This question already has answers here:
Virtual/pure virtual explained
(12 answers)
Closed 8 years ago.
With C++, virtual is used like this? What's the difference between the both?
class Animal
{
public:
virtual void something();
virtual void something() = 0;
}
I might be fuzzy on this but I think the first says: You can override me, and the second says, You must override me.
virtual void something() = 0; // says it is a pure virtual function
virtual void something(); // is a virtual function
And classes that contain atleast one pure virtual function are called abstract base classes.The main difference between an abstract base class and a regular polymorphic class is that because in abstract base classes at least one of its members lacks implementation, we cannot create instances (objects) of it.
The first states that the function is virtual: that subclasses can override the behaviour. However, the function still has an implementation in the base class. The second is pure virtual: it doesn't have an implementation, and must be overridden by a subclass. It's similar to the abstract keyword in Java and C#. It makes the class abstract too, so it cannot be instantiated.
The second "something" must be implemented by subclasses and any class containing "xxx() = 0" cannot be directly instantiated. It's called a "pure virtual" function and classes containing them are "abstract". A class containing nothing but pure virtuals is "pure abstract".
first one is just declaring a virtual method. if you extend the class and override that method then the child class' implementation is called.
2nd one is a pure virtual method. in other words neither your class or any class that extends it be instantiated (abstract) without first providing a definition for something().
Consider:
class Sloth : public Animal { void something() { ... } };
Animal animal;
Sloth sloth;
Here, we're trying to create two objects - an Animal and a Sloth. But, should we be allowed to create an animal? Perhaps the programmer has created Animal just so it can be used to refer polymorphically to derived types, as in:
std::vector<Animal*> zoo;
zoo.push_back(new Sloth());
zoo.push_back(new Sea_Eagle());
They might then expect p_animal->something() to do something useful. Given Animal is only inteded for this polymorphic abstraction, it would be wrong for anyone to put an actual "new Animal()" object directly into the zoo (or operate on one anywhere else in the program).
Say it was possible - what should then happen if a programmer using this Animal class creates an instance and calls the something() function? Perhaps nothing, or perhaps it's an error condition and should never appear in good code - rather than having Animal's something() function print an error message or throws an exception at run time. That's ugly - run-time errors mean the program is failing when the client's trying to use it.
C++ supports the "= 0" notation so that the compiler knows to prevent (base-class) Animal objects being created at compile time, so you can ship software that always works for the user.
Related
Suppose we have a class, "Animal", and subclasses, "Cat" and "Dog".
Let's say we want to allow both "Cat" and "Dog" to make a noise (cat: "meow" - dog: "woof") when we pass their objects into an intermediate function for any "Animal".
Why must we use a virtual method to do this? Couldn't we just do Animal->makeNoise() without defining a virtual method in "Animal"? As "Cat" and "Dog" are both animals, wouldn't it be clear that "makeNoise()" is referring to the Animal which has been passed to the function?
Is this just a matter of syntax or something more? I'm pretty sure in Java we don't have to do this.
In Java, all member functions are virtual by default (except static, private, and final ones).
In C++, all member functions are non-virtual by default. Making functions virtual adds overhead - both runtime and to object size - and the C++ philosophy is don't pay for what you don't use. Most of my objects are not polymorphic, so I shouldn't have to pay for polymorphism unless I need it. Since you need Animal::makeNoise() to be virtual, you must explicitly specify it as such.
C++ is designed to run with as little overhead as possible, trusting the programmer to make the correct call. Essentially, it 'gives you the gun and the option to shoot yourself in the foot', as one of my friends likes to say often. Speed and flexibility are paramount.
To correctly cause true polymorphic behavior, C++ requires it be specified. However! It is only required to be specified in the base class, as all derived class will inherit the virtual member functions. If a member inherits a virtual member function, it is good practice to place 'virtual' in the declaration, but not required.
ADTs usually implement pure virtual functions to indicate that the derived classes MUST implement the function. Such as:
animal makeNoise() = 0; /*Indicates this function contains no implementation.
and must be implemented by derived classes in order to function.*/
Again, it is not required the derived classes include 'virtual' in their inherited members so long as the base class includes this.
If you want to deduce the type of the Animal and then call make_sound(), then you would have to do a dynamic_cast on the animal object for each and every child of animal. This would include any class that is directly or indirectly a child of the Animal class.
This is both difficult to maintain and very painful to any change eg. Adding new class as a child to the Animal class.
Since c++ philosophy is efficiency, you will have to ask the compiler to provide you with run-time polymorphism as it is costly. How would you do that? By stating the make_sound() function as virtual. This creates a vtable ( a table of functions pointers ) which refers to an address of make_sound() which differs to based on the type of the object.
No need to downcast as indirection handles everything for you. What could be hundreds of lines of code is just a single line of code. That is the power of indirection!!
You could say that you have to do it because that's one of the rules of the language.
There's a reason its helpful though.
When trying to validate the code that uses an Animal, the complier knows what functions exist on an Animal. Its possible to tell whether the code is correct without checking all classes that derive from animal. So that code doesn't need to depend on all those derived classes. If you derive a new class from Animal but forget to implement the makeNoise function that's an error in the new class not the code that uses the Animal base class and the complier can point you towards that error. Without the virtual function declared in Animal there would no way to tell if its the calling code or the new class that is in error.
A key point here is that these errors would be caught at compile-time for C++ because of its static typing. Other languages can allow dynamic typing, which can make some things easier but, the errors would only be spotted at runtime.
In Java, all functions are virtual by default. In C++ they are not, so when you call a non-virtual function on a pointer of a given type, that type's implementation of that function is invoked with the object's address as this.
class Animal {
public:
void sound() { std::cout << "splat\n"; }
virtual void appearance() { std::cout << "animaly\n"; }
};
class Cat {
public:
void sound() { std::cout << "meow\n"; }
virtual void appearance() { std::cout << "furry\n"; }
};
int main() {
Animal a;
Cat c;
Animal* ac = new Cat;
a.sound(); // splat
a.appearance(); // animaly
c.sound(); // meow
c.appearance(); // furry
ac->sound(); // splat
ac->appearance(); // furry
}
This would occur when you wanted to write a function that generalized on "Animal" rather than requiring a specific derived class pointer.
In java you use virtual methods too.
It improves the loosely coupling of your software.
For example, you can use a library and don't know which animal they use internally. There is a animal implementation you don't know and you can use it because it's an animal. You get the animal with a library.getAnimal method. Now you can use their animal without to know which noise it makes, because they have to implement the method makeNoise.
Edit: So to answer your question, c++ wants a explicit declaration and in java it is implicit. so yes it is a kind of language specific idiosyncracy.
I have a class declared as follow:
class TestFoo {
public:
TestFoo();
virtual void virtualFunction();
void nonVirtualFunction();
};
that I try to implement this way
TestFoo::TestFoo(){}
void TestFoo::nonVirtualFunction(){}
which on compilation returns the error:
undefined reference to vtable for TestFoo
I tried :
TestFoo::TestFoo(){}
void TestFoo::nonVirtualFunction(){}
void TestFoo::virtualFunction(){}
which compiles ok which is consistent to the answers to these posts:
Undefined reference to vtable
undefined reference to vtable
What confuses me is that I thought the whole point of declaring a virtual function is that I would not need to define it. In this example, I am not planning to create any instance of TestFoo, but to create instances of (concrete) classes inheriting from TestFoo. But still, I want to define the function nonVirtualFunction for every subclasses of TestFoo.
Something I did not get right ?
Thanks !
the whole point of declaring a virtual function is that I would not
need to define it
Not quite, it says "I may want to replace the implementation of this function by something else in a derived class."
I may have misunderstood your question, but you seem to imply that you don't think you can define pure virtual member function in C++, which you can. You can declare one as follows.
virtual void virtualFunction() = 0;
Normally, a pure virtual function won't be defined, but of course you can. That says "There is no default implementation of this function because it won't always make sense, but I'll provide you with an implementation that you can opt into."
By the way, if a class has any virtual functions, you should also define a virtual destructor, as it is perfectly legal (and often recommended) to have a base class (smart) pointer to a derived class - without a virtual destructor, the object may not be deleted correctly.
... I thought the whole point of declaring a
virtual function is that I would not need to define it ...
For that facility you have a feature called pure virtual methods:
virtual void virtualFunction() = 0; // no linking error now
Note that, a virtual method cannot remain unimplemented. The reason is that for every virtual method declared inside a class body there has to be a vtable entry. Failing to find its body results in linking error.
Purpose of this restriction:
Unless a class is abstract - that is it has at least one virtual function - there is no way you can guarantee to the compiler that you are not going to declare an object of TestFoo. What happens when you do following:
DerivedOfTestFoo obj1;
TestFoo obj2 = obj1, *p = &obj2; // object slicing
p->virtualFunction(); // where is the body?
Other situation; in constructor there is no virtual mechanism:
TestFoo::TestFoo () {
this->virtualFunction(); // where is the body?
}
We can conclude that, compilers follow the rule, "Better to be safe than sorry". :)
Your description matches perfectly with the case of an abstract class. Declare your virtual function as:
virtual void VirtualFunction () = 0;
This means that you are not implementing the function in this class. As a result, the class becomes abstract. That is, no bare objects of this class can be instantiated.
Also, you should provide a virtual destructor.
Update: Some clarifications...
The language allows you to redefine a non-virtual function. Though, the wrong version might be called in some cases:
derived D; // rB is a reference to base class but it
base & rB=D; // points to an object of the derived class
rB.NonVirtualFunction (); // The base-class version is called
For this reason, redefining a non-virtual function is strongly discouraged nowadays. See Scott Meyers' "Effective C++, Third Edition: 55 Specific Ways to Improve Your Programs and Designs", item 36: "Never redefine an inherited non-virtual function."
See also item 7: "Declare destructors virtual in polymorphic base classes". An example:
base * pB = new derived;
delete pB; // If base's destructor is not virtual,
// ~derived() will not be called.
In case you wonder why isn't everything virtual by default, the reason is that calling a virtual function is slightly slower than calling a non-virtual one. Oh, and objects of classes with virtual functions occupy a few more bytes each.
If you want make this virtual function as pure virtual function,do not want to define it then, virtual void virtualFunction() = 0;
I'm now learning C++, the OO side, and I see this all the time:
class SomeClass{
virtual void aMethod()=0;
}
class AnotherClass{
void anotherMethod(){/*Empty*/}
}
class SomeClassSon : public SomeClass{
void aMethod(){/*Also Empty*/}
}
what is the difference between the 3 methods? The virtual equals zero, the empty one, and the virtual, since it is inherited, empty one.
Why can't I just make the SomeClassSon method like the father, virtual void equals zero?
For your
class SomeClass{
virtual void aMethod()=0;
}
the presence of a pure virtual method makes your class abstract. Once you have one such pure virtual method, =0, in your class, you cannot instantiate the class. What is more, any derived class must implement the pure virtual aMethod(), or it becomes an abstract class as well.
In your derived class, you overwrite the pure virtual method from above, and this makes the derived class non abstract. You can instantiate this derived class.
But, in derived class, method's body is empty, right? That's why your question makes sense: why not make the class pure virtual as well. Well, your class may entail other methods. If so, SomeClass cannot be instantiated (there is a pure virtual method), whereas child class SomeClassSon can be.
Same applies to your AnotherClass, which can be instantiated, contrary to SomeClass.
The difference is that virtual void aMethod() = 0 is a pure virtual function, meaning that:
SomeClass becomes an abstract base class,
meaning it cannot be instantiated.
Any class which inherits from SomeClass must implement aMethod, or it too becomes an abstract base class which cannot be instantiated
Note that any class with one or more pure virtual functions is automatically an abstract base class.
The "equals 0" you're referring to is called "pure virtual". It's a function that the child that wants to be instantiated HAS to implement as opposed to providing base functionality meaning that the parent class is going to define functionality that has to exist but that the parent has no knowledge of how the child will do it. Note that this makes the class abstract in that it cannot be instantiated. For example I may want to define a "Mammal" class I can inherit from and I want its children to act a certain way - but I can't simply make a "Mammal". Instead I would create a "Giraffe" class and make sure it acts like it's supposed to.
It's also explained at this SO question.
The "Empty" function you're referring to is instead functionality where the function is defined and can be called - but does nothing.
the declaration aMethod()=0 tells the compiler that this method must be provided for in subclasses. Any subclass that does not implement the method can not be instantiated. This helps you ensure any objects of the base class will have the method implemented.
A pure virtual function (your first example, with the =0) means that function must be overridden in a derived class for an object of that class to be instantiated.
The second is basically just a member function that does nothing. Since the function has a different name and the class isn't related to SomeClass, the two don't affect each other at all.
The third overrides the pure virtual function, so it's possible to instantiate SomeClassSon, but in the derived class the overridden function does nothing.
A pure virtual makes the class abstract. An empty non-virtual method doesn't do anything - it just leads to a linker error if you attempt to call it. By contrast, you can't attempt to call a pure virtual (unless you attempt to call it from a constructor, which is bad anyway) because the compiler won't let you create that object.
There's also a logical difference - the method marked virtual will be virtual through the inheritance chain - the others are just regular methods.
This question already has answers here:
Closed 11 years ago.
Possible Duplicate:
What is the difference between a concrete class and an abstract class?
I was coding something in Visual C++ 2008 doing an exercise in a book I am reading when I held my cursor over one of the classes and it told me it was an abstract class. Now I know it is an abstract class as that is what this exercise is about but I was curious as to what lets the Intelisense thing know that it is an abstract class.
I did a little homework and found that it may have been the fact that I have two virtual functions in this class and one of them is a pure virtual.
Is the pure virtual a dead giveaway or are there other things that would tell you you are dealing with or looking at an abstract class?
Is the pure virtual a dead giveaway
In C++, yes. Since it has no abstract keyword or equivalent, the common idiom to make a class abstract is to declare a pure virtual function in it (which prevents instantiation).*
For this same reason, in C++ there is not much difference between an interface and an abstract class - a C++ interface is simply a class containing only pure virtual functions.
*Update: Another way to prevent instantiation of a class is to declare its constructor(s) protected or private. The latter means it can't be subclassed either, but the former doesn't prevent subclassing, so in theory a protected constructor could also be a sign of an abstract class a way to force subclassing. However, I have never seen this in practice. I believe this is because an abstract class is designed to be extended, which in practice almost always means it has virtual functions. And the reason why we want it to be abstract is because some part of its implementation is not yet known, hence it is to be defined in its subclass(es). Which is exactly what a pure virtual function is meant for.
The only thing that makes a class abstract is if it has one or more unimplemented pure virtual functions, possibly derived ones.
struct A { // abstract
virtual void f() = 0;
};
struct B : public A { // abstract
};
struct C : public B { // not abstract
void f() {}
};
Note that A should also have a (probably not pure) virtual destructor.
It is also possible to implement pure virtual functions, but this is pretty rare:
struct D {
virtual void f() = 0;
};
The function has to be defined outside the class:
void D :: f() {}
but the class remains abstract.
I wouldn't say I'm an expert in C++ but I would say you've nailed it. The presence of a pure virtual method means that the class cannot be instantiated and is therefore abstract. If all of your methods are pure virtual (i.e. there is no implementation) it would be an interface and in this case you only need the header file.
I understand the need for a virtual destructor. But why do we need a pure virtual destructor? In one of the C++ articles, the author has mentioned that we use pure virtual destructor when we want to make a class abstract.
But we can make a class abstract by making any of the member functions as pure virtual.
So my questions are
When do we really make a destructor pure virtual? Can anybody give a good real time example?
When we are creating abstract classes is it a good practice to make the destructor also pure virtual? If yes..then why?
Probably the real reason that pure virtual destructors are allowed is that to prohibit them would mean adding another rule to the language and there's no need for this rule since no ill-effects can come from allowing a pure virtual destructor.
Nope, plain old virtual is enough.
If you create an object with default implementations for its virtual methods and want to make it abstract without forcing anyone to override any specific method, you can make the destructor pure virtual. I don't see much point in it but it's possible.
Note that since the compiler will generate an implicit destructor for derived classes, if the class's author does not do so, any derived classes will not be abstract. Therefore having the pure virtual destructor in the base class will not make any difference for the derived classes. It will only make the base class abstract (thanks for #kappa's comment).
One may also assume that every deriving class would probably need to have specific clean-up code and use the pure virtual destructor as a reminder to write one but this seems contrived (and unenforced).
Note: The destructor is the only method that even if it is pure virtual has to have an implementation in order to instantiate derived classes (yes pure virtual functions can have implementations, being pure virtual means derived classes must override this method, this is orthogonal to having an implementation).
struct foo {
virtual void bar() = 0;
};
void foo::bar() { /* default implementation */ }
class foof : public foo {
void bar() { foo::bar(); } // have to explicitly call default implementation.
};
All you need for an abstract class is at least one pure virtual function. Any function will do; but as it happens, the destructor is something that any class will have—so it's always there as a candidate. Furthermore, making the destructor pure virtual (as opposed to just virtual) has no behavioral side effects other than to make the class abstract. As such, a lot of style guides recommend that the pure virtual destuctor be used consistently to indicate that a class is abstract—if for no other reason than it provides a consistent place someone reading the code can look to see if the class is abstract.
If you want to create an abstract base class:
that can't be instantiated (yep, this is redundant with the term "abstract"!)
but needs virtual destructor behavior (you intend to carry around pointers to the ABC rather than pointers to the derived types, and delete through them)
but does not need any other virtual dispatch behavior for other methods (maybe there are no other methods? consider a simple protected "resource" container that needs a constructors/destructor/assignment but not much else)
...it's easiest to make the class abstract by making the destructor pure virtual and providing a definition (method body) for it.
For our hypothetical ABC:
You guarantee that it cannot be instantiated (even internal to the class itself, this is why private constructors may not be enough), you get the virtual behavior you want for the destructor, and you do not have to find and tag another method that doesn't need virtual dispatch as "virtual".
Here I want to tell when we need virtual destructor and when we need pure virtual destructor
class Base
{
public:
Base();
virtual ~Base() = 0; // Pure virtual, now no one can create the Base Object directly
};
Base::Base() { cout << "Base Constructor" << endl; }
Base::~Base() { cout << "Base Destructor" << endl; }
class Derived : public Base
{
public:
Derived();
~Derived();
};
Derived::Derived() { cout << "Derived Constructor" << endl; }
Derived::~Derived() { cout << "Derived Destructor" << endl; }
int _tmain(int argc, _TCHAR* argv[])
{
Base* pBase = new Derived();
delete pBase;
Base* pBase2 = new Base(); // Error 1 error C2259: 'Base' : cannot instantiate abstract class
}
When you want that no one should be able to create the object of Base class directly, use pure virtual destructor virtual ~Base() = 0. Usually at-least one pure virtual function is required, let's take virtual ~Base() = 0, as this function.
When you do not need above thing, only you need the safe destruction of Derived class object
Base* pBase = new Derived();
delete pBase;
pure virtual destructor is not required, only virtual destructor will do the job.
From the answers I have read to your question, I couldn't deduce a good reason to actually use a pure virtual destructor. For example, the following reason doesn't convince me at all:
Probably the real reason that pure virtual destructors are allowed is that to prohibit them would mean adding another rule to the language and there's no need for this rule since no ill-effects can come from allowing a pure virtual destructor.
In my opinion, pure virtual destructors can be useful. For example, assume you have two classes myClassA and myClassB in your code, and that myClassB inherits from myClassA. For the reasons mentioned by Scott Meyers in his book "More Effective C++", Item 33 "Making non-leaf classes abstract", it is better practice to actually create an abstract class myAbstractClass from which myClassA and myClassB inherit. This provides better abstraction and prevents some problems arising with, for example, object copies.
In the abstraction process (of creating class myAbstractClass), it can be that no method of myClassA or myClassB is a good candidate for being a pure virtual method (which is a prerequisite for myAbstractClass to be abstract). In this case, you define the abstract class's destructor pure virtual.
Hereafter a concrete example from some code I have myself written. I have two classes, Numerics/PhysicsParams which share common properties. I therefore let them inherit from the abstract class IParams. In this case, I had absolutely no method at hand that could be purely virtual. The setParameter method, for example, must have the same body for every subclass. The only choice that I have had was to make IParams' destructor pure virtual.
struct IParams
{
IParams(const ModelConfiguration& aModelConf);
virtual ~IParams() = 0;
void setParameter(const N_Configuration::Parameter& aParam);
std::map<std::string, std::string> m_Parameters;
};
struct NumericsParams : IParams
{
NumericsParams(const ModelConfiguration& aNumericsConf);
virtual ~NumericsParams();
double dt() const;
double ti() const;
double tf() const;
};
struct PhysicsParams : IParams
{
PhysicsParams(const N_Configuration::ModelConfiguration& aPhysicsConf);
virtual ~PhysicsParams();
double g() const;
double rho_i() const;
double rho_w() const;
};
If you want to stop instantiating of base class without making any change in your already implemented and tested derive class, you implement a pure virtual destructor in your base class.
You are getting into hypotheticals with these answers, so I will try to make a simpler, more down to earth explanation for clarity's sake.
The basic relationships of object oriented design are two:
IS-A and HAS-A. I did not make those up. That is what they are called.
IS-A indicates that a particular object identifies as being of the class that is above it in a class hierarchy. A banana object is a fruit object if it is a subclass of the fruit class. This means that anywhere a fruit class can be used, a banana can be used. It is not reflexive , though. You can not substitute a base class for a specific class if that specific class is called for.
Has-a indicated that an object is part of a composite class and that there is an ownership relationship. It means in C++ that it is a member object and as such the onus is on the owning class to dispose of it or hand ownership off before destructing itself.
These two concepts are easier to realize in single-inheritance languages than in a multiple inheritance model like c++, but the rules are essentially the same. The complication comes when the class identity is ambiguous, such as passing a Banana class pointer into a function that takes a Fruit class pointer.
Virtual functions are, firstly, a run-time thing. It is part of polymorphism in that it is used to decide which function to run at the time it is called in the running program.
The virtual keyword is a compiler directive to bind functions in a certain order if there is ambiguity about the class identity. Virtual functions are always in parent classes (as far as I know) and indicate to the compiler that binding of member functions to their names should take place with the subclass function first and the parent class function after.
A Fruit class could have a virtual function color() that returns "NONE" by default.
The Banana class color() function returns "YELLOW" or "BROWN".
But if the function taking a Fruit pointer calls color() on the Banana class sent to it -- which color() function gets invoked?
The function would normally call Fruit::color() for a Fruit object.
That would 99% of the time not be what was intended.
But if Fruit::color() was declared virtual then Banana:color() would be called for the object because the correct color() function would be bound to the Fruit pointer at the time of the call.
The runtime will check what object the pointer points to because it was marked virtual in the Fruit class definition.
This is different than overriding a function in a subclass. In that case
the Fruit pointer will call Fruit::color() if all it knows is that it IS-A pointer to Fruit.
So now to the idea of a "pure virtual function" comes up.
It is a rather unfortunate phrase as purity has nothing to do with it. It means that it is intended that the base class method is never to be called.
Indeed a pure virtual function can not be called. It must still be defined, however. A function signature must exist. Many coders make an empty implementation {} for completeness, but the compiler will generate one internally if not. In that case when the function is called even if the pointer is to Fruit , Banana::color() will be called as it is the only implementation of color() there is.
Now the final piece of the puzzle: constructors and destructors.
Pure virtual constructors are illegal, completely. That is just out.
But pure virtual destructors do work in the case that you want to forbid the creation of a base class instance. Only sub classes can be instantiated if the destructor of the base class is pure virtual.
the convention is to assign it to 0.
virtual ~Fruit() = 0; // pure virtual
Fruit::~Fruit(){} // destructor implementation
You do have to create an implementation in this case. The compiler knows this is what you are doing and makes sure you do it right, or it complains mightily that it can not link to all the functions it needs to compile. The errors can be confusing if you are not on the right track as to how you are modeling your class hierarchy.
So you are forbidden in this case to create instances of Fruit, but allowed to create instances of Banana.
A call to delete of the Fruit pointer that points to an instance of Banana
will call Banana::~Banana() first and then call Fuit::~Fruit(), always.
Because no matter what, when you call a subclass destructor, the base class destructor must follow.
Is it a bad model? It is more complicated in the design phase, yes, but it can ensure that correct linking is performed at run-time and that a subclass function is performed where there is ambiguity as to exactly which subclass is being accessed.
If you write C++ so that you only pass around exact class pointers with no generic nor ambiguous pointers, then virtual functions are not really needed.
But if you require run-time flexibility of types (as in Apple Banana Orange ==> Fruit ) functions become easier and more versatile with less redundant code.
You no longer have to write a function for each type of fruit, and you know that every fruit will respond to color() with its own correct function.
I hope this long-winded explanation solidifies the concept rather than confuses things. There are a lot of good examples out there to look at,
and look at enough and actually run them and mess with them and you will get it.
You asked for an example, and I believe the following provides a reason for a pure virtual destructor. I look forward to replies as to whether this is a good reason...
I do not want anyone to be able to throw the error_base type, but the exception types error_oh_shucks and error_oh_blast have identical functionality and I don't want to write it twice. The pImpl complexity is necessary to avoid exposing std::string to my clients, and the use of std::auto_ptr necessitates the copy constructor.
The public header contains the exception specifications that will be available to the client to distinguish different types of exception being thrown by my library:
// error.h
#include <exception>
#include <memory>
class exception_string;
class error_base : public std::exception {
public:
error_base(const char* error_message);
error_base(const error_base& other);
virtual ~error_base() = 0; // Not directly usable
virtual const char* what() const;
private:
std::auto_ptr<exception_string> error_message_;
};
template<class error_type>
class error : public error_base {
public:
error(const char* error_message) : error_base(error_message) {}
error(const error& other) : error_base(other) {}
~error() {}
};
// Neither should these classes be usable
class error_oh_shucks { virtual ~error_oh_shucks() = 0; }
class error_oh_blast { virtual ~error_oh_blast() = 0; }
And here is the shared implementation:
// error.cpp
#include "error.h"
#include "exception_string.h"
error_base::error_base(const char* error_message)
: error_message_(new exception_string(error_message)) {}
error_base::error_base(const error_base& other)
: error_message_(new exception_string(other.error_message_->get())) {}
error_base::~error_base() {}
const char* error_base::what() const {
return error_message_->get();
}
The exception_string class, kept private, hides std::string from my public interface:
// exception_string.h
#include <string>
class exception_string {
public:
exception_string(const char* message) : message_(message) {}
const char* get() const { return message_.c_str(); }
private:
std::string message_;
};
My code then throws an error as:
#include "error.h"
throw error<error_oh_shucks>("That didn't work");
The use of a template for error is a little gratuitous. It saves a bit of code at the expense of requiring clients to catch errors as:
// client.cpp
#include <error.h>
try {
} catch (const error<error_oh_shucks>&) {
} catch (const error<error_oh_blast>&) {
}
Maybe there is another REAL USE-CASE of pure virtual destructor which I actually can't see in other answers :)
At first, I completely agree with marked answer: It is because forbidding pure virtual destructor would need an extra rule in language specification. But it's still not the use case that Mark is calling for :)
First imagine this:
class Printable {
virtual void print() const = 0;
// virtual destructor should be here, but not to confuse with another problem
};
and something like:
class Printer {
void queDocument(unique_ptr<Printable> doc);
void printAll();
};
Simply - we have interface Printable and some "container" holding anything with this interface. I think here it is quite clear why print() method is pure virtual. It could have some body but in case there is no default implementation, pure virtual is an ideal "implementation" (="must be provided by a descendant class").
And now imagine exactly the same except it is not for printing but for destruction:
class Destroyable {
virtual ~Destroyable() = 0;
};
And also there could be a similar container:
class PostponedDestructor {
// Queues an object to be destroyed later.
void queObjectForDestruction(unique_ptr<Destroyable> obj);
// Destroys all already queued objects.
void destroyAll();
};
It's simplified use-case from my real application. The only difference here is that "special" method (destructor) was used instead of "normal" print(). But the reason why it is pure virtual is still the same - there is no default code for the method.
A bit confusing could be the fact that there MUST be some destructor effectively and compiler actually generates an empty code for it. But from the perspective of a programmer pure virtuality still means: "I don't have any default code, it must be provided by derived classes."
I think it's no any big idea here, just more explanation that pure virtuality works really uniformly - also for destructors.
This is a decade old topic :)
Read last 5 paragraphs of Item #7 on "Effective C++" book for details, starts from "Occasionally it can be convenient to give a class a pure virtual destructor...."
we need to make destructor virtual bacause of the fact that , if we dont make the destructor virtual then compiler will only destruct the contents of base class , n all the derived classes will remain un changed , bacuse compiler will not call the destructor of any other class except the base class.