Goal:
Sharing a base class with the same functionality except only a few descendants which merely hides methods (getters and setters) as they instantiate and free a protected or private pointer.
Problem:
When re-declaring the setter only I get this error when invoking the getter.
1>c:\projectengine\problem\main.cpp(8): error C2660: 'cDerived::SomeClass'
: function does not take 0 arguments
The getter isn't really needed but why disclose functionality if it's there already.
Conclusion:
cBase::SomeClass() is untouched by cDerived.
When commenting out the next line, no compile error (of course):
virtual void SomeClass( cSomeClass* value ) override {}; // setter, C2660
However that would work, and thus not changing anything, there rises a risk for a memory leak. On the other hand, this accidental derivative should inherited the rest of the functionality of cBase which is a descendant by itself.
Should I rewrite cBase or is it possible to hide only the setter in cDerived?
The code with the error:
main.cpp
class cSomeClass
{
};
class cBase : public cAbsoluteBase
{
public:
cBase() : m_pSomeClass( 0 ){}
virtual cSomeClass* SomeClass(){ return m_pSomeClass; }
virtual void SomeClass( cSomeClass* value ){ m_pSomeClass = value; };
protected:
cSomeClass* m_pSomeClass;
};
class cDerived : public cBase
{
private:
// hide these
//virtual cSomeClass* SomeClass() override { return m_pSomeClass; };
virtual void SomeClass( cSomeClass* value ) override {};
};
cDerived derived;
int main()
{
cSomeClass* someClass = derived.SomeClass(); // C2660
return 0;
}
You need a using declaration. Otherwise, overloads in the derived class hide the functions in the base class. (Because usually this is an error. If you had to specialize one overload, usually you have to specialize all of them.)
I'm not sure I could follow your description accurately, but from the code snippets, it looks like you want to publicly inherit from a class, but remove one of the methods of the base classes interface.
That is possible, it is even possible to do accidentally, but it is a very bad idea™. Public inheritance is an “is-a” relationship. If you have a class B publicly inheriting from class A, then what declaring that relationship means is “objects of type B may be larger than those of type A, but apart from that, you can use a B wherever an A is expected.” Obviously Bs may have additional behavior, but public inheritance promises that they work as fully functional As. (This is known as the “Liskov Substitution Principle,” by the way.)
Note that this is also why one of the early textbook examples for inheritance was utterly broken: Having Square inherit from Rectangle. Yes, a square is a rectangle, but that doesn't mean it is sound OO design to have Square publicly inherit from Rectangle. Because sub-classes having additional restrictions over their parent classes breaks a lot of otherwise perfectly valid and useful code, by not adhering to the established interface contract. (In the case of squares, they break the contract clause that you should be able to change the sides of a rectangle independently from one another. In your case, you seem to be removing parts of the interface altogether.)
Related
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?.
I am confused about the concepts of inheritance and polymorphism. I mean, what is the difference between code re-usability and function overriding? Is it impossible to reuse parent class function using inheritance concept or else is it impossible to override parent class variables using Polymorphism. There seems little difference for me.
class A
{
public:
int a;
virtual void get()
{
cout<<"welcome";
}
};
class B:public A
{
a =a+1; //why it is called code reuse
void get() //why it is called overriding
{
cout<<"hi";
}
};
My doubt is about the difference between the code reuse and function overriding.
Lets start with your example.
class A
{
public:
int a;
virtual void get()
{
cout<<"welcome";
}
};
class B:public A
{
a =a+1; //why it is called code reuse
void get() //why it is called overriding
{
cout<<"hi";
}
};
Inheritance: Here you are deriving class B from class A, this means that you can access all of its public variables and method.
a = a + 1
Here you are using variable a of class A, you are reusing the variable a in class B thereby achieving code reusability.
Polymorphism deals with how a program invokes a method depending on the things it has to perform: in your example you are overriding the method get() of class A with method get() of class B. So when you create an instance of Class B and call method get you'll get 'hi' in the console not 'welcome'
Function inheritance allows for abstraction of behaviour from a "more concrete" derived class(es) to a "more abstract" base class. (This is analogous to factoring in basic math and algebra.) In this context, more abstract simply means that less details are specified. It is expected that derived classes will extend (or add to) what is specified in the base class. For example:
class CommonBase
{
public:
int getCommonProperty(void) const { return m_commonProperty; }
void setCommonProperty(int value) { m_commonProperty = value; }
private:
int m_commonProperty;
};
class Subtype1 : public CommonBase
{
// Add more specific stuff in addition to inherited stuff here...
public:
char getProperty(void) const { return m_specificProperty1; }
private:
char m_specificProperty1;
};
class Subtype2 : public CommonBase
{
// Add more specific stuff in addition to inherited stuff here...
public:
float getProperty(void) const { return m_specificProperty2; }
private:
float m_specificProperty2;
};
Note that in the above example, getCommonProperty() and setCommonProperty(int) are inherited from the CommonBase class, and can be used in instances of objects of type Subtype1 and Subtype2. So we have inheritance here, but we don't really have polymorphism yet (as will be explained below).
You may or may not want to instantiate objects of the base class, but you can still use it to collect/specify behaviour (methods) and properties (fields) that all derived classes will inherit. So with respect to code reuse, if you have more than one type of derived class that shares some common behaviour, you can specify that behaviour only once in the base class and then "reuse" that in all derived classes without having to copy it. For example, in the above code, the specifications of getCommmonProperty() and setCommonProperty(int) can be said to be reused by each Subtype# class because the methods do not need to be rewritten for each.
Polymorphism is related, but it implies more. It basically means that you can treat objects that happen to be from different classes the same way because they all happen to be derived from (extend) a common base class. For this to be really useful, the language should support virtual inheritance. That means that the function signatures can be the same across multiple derived classes (i.e., the signature is part of the common, abstract base class), but will do different things depending on specific type of object.
So modifying the above example to add to CommonBase (but keeping Subtype1 and Subtype2 the same as before):
class CommonBase
{
public:
int getCommonProperty(void) const { return m_commonProperty; }
void setCommonProperty(int value) { m_commonProperty = value; }
virtual void doSomething(void) = 0;
virtual ~CommonBase() { }
private:
int m_commonProperty;
};
Note that doSomething() is declared here as a pure virtual function in CommonBase (which means that you can never instantiate a CommonBase object directly -- it didn't have to be this way, I just did that to keep things simple). But now, if you have a pointer to a CommonBase object, which can be either a Subtype1 or a Subtype2, you can call doSomething() on it. This will do something different depending on the type of the object. This is polymorphism.
void foo(void)
{
CommonBase * pCB = new Subtype1;
pCB->doSomething();
pCB = new Subtype2;
pCB->doSomething(); // Does something different...
}
In terms of the code sample you provided in the question, the reason get() is called "overriding" is because the behaviour specified in the B::get() version of the method takes precedence over ("overrides") the behaviour specified in the A::get() version of the method if you call get() on an instance of a B object (even if you do it via an A*, because the method was declared virtual in class A).
Finally, your other comment/question about "code reuse" there doesn't quite work as you specified it (since it's not in a method), but I hope it will be clear if you refer to what I wrote above. When you are inheriting behaviour from a common base class and you only have to write the code for that behaviour once (in the base class) and then all derived classes can use it, then that can be considered a type of "code reuse".
You can have parametric polymorphism without inheritance. In C++, this is implemented using templates. Wiki article:
http://en.wikipedia.org/wiki/Polymorphism_%28computer_science%29#Parametric_polymorphism
Suppose I have a pure virtual method in the base interface that returns to me a list of something:
class base
{
public:
virtual std::list<something> get() = 0;
};
Suppose I have two classes that inherit the base class:
class A : public base
{
public:
std::list<something> get();
};
class B : public base
{
public:
std::list<something> get();
};
I want that only the A class can return a list<something>, but I need also to have the possibility to get the list using a base pointer, like for example:
base* base_ptr = new A();
base_ptr->get();
What I have to do?
Have I to return a pointer to this list? A reference?
Have I to return a null pointer from the method of class B? Or have I to throw an exception when I try to get the list using a B object? Or have I to change the base class method get, making it not pure and do this work in the base class?
Have I to do something else?
You have nothing else to do. The code you provide does exactly that.
When you get a pointer to the base class, since the method was declared in the base class, and is virtual, the actual implementation will be looked up in the class virtual function table and called appropriately.
So
base* base_ptr = new A();
base_ptr->get();
Will call A::get(). You should not return null from the implementation (well you can't, since null is not convertible to std::list< something > anyway). You have to provide an implementation in A/B since the base class method is declared pure virtual.
EDIT:
you cannot have only A return an std::list< something > and not B since B also inherits the base class, and the base class has a pure virtual method that must be overriden in the derived class. Inheriting from a base class is a "is-a" relationship. The only other way around I could see would be to inherit privately from the class, but that would prevent derived to base conversion.
If you really don't want B to have the get method, don't inherit from base.
Some alternatives are:
Throwing an exception in B::get():
You could throw an exception in B::get() but make sure you explain your rationale well as it is counter-intuitive. IMHO this is pretty bad design, and you risk confusing people using your base class. It is a leaky abstraction and is best avoided.
Separate interface:
You could break base into separate interface for that matter:
class IGetSomething
{
public:
virtual ~IGetSomething() {}
virtual std::list<something> Get() = 0;
};
class base
{
public:
// ...
};
class A : public base, public IGetSomething
{
public:
virtual std::list<something> Get()
{
// Implementation
return std::list<something>();
}
};
class B : public base
{
};
The multiple inheritance in that case is OK because IGetSomething is a pure interface (it does not have member variables or non-pure methods).
EDIT2:
Based on the comments it seems you want to be able to have a common interface between the two classes, yet be able to perform some operation that one implementation do, but the other doesn't provide. It is quite a convoluted scenario but we can take inspiration from COM (don't shoot me yet):
class base
{
public:
virtual ~base() {}
// ... common interface
// TODO: give me a better name
virtual IGetSomething *GetSomething() = 0;
};
class A : public Base
{
public:
virtual IGetSomething *GetSomething()
{
return NULL;
}
};
class B : public Base, public IGetSomething
{
public:
virtual IGetSomething *GetSomething()
{
// Derived-to-base conversion OK
return this;
}
};
Now what you can do is this:
base* base_ptr = new A();
IGetSomething *getSmthing = base_ptr->GetSomething();
if (getSmthing != NULL)
{
std::list<something> listOfSmthing = getSmthing->Get();
}
It is convoluted, but there are several advantages of this method:
You return public interfaces, not concrete implementation classes.
You use inheritance for what it's designed for.
It is hard to use mistakenly: base does not provide std::list get() because it is not a common operation between the concrete implementation.
You are explicit about the semantics of GetSomething(): it allows you to return an interface that can be use to retrieve a list of something.
What about just returning an empty std::list ?
That would be possible but bad design, it's like having a vending machine that can give Coke and Pepsi, except it never serves Pepsi; it's misleading and best avoided.
What about just returning a boost::optional< std::list< something > > ? (as suggested by Andrew)
I think that's a better solution, better than returning and interface that sometimes could be NULL and sometimes not, because then you explicitly know that it's optional, and there would be no mistake about it.
The downside is that it puts boost inside your interface, which I prefer to avoid (it's up to me to use boost, but clients of the interface shouldn't have to be forced to use boost).
return boost::optional in case you need an ability to not return (in B class)
class base
{
public:
virtual boost::optional<std::list<something> > get() = 0;
};
What you are doing is wrong. If it is not common to both the derived classes, you should probably not have it in the base class.
That aside, there is no way to achieve what you want. You have to implement the method in B also - which is precisely the meaning of a pure virtual function. However, you can add a special fail case - such as returning an empty list, or a list with one element containing a predetermined invalid value.
Consider the following code:
class Base
{
public:
virtual void Foo() {}
};
class Derived : public Base
{
private:
void Foo() {}
};
void func()
{
Base* a = new Derived;
a->Foo(); //fine, calls Derived::Foo()
Derived* b = new Derived;
// b->Foo(); //error
static_cast<Base*>(b)->Foo(); //fine, calls Derived::Foo()
}
I've heard two different schools of thought on the matter:
Leave accessibility the same as the base class, since users could use static_cast<Derived*> to gain access anyway.
Make functions as private as possible. If users require a->Foo() but not b->Foo(), then Derived::Foo should be private. It can always be made public if and when that's required.
Is there a reason to prefer one or the other?
Restricting access to a member in a subtype breaks the Liskov substitution principle (the L in SOLID). I would advice against it in general.
Update: It may "work," as in the code compiles and runs and produces the expected output, but if you are hiding a member your intention is probably making the subtype less general than the original. This is what breaks the principle. If instead your intention is to clean up the subtype interface by leaving only what's interesting to the user of the API, go ahead and do it.
Not an answer to your original question, but if we are talking about classes design...
As Alexandrescu and Sutter recommend in their 39th rule, you should prefer using public non-virtual functions and private/protected virtual:
class Base {
public:
void Foo(); // fixed API function
private:
virtual void FooImpl(); // implementation details
};
class Derived {
private:
virtual void FooImpl(); // special implementation
};
This depends on your design, whether you want to access the virtual function with a derived class object or not.
If not then yes, it's always better to make them private or protected.
There is no code execution difference based on access specifier, but the code becomes cleaner.
Once you have restricted the access of the class's virtual function; the reader of that class can be sure that this is not going to be called with any object or pointer of derived class.
In C++ why the pure virtual method mandates its compulsory overriding only to its immediate children (for object creation), but not to the grand children and so on ?
struct B {
virtual void foo () = 0;
};
struct D : B {
virtual void foo () { ... };
};
struct DD : D {
// ok! ... if 'B::foo' is not overridden; it will use 'D::foo' implicitly
};
I don't see any big deal in leaving this feature out.
For example, at language design point of view, it could have been possible that, struct DD is allowed to use D::foo only if it has some explicit statement like using D::foo;. Otherwise it has to override foo compulsory.
Is there any practical way of having this effect in C++?
I found one mechanism, where at least we are prompted to announce the overridden method explicitly. It's not the perfect way though.
Suppose, we have few pure virtual methods in the base class B:
class B {
virtual void foo () = 0;
virtual void bar (int) = 0;
};
Among them, suppose we want only foo() to be overridden by the whole hierarchy. For simplicity, we have to have a virtual base class, which contains that particular method. It has a template constructor, which just accepts the type same as that method.
class Register_foo {
virtual void foo () = 0; // declare here
template<typename T> // this matches the signature of 'foo'
Register_foo (void (T::*)()) {}
};
class B : public virtual Register_foo { // <---- virtual inheritance
virtual void bar (int) = 0;
Base () : Register_foo(&Base::foo) {} // <--- explicitly pass the function name
};
Every subsequent child class in the hierarchy would have to register a foo inside its every constructor explicitly. e.g.:
struct D : B {
D () : Register_foo(&D::foo) {}
virtual void foo () {};
};
This registration mechanism has nothing to do with the business logic. Though, the child class can choose to register using its own foo or its parent's foo or even some similar syntax method, but at least that is announced explicitly.
In your example, you have not declared D::foo pure; that is why it does not need to be overridden. If you want to require that it be overridden again, then declare it pure.
If you want to be able to instantiate D, but force any further derived classes to override foo, then you can't. However, you could derive yet another class from D that redeclares it pure, and then classes derived from that must override it again.
What you're basically asking for is to require that the most derived
class implement the functiom. And my question is: why? About the only
time I can imagine this to be relevant is a function like clone() or
another(), which returns a new instance of the same type. And that's
what you really want to enforce, that the new instance has the same
type; even there, where the function is actually implemented is
irrelevant. And you can enforce that:
class Base
{
virtual Base* doClone() const = 0;
public:
Base* clone() const
{
Base* results = doClone();
assert( typeid(*results) == typeid(*this) );
return results;
}
}
(In practice, I've never found people forgetting to override clone to
be a real problem, so I've never bothered with something like the above.
It's a generally useful technique, however, anytime you want to enforce
post-conditions.)
A pure virtual means that to be instantiated, the pure virtual must be overridden in some descendant of the class that declares the pure virtual function. That can be in the class being instantiated or any intermediate class between the base that declares the pure virtual, and the one being instantiated.
It's still possible, however, to have intermediate classes that derive from one with a pure virtual without overriding that pure virtual. Like the class that declares the pure virtual, those classes can only be used as based classes; you can't create instances of those classes, only of classes that derive from them, in which every pure virtual has been implemented.
As far as requiring that a descendant override a virtual, even if an intermediate class has already done so, the answer is no, C++ doesn't provide anything that's at least intended to do that. It almost seems like you might be able to hack something together using multiple (probably virtual) inheritance so the implementation in the intermediate class would be present but attempting to use it would be ambiguous, but I haven't thought that through enough to be sure how (or if) it would work -- and even if it did, it would only do its trick when trying to call the function in question, not just instantiate an object.
Is there any practical way of having this effect in C++?
No, and for good reason. Imagine maintenance in a large project if this were part of the standard. Some base class or intermediate base class needs to add some public interface, an abstract interface. Now, every single child and grandchild thereof would need to changed and recompiled (even if it were as simple as adding using D::foo() as you suggested), you probably see where this is heading, hells kitchen.
If you really want to enforce implementation you can force implementation of some other pure virtual in the child class(s). This can also be done using the CRTP pattern as well.