected is created, at that time you cannot be creating a sub-class of the class, you must be creating the class so there would never be any need to declare a constructor virtual.
And another reason is, the constructors have the same name as its class name and if we declare constructor as virtual, then it should be redefined in its derived class with the same name, but you can not have the same name of two classes. So it is not possible to have a virtual constructor.
When a constructor is invoked, although there is no object created till that point, we still know the kind of object that is gonna be created because the specific constructor of the class to which the object belongs to has already been called.
Virtual keyword associated with a function means the function of a particular object type is gonna be called.
So, my thinking says that there is no need to make the virtual constructor because already the desired constructor whose object is gonna be created has been invoked and making constructor virtual is just a redundant thing to do because the object-specific constructor has already been invoked and this is same as calling class-specific function which is achieved through the virtual keyword.
Although the inner implementation won’t allow virtual constructor for vptr and vtable related reasons.
Another reason is that C++ is a statically typed language and we need to know the type of a variable at compile-time.
The compiler must be aware of the class type to create the object. The type of object to be created is a compile-time decision.
If we make the constructor virtual then it means that we don’t need to know the type of the object at compile-time(that’s what virtual function provide. We don’t need to know the actual object and just need the base pointer to point an actual object call the pointed object’s virtual functions without knowing the type of the object) and if we don’t know the type of the object at compile time then it is against the statically typed languages. And hence, run-time polymorphism cannot be achieved.
Hence, Constructor won’t be called without knowing the type of the object at compile-time. And so the idea of making a virtual constructor fails.
"A constructor can not be virtual"
there are some valid reasons that justify this statement.
to create an object the constructor of the object class must be of the same type as the class. But, this is not possible with a virtually implemented constructor.
at the time of calling the constructor, the virtual table would not have been created to resolve any virtual function calls. Thus, a virtual constructor itself would not have anywhere to look up to.
As a result, it is not possible to declare a constructor to be virtual.
The Vpointer is created at the time of object creation. vpointer wont exists before object creation. so there is no point of making the constructor as virtual.