Is it dangerous to create pure virtual function of a virtual function? - c++

Let us suppose that we have an abstract class NonEditableSuperBase from which we create another abstract class MyBase.
The first class NonEditableSuperBase has a virtual function (non pure virtual). However, I want to force that if someone creates a class that derives from MyBase, he/she must provide an implementation to the mentioned function.
Hence, my idea is to define the function as pure virtual in MyBase.
My question: Is it a bad idea given that it was just virtual in NonEditableSuperBase?
Example:
//NonEditableSuperBase.h
class NonEditableSuperBase
{
...
public:
virtual int someMethod(); //It has an implementation, suppose {return 42;}
};
//MyBase.h
class MyBase: public NonEditableSuperBase
{
public:
explicit MyBase();
virtual ~MyBase() = default;
virtual int someMethod() = 0; //I make it pure virtual
};
//MyBase.cpp
MyBase::MyBase() : NonEditableSuperBase() { }
//Now someone creates a derived class from MyBase.
class SuperDerived : public MyBase
{
public:
explicit SuperDerived();
int someMethod(); //The user must create an implementation of the function
};
Update:
As an example, in my case I want to create some derived classes from the QAbstractTableModel class of the Qt framework. To reuse some code I want to create an intermediate abstract class.
QAbstractTableModel <- MyAbstractModel <- MyModelA (or MyModelB ... etc).
However, I want to ensure that the models (MyModelA, MyModelB) re-implement some of the virtual functions of QAbstractTableModel (like the ::index() function) because some of the additional methods of MyAbstractModel requires specific implementations of the primer functions.

From ISO IEC 14882 2014:
§ 10.4 says:
5 [ Note: An abstract class can be derived from a class that is not abstract, and a pure virtual function may
override a virtual function which is not pure. —end note ]
So It's perfectly possible to do that.
Example of use case:
You can have a basic type, that is implemented as a concrete type (base class). Now, for subtypes, we might be in a need of further additional information. So, we can have an abstract intermediate object to meet our needs.

[I'm assuming that the intent was for MyBase to publicly derive from NonEditableSuperBase, which is not what the code sample in the question actually does.]
It doesn't seem inherently dangerous, but consider that a class like SuperDerived that derives from MyBase could explicitly choose to use the NonEditableSuperBase implementation.
class SuperDerived : public MyBase {
public:
using NonEditableSuperBase::someMethod;
// Or, explicitly:
// int someMethod() override { return NonEditableSuperBase::someMethod(); }
};
This satisfies the pure-virtual requirement imposed by MyBase but acts exactly as if MyBase didn't have that requirement. You've made the author of SuperDerived do something, but you haven't actually prevented them from using the ultimate base class's implementation.

Related

How to enforce overriding a virtual function in subclasses

Let's imagine I have the following code:
class Base {
virtual void myFunc();
};
class Subclass1 : public Base { // OK
virtual void myFunc() override; // overrides the base function
};
class Subclass2 : public Base { // How to make this fail ?
// does not declare an override for myFunc
};
Is there any way in C++ to make compilation fail because Subclass2 does not implement its variant of myFunc() ? I know how to make the opposite, that is to make compilation fail because Subclass1 overrides myFunc() (by using final), but I'm not aware of any mechanic to achieve what I want in C++11.
I'm curious about possibilities in later C++ standards as well.
You can force one subclass level to override a method, by making it 'pure' (virtual method()=0;), but you cannot force multiple levels to each override it again - and that should really not be your concern.
In other words, if you provide a base (or derived) class to an enduser, you can force him to override (define) a method in his derived class. If that user wants to derive from his derived class again, it’s his business to decide if the next (further) derived class needs a different override or not.

Why in C++ 'virtual' and '=0' is both needed to describe a method is abstract?

As it is explained in The C++ programming language:
virtual void push(char c) = 0;
virtual void pop() = 0;
The word virtual means 'may be redefined later in a class derived from this one'
The =0 syntax says that some class derived from Stack must define the function.
So why =0 symbol is needed? Does it means that a derived class must define this function, and that's to say when there is no =0, some derived classes are not forced to define this method?
I'm confusing about this, need some help.
Your thoughts were right.
So why =0 symbol is needed? Does it means that a child class must
define this function, and that's to say when there is no =0, some
child classes are not forced to define this method?
Basically you can:
Make a method non-virtual
This doesn't allow any class deriving from the class that implements the method (through either public or protected) to change the method's behavior.
Make a method virtual
This allows (but doesn't enforce) any class deriving from the class that implements the method (through either public or protected) to change the behavior of the method in the base class. You don't even have to call the original base class method anymore so you can make severe changes if needed.
Make a method pure virtual ( virtual = 0 )
This enforces that any class deriving from the class that implements the method (through either public or protected) to implement some kind of behavior/body for this method. If the deriving class does not provide an implementation then this class will instantly become abstract itself. This allows to omit the behavior/body of the method in the base class and because of this it is not allowed to directly instantiate a class that has one or more pure virtual methods (abstract class).
So why =0 symbol is needed?
Consider the following:
struct foo
{
virtual void some() const { cout << "foo" << endl; }
};
struct bar : public foo
{
virtual void some() const { cout << "bar << endl; }
};
struct baz : public foo
{
}
Suppose you have a pointer foo *p pointing to some object, and you call p->some().
If p points to a bar object, it will print "bar".
If p points to a baz object, it will print "foo".
In some cases, this might not be what you want. You might want to specify that any derived class needs to override it. The =0 does that.
The purpose of an abstract class (those classes have pure virtual method, =0) is to provide an appropriate base class from which other classes can inherit. Abstract classes cannot be used to instantiate objects and serves only as an interface.
Thus, if a subclass of an abstract class needs to be instantiated, it has to implement each of the virtual functions, which means that it supports the interface declared by the abstract class.
That is the base concept for interface.
In short, it a way to be sure a derivative class will implement those methods from the base class.
So why =0 symbol is needed?
Virtual function with sequence = 0 is known as pure virtual function, (the sequence = 0 is known as pure-specifier), it makes the class an abstract class, which can't be instantiated. For the derived classes, if they want to make it possible to be instantiated, they have to implement the pure virtual function.
No objects of an abstract class can be created. Abstract types cannot be used as parameter types, as function return types, or as the type of an explicit conversion. Pointers and references to an abstract class can be declared.
For example,
class Stack {
virtual void push(char c) = 0;
};
...
Stack s; // Fail, Stack is an abstract class
and if
class Stack {
virtual void push(char c);
};
...
Stack s; // Fine, if you won't call `push()` on it.
First consider what is the reason to use virtual methods.
1.Interface(polymorphism).
Interface in C++ is a pure virtual class, which means, that all of its methods are pure virtual (like these two you've mentioned above), it has an virtual destructor and has no constructor (it's obvious, because we can't create its instances). It also shouldn't have any data.
Let's define an interface (pure abstract class in C++):
class Interface
{
public:
virtual ~Interface(){}
virtual void somePublicMethod() = 0;
};
And define the class, which is an implementation of interface:
class Implementation : public Interface
{
public:
~Implementation() override {}
void somePublicMethod() override {}
};
If you define another pure virtual method in interface:
virtual void anotherPublicMethod() = 0;
And you don't override it in implementation you will receive compilation error, when you will declare object of Implementation type, because real Implementation object must have definitions (bodies) for all derived methods.
Yo can also define default behavior of some interface's method:
void Interface::somePublicMethod()
{
//define default behavior here
}
And call it in derived class:
void Implementation::somePublicMethod()
{
Interface::somePublicMethod();
}
How interface is used in polymorphism you will read in other topics.
2."Ordinary" Inheritance.
In "ordinary" inheritance you should use virtual methods instead of pure virtual ones, because you want to have instances of both Base class and Derived class(es). Virtual methods only indicates, that they could be overridden in derived class(es).
Conclusion
In general, if you want to have an instance of any class, this class must have all methods defined (so the must have bodies and it's obvious that they mustn't be pure virtual).

Can a C++ class without any pure virtual methods be considered abstract?

My understanding is that abstract classes must have one or more pure virtual methods.
Can this class be considered abstract?
class B {
protected:
B() { }
public:
virtual ~B() { }
};
Finally, is the term abstract class defined in any of the recent C++ standards?
No, such a class cannot be considered abstract because (as mentioned in the comments, excerpt from the working draft):
A class is abstract if it has at least one pure virtual function.
Unfortunately, there are cases when one cannot add a pure virtual method to a class to turn it in an abstract one and still he doesn't want users to be able to instantiate that class.
For the sake of curiosity, I'm adding this answer to mention an often unknown technique to work around the issue.
Actually, you can easily turn such a class in an abstract one, even if you don't have any virtual method to be added to it.
The basic idea is to exploit the destructor declaration to do that.
A minimal, (not) working example follows:
struct B { virtual ~B() = 0; };
// keep in mind the ODR
B::~B() { }
int main() { B b{}; }
The code above won't compile with the error:
cannot declare variable 'b' to be of abstract type 'B'
Please, note that the definition of the destructor should be placed in a .cpp file, so as not to violate the ODR.
To be honest, I haven't found any case in which this technique can be used till now. Anyway, it's worth mentioning it for future readers.
abstract classes must have one or more pure virtual methods.
Exactly, and you class don't have it.
In accordance with this, a abstract class is a type which cannot be instantiated, but can be used as a base class (note: not "by"). In C++ this can be achieved with the usage of pure virtual method.
A pure virtual method is a virtual function whose declarator has the following syntax:
class B {
virtual void foo() = 0;
}
Note: the syntax = 0 which indicates a pure virtual method. That simply means you don't have to specify an implementation for that method, and it cannot be possible to create any instance of that class (that is, a abstract class).
In conclusion your class B is not an abstract class.
Finally, is the term abstract class defined in any of the recent C++ standards?
The abstract class is a definition itself, and it's define as I've just mentioned before.
If you mean a specific defined syntax as, for example in Java (abstract class ...), then the answer is no. Again an abstract class in C++ is defined just with a class which has a pure virtual method.
No, class B can not be considered as abstract.
class A {
public:
virtual void method() = 0;
virtual ~A() = 0;
}
A is pure virtual, you cannot create object of A. You must create children class B which implements method after A.

C++: where to place constructor in inheritance chain?

I have following dilemma:
I have a full abstract class. Each inheriting class will need 3 same parameters. Each of them will additionally need other specific parameters.
I could:
1) implement a common constructor for initializing 3 common parameters in my base class, but then I have to make non-abstract getters for corresponding fields (they are private).
OR
2) leave my base class abstract and implement constructors in inherited classes, but then I have to make it in each class fields for common parameters.
Which is a better approach? I don't want to use protected members.
An abstract class is one who has at least one pure virtual (or, as you call it, abstract) function. Having non-abstract, non-virtual functions does not change the fact that your class is abstract as long as it has at least one pure virtual function. Go for having the common functionality in your base class, even if it is abstract.
One way to avoid code duplication without polluting your abstract interface with data members, is by introducing an additional level of inheritance:
// Only pure virtual functions here
class Interface {
public:
virtual void foo() = 0;
};
// Things shared between implementations
class AbstractBase : public Interface {
};
class ImplementationA : public AbstractBase {
};
class ImplementationB : public AbstractBase {
};
If your class looks like this, a pure abstract class:
class IFoo {
public:
virtual void doThings() = 0;
}
class Foo {
public:
Foo(std::string str);
void doThings() override;
}
The value your inheritance has is to provide you with the oppurtunity to subsitute Foo with another at runtime, but hiding concrete implementations behind interfaces. You can't use that advantage with Constructors, there's no such thing as a virtual constructor (that's why things like the Abstract Factory Pattern exist). All your implementations of Foo take a std::string and all your implementations of doThings use that string? Great, but that's a coincidence not a contract and doesn't belong in IFoo.
Lets talk about if you've created concrete implementations in IFoo, so that it's a abstract class and not a pure abstract class (IFoo would be a bad name now btw). (*1) Lets assume using inheritance to share behaviour was the correct choice for you class, which is sometimes true. If the class has fields that need to be initialised create a protected constructor (to be called from every child implementation) and remove/ommit the default one.
class BaseFoo {
private:
std::string _fooo;
protected:
BaseFoo(std::string fooo) : _fooo(fooo) {}
public:
virtual void doThings() = 0;
std::string whatsTheBaseString() { return _fooo;}
}
Above is the way you correctly pass fields needed by a base class from the child constructor. This is a compile time guarantee that a child class will 'remember' to initialize _fooo correctly and avoids exposing the actual member fooo to child classes. Trying to initialize _fooo in all the child constructors directly would be incorrect here.
*1) Quickly, why? Composition may be a better tool here?.

Ambiguity in multiple inheritance of interfaces in C++

I made a test code as following:
#include <iostream>
using namespace std;
#ifndef interface
#define interface struct
#endif
interface Base
{
virtual void funcBase() = 0;
};
interface Derived1 : public Base
{
virtual void funcDerived1() = 0;
};
interface Derived2 : public Base
{
virtual void funcDerived2() = 0;
};
interface DDerived : public Derived1, public Derived2
{
virtual void funcDDerived() = 0;
};
class Implementation : public DDerived
{
public:
void funcBase() { cout << "base" << endl; }
void funcDerived1() { cout << "derived1" << endl; }
void funcDerived2() { cout << "derived2" << endl; }
void funcDDerived() { cout << "dderived" << endl; }
};
int main()
{
DDerived *pObject = new Implementation;
pObject->funcBase();
return 0;
}
The reason I wrote this code is to test if the function funcBase() can be called in an instance of DDerived or not. My C++ complier (Visual Studio 2010) gave me a compile error message when I tried to compile this code. In my opinion, there is no problem in this code because it is certain that the function funcBase() will be implemented (thus overriden) in some derived class of the interface DDerived, because it is pure virtual. In other words, any pointer variable of type Implementation * should be associated with an instance of a class deriving Implentation and overriding the function funcBase().
My question is, why the compiler give me such an error message? Why the C++ syntax is defined like that; i.e., to treat this case as an error? How can I make the code runs? I want to allow multiple inheritance of interfaces. Of course, if I use "virtual public" or re-declare the function funcBase() in Implementation like
interface DDerived : public Derived1, public Derived2
{
virtual void funcBase() = 0;
virtual void funcDDerived() = 0;
};
then everything runs with no problem.
But I don't want to do that and looking for more convenient method, because virtual inheritance may degrade the performance, and re-declaration is so tedious to do if inheritance relations of classes are very complex. Is there any methods to enable multiple inheritance of interfaces in C++ other than using virtual inheritance?
As you've defined it, your object structure looks like this:
The important point here is that each instance of Implementation contains two entirely separate instances of Base. You're providing an override of Base::funcBase, but it doesn't know whether you're trying to override funcBase for the Base you inherited through Derived1, or the Base you inherited through Derived2.
Yes, the clean way to deal with this is virtual inheritance. This will change your structure so there's only one instance of Base:
This is almost undoubtedly what you really want. Yes, it got a reputation for performance problems in the days of primitive compilers and 25 MHz 486's and such. With a modern compiler and processor, you're unlikely to encounter a problem.
Another possibility would be some sort of template-based alternative, but that tends to pervade the rest of your code -- i.e., instead of passing a Base *, you write a template that will work with anything that provides functions A, B, and C, then pass (the equivalent of) Implementation as a template parameter.
The C++ language is designed in such a way that in your first approach without virtual inheritance there will be two parent copies of the method and it can't figure out which one to call.
Virtual inheritance is the C++ solution to inheriting the same function from multiple bases, so I would suggest just using that approach.
Alternately have you considered just not inheriting the same function from multiple bases? Do you really have a derived class that you need to be able to treat as Derived1 or Derived2 OR Base depending on the context?
In this case elaborating on a concrete problem rather than a contrived example may help provide a better design.
DDerived *pObject = new Implementation;
pObject->funcBase();
This creates a pointer of type DDerived to a Implementation. When you are using DDerived you really just have a pointer to an interface.
DDerived does not know about the implementation of funcBase because of the ambiguity of having funcBase being defined in both Derived1 and Derived2.
This has created a inheritance diamond which is what is really causing the problem.
http://en.wikipedia.org/wiki/Diamond_problem
I also had to check on the interface "keyword" you have in there
it's an ms-specific extension that's recognised by visual studio
I think C++ Standard 10.1.4 - 10.1.5 can help you to understand the problem in your code.
class L { public: int next; /∗ ... ∗/ };
class A : public L { /∗...∗/ };
class B : public L { /∗...∗/ };
class C : public A, public B { void f(); /∗ ... ∗/ };
10.1.4 A base class specifier that does not contain the keyword virtual,
specifies a non-virtual base class. A base class specifier that
contains the keyword virtual, specifies a virtual base class. For each
distinct occurrence of a non-virtual base class in the class lattice
of the most derived class, the most derived object (1.8) shall contain
a corresponding distinct base class subobject of that type. For each
distinct base class that is specified virtual, the most derived object
shall contain a single base class subobject of that type. [ Example:
for an object of class type C, each distinct occurrence of a
(non-virtual) base class L in the class lattice of C corresponds
one-to-one with a distinct L subobject within the object of type C.
Given the class C defined above, an object of class C will have two
subobjects of class L as shown below.
10.1.5 In such lattices, explicit qualification can be used to specify which
subobject is meant. The body of function C::f could refer to the
member next of each L subobject: void C::f() { A::next = B::next; } //
well-formed. Without the A:: or B:: qualifiers, the definition of C::f
above would be ill-formed because of ambiguity
So just add qualifiers when calling pObject->funcBase() or solve ambiguity in another way.
pObject->Derived1::funcBase();
Updated: Also very helpful reading will be 10.3 Virtual Functions of Standard.
Have a nice weekend :)