Interfaces in c++ - c++

I would like to use interfaces in c++ like in java or in c#. I decided to use purely abstract classes with multiple inheritance, but something is terribly wrong when I specialize the interface:
class Interface
{
public:
virtual int method() = 0;
};
// Default implementation.
class Base: virtual public Interface
{
public:
virtual int method() {return 27;}
};
// specialized interface
class Interface2: public Interface
{
public:
virtual int method() = 0;
// some other methods here
};
// concrete class - not specialised - OK
class Class: public virtual Interface, public virtual Base
{
};
// concrete class - specialised
class Class2: public Interface2, public Base
{
};
int main()
{
Class c;
Class2 c2;
return 0;
}
Warning 1 warning C4250: 'Class' : inherits 'Base::Base::method' via dominance 30
Error 2 error C2259: 'Class2' : cannot instantiate abstract class 42
What is the proper way to do this?

Class2 inherits from an abstract class (Interface2) but does not implement the pure virtual method, so it remains as an abstract class.

Heh heh, this problem tickles something buried deep in my head somewhere. I can't quite put my finger on it but I think it's to do with defining an interface heirarchy and then inheriting both an interface and an implementation. You then avoid having to implement all functions with by forwarding calls to a base class. I think.
I think this simple example shows the same thing, but is maybe a bit easier to understand because it uses things that can be easily visualized: (please forgive the struct laziness)
#include <iostream>
using namespace std;
struct Vehicle
{
virtual void Drive() = 0;
};
struct VehicleImp : virtual public Vehicle
{
virtual void Drive()
{
cout << "VehicleImp::Drive\n";
}
};
struct Tank : virtual public Vehicle
{
virtual void RotateTurret() = 0;
};
struct TankImp : public Tank, public VehicleImp
{
virtual void RotateTurret()
{
cout << "TankImp::RotateTurret\n";
}
// Could override Drive if we wanted
};
int _tmain(int argc, _TCHAR* argv[])
{
TankImp myTank;
myTank.Drive(); // VehicleImp::Drive
myTank.RotateTurret(); // TankImp::RotateTurret
return 0;
}
TankImp has essentially inherited the Tank interface and the Vehicle implementation.
Now, I'm pretty sure this is a well known and acceptable thing in OO circles (but I don't know if it has a fancy name), so the dreaded diamond thing is ok in this case, and you can safely suppress the dominance warning because it's what you want to happen in this case.
Hope that somehow helps point you in the right direction!
BTW, your code didn't compile because you hadn't implemented the pure virtual "method" in Class2.
EDIT:
Ok I think I understand your problem better now and I think the mistake is in Interface2. Try changing it to this:
// specialized interface
class Interface2: public virtual Interface // ADDED VIRTUAL
{
public:
//virtual int method() = 0; COMMENTED THIS OUT
// some other methods here
};
Interface2 should not have the pure virtual defintion of method, since that is already in Interface.
The inheritance of Interface needs to be virtual otherwise you will have an ambiguity with Base::method when you derive from Interface2 and Base in Class2.
Now you should find it will compile, possibly with dominance warnings, and when you call c2.method(), you get 27.

Based on this comment
If the method is not reimplemented in Class2 or Class (it is not in
this case) Base::method() will be called. Otherwise the reimplementation
will be called. There is an interface hierarchy with a common base
dumb implementation.
– danatel 16 mins ago
That's not what you got, you don't have a common base, you've got
Interface -> Interface2 -> Class2
Interface -> Base -> Class2
The interface is not 'merged' in the derivation tree, interface2 does not inherit virtually from interface, so it'll have its own interface super class.
It's like the pure virtual method() exists twice in Class2, once implemented via Class, and once not-implemented.
And even if you had inherited virtually, the common base (Interface) still would not have an implementation
If Base contains trivial operations that should be usuable in the whole hierarchy, then why not have Base as your startpoint? (even if still pure virtual with an implementation).
If this was just a very simple example to make the question short, something like the Bridge Pattern might be more usefull. But it's hard to guide you further without knowing more.

You should also look at defining a virtual destructor in your Interface if you might be deleting using an Interface or Base pointer.
Without a virtual destructor you will have problems if you do something like:
Base *b = new Class2();
delete b;

Regarding Class: All you need to do is derive Class from Base -- the fact that it implements Interface is implied, and in fact, inescapable:
class Class: public Base // virtual inheritance is unnecessary here
{
};
Class will inherit method() from Base as desired.
Regarding Class2:
Disclaimer: Negative result ahead
Based on your comment on Tom's answer, I thought I had the answer for Class2:
// concrete class - specialised
class Class2: public Interface2, public Base
{
public:
using Base::method; // "Imports" all members named "method" from Base
};
But actually, this doesn't work. Grovelling through the C++ standard reveals that
section 7.3.3, paragraph 14 explains that using can't be used to resolve ambiguous accesses to inherited members:
... [Note: because a using-declaration designates a base class member (and not a member subobject or a member function of a base class subobject), a using-declaration cannot be used to resolve inherited member ambiguities. ...]
It seems that the only way to get the desired behaviour in Class2 is to manually forward the method:
// concrete class - specialised
class Class2: public Interface2, public Base
{
public:
virtual int method() { return Base::method(); }
};
Regarding virtual inheritance: You don't need it for Class's declaration, but you probably do need it for Interface2's declaration to ensure that Class2 only has a single subobject of type Interface -- as it stands, every Class2 object has two subobjects of this type. (Although that won't cause problems if Interface is in fact a pure interface, lacking member variables.) If it helps, draw a diagram: every time a base class appears without the keyword virtual, it appears as a distinct object; all base classes that appear with the keyword virtual are condensed into one object.
[UPDATE: markh44's excellent answer shows that the above approach (of making Interface2 inherit virtually from Interface) will in fact allow Class2 to automatically inherit the implementation of method() from Base! Problem solved!]

This answer in a different forum seems to tackle the exact problem you mention.

In general, you should avoid the diamond inhertance pattern:
Interface
/ \
Base Interface2
\ /
Class2
This will cause you call kinds of grief down the road if you're not careful. Ambiguity will bite you.
In your specific instance there's no need for Interface2 to inherit from Interface. Interface2 doesn't need to specify "method" since it's abstract. Remove the inheritance between Interface and Interface2 to break the diamond. Then you're hierarchy looks like:
Interface Interface Interface2
| | |
Base Base |
| \ /
Class Class2
And your implementation looks like:
// concrete class - not specialised - OK
class Class: public Base
{
};
// concrete class - specialised
class Class2: public Base, public Interface2
{
virtual int method() {return 35;}
virtual void Inteface2Method { ... }
};

Related

create an abstract class from which bass class will be derived c++

I have to make a base class with one derived class in C++. In the end I need to evidence polymorphism between 2 methodes void Read and void Show.
After I did this, I have to create an abstract class Object from which the base class will be derived.This class will contain just pure virtual methods Read and Show.
I don't think I fully understand this. So I make an abstract class Object where I put these 2 methods with "=0" to mke them pure. But this means that I have to edit the base class also, to make it derived from class Object? Like, before:
class Base {
}
after
Class Base : public Object
Can someone make me understand?
You're on the right track. In your Base class, you can mark methods virtual that you intend to be polymorphic. You can set the functions = 0 if they are "pure virtual" and have no implementation in the base class, and are required in any concrete derived class.
class Base
{
public:
virtual void Read() = 0;
virtual void Show() = 0;
};
Then for your derived class you had the syntax backwards. Note that the override keyword indicates that you are overriding a virtual method from the base class. This keyword enforces that the function signature matches the one from a base class.
class Object : public Base
{
public:
void Read() override; // implement this
void Show() override; // implement this
};
The usage could look like the following. Note that you instantiate an Object but carry a pointer to a Base*. Due to polymorphism, the derived implementations are invoked.
int main()
{
Object obj;
Base* p = &obj;
p->Read(); // due to polymorphism will call Object::Read
p->Show(); // due to polymorphism will call Object::Show
}

C++: Are common ancestry and an interface/prototype mutually exclusive features of a C++ class?

In C++, does inheriting a common ancestor and inheriting an interface (and requiring definition of a method in derived classes) require multiple inheritance? Eg. do I have to do the following (instead of merging MyInterface and ParentClass):
class MyInterface;
class ParentClass;
class DerivedClass1;
class DerivedClass2;
class SomeOtherType;
class YetAnotherType;
class MyInterface {
public:
// Must be defined in all derived classes
virtual SomeOtherType my_common_fxn(...) = 0;
...
};
class ParentClass {
private:
// Common ancestor
YetAnotherType _useful_member;
}
class DerivedClass1 : MyInterface, ParentClass {
public:
// Do some things with _useful_member, using approach #1
SomeOtherType my_common_fxn(...);
...
}
class DerivedClass2 : MyInterface, ParentClass {
public:
// Do some things with _useful_member, using approach #2
SomeOtherType my_common_fxn(...);
...
}
void fxn_or_method_using(ParentClass);
Is it possible to (elegantly) merge the functionality of MyInterface and ParentClass into a single class? (I believe that as MyInterface is an ABC I cannot use this type as a parameter to fxn_or_method_using.)
Apologies in advance if this is a duplicate- I've searched but none of the existing C++ questions appeared to line up. Q's and/or A's may have been over my (untrained) head.
There's nothing wrong with your inheritance model.
But in C++ you need a pointer or reference for polymorphism. Your fxn_or_method_using takes its parameter by value. That has several problems. It causes slicing, it prevents polymorphic function calls, and it can't work for an abstract type because you can't create instances of those.
If you change fxn_or_method_using to take its parameter by reference not value, then you can declare it as referring to MyInterface if you wish. All the disadvantages disappear and you get the polymorphic behaviour you want.
No. You can mix virtual and pure virtual and concrete inheritance all from the same class in C++ with no problems.
class baseClass{
public:
blah1(){
//stuff
}
virtual blah2();
virtual blah3() = 0;
};
class derivedClass : baseClass
{
};

Virtual member functions in most-derived class?

Consider the following base class:
class Base
{
public:
virtual ~Base(void);
virtual void foo(void);
virtual void bar(void) = 0;
}
Now suppose I know that a given class should be the most derived class of Base. Should I declare the functions virtual? The most derived class can/will be used polymorphically with Base.
For example, should I use MostDerived1 or MostDerived2?
class MostDerived1 : public Base
{
public:
~MostDerived1(void);
void foo(void);
void bar(void);
}
class MostDerived2 : public Base
{
public:
virtual ~MostDerived2(void);
virtual void foo(void);
virtual void bar(void);
}
I'm leaning towards MostDerived1 because it most closely models the intent of the programmer: I don't want another child class of MostDerived1 to be used polymorphically with MostDerived1.
Is this reasoning correct? Are there any good reasons why I should pick MostDerived2, aside from the obvious there could be a >0% chance MostDerived2 should be used polymorphically with any deriving classes (class OriginalAssumptionWrong : public MostDerived2)?
Keep in mind MostDerived1/MostDerived2 can both be used polymorphically with Base.
Adding virtual to derived classes doesn't change their behavior, MostDerived and MostDerived2 are have exactly the same behavior.
It does however document your intention, however. I would recommend it for that purpose. The override keyword also helps with this, assuming its available on your platform.
You can't turn off virtualness. Another class derived from either MostDerived1 or MostDerived2 can also override any of the virtual functions regardless of whether you omit the virtual keyword somewhere in the class hierarchy or not.
If you want to enforce that no other class derives from MostDerived1, define it as
class MostDerived1 final : public Base
{
// ...
};
The final keyword can also be used for individual virtual member functions, ensuring no derived class overrides that specific function.
Once function declear somewhere at the hierarchy as a virtual, it's virtual for ever.
You can use final or override if you using C++11

Can an abstract class be implemented without pure virtual functions in C++?

I thought of using protected constructor, but it couldn't completely solve the purpose since the class inheriting from it would be able to instantiate the base class.
As for private constructor, the derived classes too would not be instantiated.
So, any suitable technique would be appreciated.
It is unclear what you are really asking for. So let me try to clear some points:
Pure virtual functions can have definitions
If your concern is that you want to provide definitions for all of the virtual functions in your base you can provide the definitions for the pure virtual functions, and they will be available for static dispatch.
Protected grants access to your base subobject, not to every instance of base
There is a common misconception that protected allows a particular derived type accessing any instance of base. That is not true. The keyword protected grants access to the base subobject within the derived type.
class base {
protected: base() {}
};
class derived : public base {
derived() : base() { // fine our subobject
base b; // error, `b` is not your subobject
}
};
The definition of an abstract class is one that has at least one pure virtual function (virtual function-signature = 0; You can't create an abstract class without them.
Can an abstract class be implemented without pure virtual functions in C++?
If you choose the point of view from Static Polymorphism, you can do that!
An abstract base class would be simply missing a default method implementation for an interface method from the deriving class.
Additionally you can use protected constructors for those CRTP base class templates, to require inheritance for instantiation.
UPDATE:
I found a nice slide show, that explains static vs dynamic polymorphism comprehensively. Each technique has it's pros and cons and certain fields of usage, additionally you can mix both techniques (wisely of course).
To elaborate a bit, I'll give a sample:
template<class Derived>
class AbstractBase
{
public:
// Equivalent for a pure virtual function
void foo()
{
// static_cast<> enforces an 'Is a' relation from Derived to AbstractBase
static_cast<Derived*>(this)->fooImpl();
}
// Equivalent for simple virtual function (overidable from Derived)
void bar()
{
static_cast<Derived*>(this)->barImpl();
}
// Default implementation for any call to bar()
void barImpl()
{
}
protected:
AbstractBase() {}
};
// Compilation will fail, since ConcreteClass1 doesn't provide
// a declaration for fooImpl()
class ConcreteClass1
: public AbstractBase<ConcreteClass1>
{
}
// Compiles fine
class ConcreteClass2
: public AbstractBase<ConcreteClass2>
{
public:
void fooImpl()
{
// Concrete implementation ...
}
}
The following sample shows that the pattern introduced above enforces an 'Is a' relationship between abstract class and inheriting class (the template parameter)
class ConcreteClass3
{
public:
void fooImpl()
{
// Concrete implementation ...
}
}
// Instantiation will fail, because
// * the constructor is protected
// * at least the static cast will fail
AbstractBase<ConcreteClass3> instance;
I read it in my book
An abstract class is a class that is designed to be specifically used as a base class. An abstract class contains at least one pure virtual function. You declare a pure virtual function by using a pure specifier (= 0) in the declaration of a virtual member function in the class declaration.

Prevent class inheritance in C++

Recently one of my friend asked me how to prevent class inheritance in C++. He wanted the compilation to fail.
I was thinking about it and found 3 answers. Not sure which is the best one.
1) Private Constructor(s)
class CBase
{
public:
static CBase* CreateInstance()
{
CBase* b1 = new CBase();
return b1;
}
private:
CBase() { }
CBase(CBase3) { }
CBase& operator=(CBase&) { }
};
2) Using CSealed base class, private ctor & virtual inheritance
class CSealed
{
private:
CSealed() {
}
friend class CBase;
};
class CBase : virtual CSealed
{
public:
CBase() {
}
};
3) Using a CSealed base class, protected ctor & virtual inheritance
class CSealed
{
protected:
CSealed() {
}
};
class CBase : virtual CSealed
{
public:
CBase() {
}
};
All the above methods make sure that CBase class cannot be inherited further.
My Question is:
Which is the best method ? Any other methods available ?
Method 2 & 3 will not work unless the CSealed class is inherited virutally. Why is that ? Does it have anything to do with vdisp ptr ??
PS:
The above program was compiled in MS C++ compiler (Visual Studio).
reference : http://www.codeguru.com/forum/archive/index.php/t-321146.html
As of C++11, you can add the final keyword to your class, eg
class CBase final
{
...
The main reason I can see for wanting to do this (and the reason I came looking for this question) is to mark a class as non subclassable so you can safely use a non-virtual destructor and avoid a vtable altogether.
You can't prevent inheritance (before C++11's final keyword) - you can only prevent instantiation of inherited classes. In other words, there is no way of preventing:
class A { ... };
class B : public A { ... };
The best you can do is prevent objects of type B from being instantiated. That being the case, I suggest you take kts's advice and document the fact that A (or whatever) is not intended to be used for inheritance, give it a non-virtual destructor, and no other virtual functions, and leave it at that.
You are going through contortions to prevent further subclassing. Why? Document the fact that the class isn't extensible and make the dtor non-virtual. In the spirit of c, if someone really wants to ignore the way you intended this to be used why stop them? (I never saw the point of final classes/methods in java either).
//Note: this class is not designed to be extended. (Hence the non-virtual dtor)
struct DontExtened
{
DontExtened();
/*NOT VIRTUAL*/
~DontExtened();
...
};
1) is a matter of taste. If I see it correctly, your more fancy 2nd and 3rd solutions move the error in certain circumstances from link time to compile time, which in general should be better.
2) Virtual inheritance is needed to force the responsibility to initialize the (virtual) base class to the most derived class from where the base class ctor is no longer reachable.
To answer your question, you can't inherit from CBase because in virtual inheritance a derived class would need to have direct access to the class from which it was inherited virtually. In this case, a class that would derive from CBase would need to have direct access to CSealed which it can't since the constructor is private.
Though I don't see the usefulness of it all (ie: stopping inheritance) you can generalize using templates (I don't think it compiles on all compilers but it does with MSVC)
template<class T>
class CSealed
{
friend T; // Don't do friend class T because it won't compile
CSealed() {}
};
class CBase : private virtual CSealed<CBase>
{
};
If you can, I'd go for the first option (private constructor). The reason is that pretty much any experienced C++ programmer will see that at a glance and be able to recognize that you are trying to prevent subclassing.
There might be other more tricky methods to prevent subclassing, but in this case the simpler the better.
From C++11 onward, there is a clean solution that I am surprise not to see here. We can make the class final preventing any further inheritance.
class Foo final {};
class Bar: public Foo {}; // Fails to compile as Foo is marked as final
class myclass;
class my_lock {
friend class myclass;
private:
my_lock() {}
my_lock(const my_lock&) {}
};
class myclass : public virtual my_lock {
// ...
public:
myclass();
myclass(char*);
// ...
};
myclass m;
class Der : public myclass { };
Der dd; // error Der::dd() cannot access
// my_lock::my_lock(): private member
I found it here to give credit. I am posting here just other people can easily access
http://www.devx.com/tips/Tip/38482
To elaborate on Francis' answer: if class Bottom derives from class Middle, which virtually inherits from class Top, it is that most derived class (Bottom) that is responsible for constructing the virtually inherited base class (Top). Otherwise, in the multiple-inheritance/diamond-of-death scenario (where virtual inheritance is classically used), the compiler wouldn't know which of the two "middle" classes should construct the single base class. The Middle's constructor's call to the Top's constructor is therefore ignored when Middle is being constructed from Bottom:
class Top {
public:
Top() {}
}
class Middle: virtual public Top {
public:
Middle(): Top() {} // Top() is ignored if Middle constructed through Bottom()
}
class Bottom: public Middle {
public:
Bottom(): Middle(), Top() {}
}
So, in the the approach 2) or 3) in your question, Bottom() can't call Top() because it's inherited privately (by default, like in your code, but it's worth making it explicit) in Middle and thus is not visible in Bottom. (source)