Sometimes I accidentally forget to call the superclass's method in C++ when I override a method.
Is there any way to help figure out when I'm overriding a method with, so that I don't forget to call the superclass's method? (Something like Java's #Override, except that C++ doesn't have annotations...)
One suggestion is the Non-Virtual Inferface Idiom. I.e., make your public methods non-virtual and have them call private or protected virtual methods that derived classes can override to implement their specific behavior.
If you don't have control over the base class, you could perhaps use an intermediate class:
class Foo // Don't control this one
{
public:
virtual void action();
};
class Bar : public Foo // Intermediate base class
{
public:
virtual void action()
{
doAction();
Foo::action();
}
protected:
virtual void doAction() = 0;
};
Derive your classes from Bar and override doAction() on each. You could even have doBeforeAction() and doAfterAction() if necessary.
With regards to Java's #Override, there is a direct equivalent in C++11, namely the override special identifier.
Sadly, neither #Override nor override solve the problem since: (a) they're optional; (b) the responsibility of calling the base class's method still rests with the programmer.
Furthermore, I don't know of any widely available method that would address the problem (it's quite tricky, esp. given that you don't necessarily want to call the base class's method -- how is the machine to know?).
Unfortunately Í'm not aware of a common mechanism to do this.
In C++ if you're needing to use the base class's functionality in addition to added child functionality you should look at the template method pattern. This way the common logic always lives in the base class and there's no way to forget to execute it, and you override in the child only the piece you need to change.
Related
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.
Building a GUI system and I have a few classes for different GUI components that derive from a base "GUIcontrol" class. What I want is to have just one function to return any type of component but be able to work with the functions specific to that component type (functions of the derived class). I noticed that the polymorphism approach is going to become a problem I have to declare all the derived functions in the base which is unnecessary for this, since I will never create an object just from the base class.
class GUIcontrol {
protected:
std::string _name;
// these two methods (along with name()) will be used by all types
virtual void position(/*parameters*/)
virtual void useImage(/*parameters*/)
// these should be only in derived types
virtual void setHotSpot(/*parameters*/);
virtual void setScrollButtons(/*parameters*/);
public:
std::string name();
/*etc*/
}
class GUIbutton : public GUIcontrol {
public:
void setHotSpot(/*parameters*/);
}
class GUIscrollBar : public GUIcontrol {
public:
void setScrollButtons(/*parameters*/);
}
GUIcontrol* GUIsystem::getControl(std::string name);
The problem with this is that if I want to add more functions unique to GUIbutton or GUIscrollBar, or any functions to other derived GUI classes, I also have to declare them virtual in the base class so the compiler doesn't complain about something like "setHotSpot" not being a member of the base class it returns.
The base class does have member functions that will apply to all the derived classes, such as telling the object where it should be positioned, what image it needs to use, what it should be called, etc. But I don't want to keep stuffing the base class with other functions that need to stay exclusive to certain derived classes.
As I keep adding more virtual functions I would end up with a huge blob object for the base class. Can I design this in a cleaner way? Note that I am still not sure if I want to use static_cast/dynamic_cast for getControl() to solve this but just want to know if there are any other ways around this to clean it up.
The base class should only contain methods for functionality common to all controls.
If you're going to use functionality that only makes sense for one type of control, you should be checking that the control is of the correct type anyway, and can then cast it to that type.
The base class is exclusively common functionality. If you want your method to behave differently for different controls, use dynamic_cast. If you want it to act the same for all controls, use a virtual method.
This is your problem:
What I want is to have just one
function to return any type of
component but be able to work with the
functions specific to that component
type (functions of the derived class).
What you want is to treat them the same but differently. Huh. I wonder how you're going to make that work. You need to decide if you want to treat them all the same, or if you want to treat them differently.
Type checking and then downcasting isn't the right way to do this. What you should be doing is placing generic methods onto your base class which perform the types of operations you want, and then overriding them in subclasses. For example, if you want the GUIControl to be able to draw itself, then put a doDraw() method on the base class, then override that in each subclass to do as is needed. If you instead put a getTitleBar(), getText() etc. methods on your subclass, then have the caller downcast and calls those specific methods depending on the type, your encapsulation is broken. If you have some common code that multiple subclasses need to do their drawing, then you factor this out either through another parent class, or through composition. Using dynamic_cast, or putting specific methods on the generic subclass, will likely make your code worse.
If I have this right: You want to be able to pass around base class objects but have a clean way to call specific derived class methods where the derived class implements those methods?
Sounds like the 'mixin' pattern might help:
struct Base
{
virtual ~Base() {}
};
struct Mixin
{
virtual ~Mixin() {}
virtual void mixedMethod() = 0;
};
struct Concrete : Base, Mixin
{
virtual void mixedMethod() { std::cout << "Mixing" << std:: endl; }
};
Base* create() { return new Concrete;}
bool mixIt(Base& b)
{
Mixin* m = dynamic_cast<Mixin*>(&b);
if (m)
m->mixedMethod();
return m;
}
void test ()
{
Base* b = create();
assert(mixIt(*b));
Base base;
assert(!mixIt(base));
}
[ Yes, real code never uses struct for polymorhic classes; just keeping it compact.]
The idea here is that the availability of a given method is encapsulated in the Mixin class, which is an pure abstract base class, possibly with only a single pure virtual function.
If you want "know" your base class object is of the derived type, you can call the mixin classes method. You can wrap the test and the call in a non-member function; this allows you to keep the base calss interface itself clean.
I'm extending a class provided by a third part library. The class, let's call it Foo, has a reset() method which can be called in order to restart Foo's behavior. The reset() method is also used internally by the class.
class Foo
{
public:
void reset () {
/* ... */
}
void something () {
reset();
}
};
So far, I need to overload the reset() method in order to reset my additional features as well:
class Bar : public Foo
{
public:
void reset() {
/* ...something... */
Foo::reset();
}
};
Unfortunately, since the Foo::reset() method is not virtual, by calling Bar::something() I get the Foo::reset() method called instead of Bar::reset().
Is there a way (different from overloading Foo::something() as well) to make it backward-virtual?
You cannot extend classes that were not intended to be extended.
You can't make reset() virtual in your library in such a way that it will affect the base class without changing the base class's code. For starters, the compiler has not added the necessary bookkeeping code that allows it to make virtual calls to reset().
There is no 'clean way' of doing it using inheritance. Virtual is a compile/link time difference: using a vtable to resolve method at runtime (virtual) vs direct linking (non-virtual).
No this is not possible. The virtualness of a method is decided when the method is declared and you cannot change it later in a base class.
To allow you to do so would be nothing short of a disaster. Virtual methods simply behave differently than non-virtual methods and it must be accounted for when designing an object. With this proposal I would have to consider that all of my methods could behave in 2 distinctly different ways. That significantly adds to the design cost of the application and reduces the reliability.
If neither reset nor something are virtual, you're screwed, and that's the end of the story. If something is virtual you could override it. However, if these necessary methods aren't virtual, then the class isn't intended to be extended (or implemented properly, if it was intended).
Edit:
You COULD try composition, e.g.
class Bar {
Foo f;
// foo's methods, etc, wrapped here
void something() {
f.reset();
reset();
}
};
But if you need the whole, implicit conversion, thing, you're still stuffed.
I have an abstract class with a couple pure virtual functions, and one of the classes I derive from it does not use one of the pure virtual functions:
class derivative: public base
{
public:
int somevariable;
void somefunction();
};
anyways, when I try to compile it, I get an error (apparently, a class is still considered abstract if derive from an abstract class and don't override all pure virtual functions). Anyways, it seems pointless to define a function
int purevirtfunc(){return 0;}
just because it needs to be defined through a technicality. Is there anyway to derive a class from an abstract class and not use one of the abstract class's pure virtual functions?
If your derived class doesn't "use" the base class pure virtual function, then either the derived class should not be derived from the base, or the PVF should not be there. In either case, your design is at fault and needs to be re-thought.
And no, there is no way of deleting a PVF.
A pure virtual class is an interface, one which your code will expect to be fulfilled. What would happen if you implemented that interface and didn't implement one of the methods? How would the code calling your interface know that you didn't implement the method?
Your options are:
Implement the method as you describe (making it private would indicate that it shouldn't be used).
Change your class hierarchy to take into consideration the design change.
The purpose of deriving from abstract classes is that external code can use the abstract class and expect that all functions have been implemented properly. Being able to unimplement a method would defeat this purpose, making the code uncompilable. You're free to throw an exception, if you so choose, however.
It's not a technicality at all. If your derived class does not exhibit all of the behavior of the parent, it should not be derived from the parent. This is a major design smell, and you probably need some design refactoring.
When you inherit from a class that has pure virtual functions, you MUST implement those functions. If you don't, then your derived class is also abstract, and you can't create an object of the derived class.
No. Either provide a default implementation in the base class or a simple implementation in the derived class, as you suggested.
There were already good answers, but if you want more info from the theoretical OO design side, check out the Liskov substitution principle.
Allowing this wouldn't make any sense. What would happen if you called the function without an implementation? A runtime error (that would be silly)? You could argue that it could a compile time error in some cases, but this is not possible if the exact type is not known (e.i. you pass a pointer to an instance of the derived class to a function).
As many people have already stated, it sounds like either the base method shouldn't be pure virtual or you should rethink whether your derived class really ISA base.
However, it is possible to provide an implementation for the pure virtual method in the base class. This can act like a default implementation for derived classes, but you still require the derived class to choose the base class's implementation explicity.
I don't know if that will help you with your problem or not.
No, there isn't. The convention in late bound languages when this situation occurs (as it legitimately might, but consider your design to see whether this method can be moved elsewhere, perhaps to its abstract class), is to raise an exception, and make sure that users of that method know that some implementations may raise that exception.
Seems i faced with the same problem, when trying to hide method getVertex() in derived class.
Maybe it's help.
class Body2D
{
public:
virtual ~Body2D() = default;
virtual double getCenter() const = 0;
virtual double getVertex() const = 0;
};
class Ellipse : public Body2D
{
public:
Ellipse(){};
double getCenter() const override{};
private:
double getVertex() const override{};
};
Couldn't you just do
class Foo {
public:
virtual void foo() = 0;
};
class Bar {
public:
virtual void foo() = delete;
};
When I declare a base class, should I declare all the functions in it as virtual, or should I have a set of virtual functions and a set of non-virtual functions which I am sure are not going to be inherited?
A function only needs to be virtual iff a derived class will implement that function in a different way.
For example:
class Base {
public:
void setI (int i) // No need for it to be virtual
{
m_i = i;
}
virtual ~Base () {} // Almost always a good idea
virtual bool isDerived1 () // Is overridden - so make it virtual
{
return false;
}
private:
int m_i;
};
class Derived1 : public Base {
public:
virtual ~Derived () {}
virtual bool isDerived1 () // Is overridden - so make it virtual
{
return true;
}
};
As a result, I would error the side of not having anything virtual unless you know in advance that you intend to override it or until you discover that you require the behaviour. The only exception to this is the destructor, for which its almost always the case that you want it to be virtual in a base class.
You should only make functions you intend and design to be overridden virtual. Making a method virtual is not free in terms of both maintenance and performance (maintenance being the much bigger issue IMHO).
Once a method is virtual it becomes harder to reason about any code which uses this method. Because instead of considering what one method call would do, you must consider what N method calls would do in that scenario. N represents the number of sub classes which override that method.
The one exception to this rule is destructors. They should be virtual in any class which is intended to be derived from. It's the only way to guarantee that the proper destructor is called during deallocation.
The non-virtual interface idiom (C++ Coding Standards item 39) says that a base class should have non-virtual interface methods, allowing the base class to guarantee invariants, and non-public virtual methods for customization of the base class behaviour by derived classes. The non-virtual interface methods call the virtual methods to provide the overridable behaviour.
I tend to make only the things I want to be overridable virtual. If my initial assumptions about what I will want to override turn out to be wrong, I go back and change the base class.
Oh, and obviously always make your destructor virtual if you're working on something that will be inherited from.
If you are creating a base class ( you are sure that somebody derives the class ) then you can do following things:
Make destructor virtual (a must for base class)
Define methods which should be
derived and make them virtual.
Define methods which need not be ( or
should not be) derived as
non-virtual.
If the functions are only for derived
class and not for base class then mark
them as protected.
Compiler wouldn't know which actual piece of code will be run when pointer of base type calls a virtual function. so the actual piece of code that would be run needs to be evaluated at run-time according to which object is pointed by base class pointer. So avoid the use of virtual function if the function is not gonne be overriden in an inherited class.
TLDR version:
"you should have a set of virtual functions and a set of non-virtual functions which you are sure are not going to be inherited." Because virtual functions causes a performance decrease at run-time.
The interface functions should be, in general, virtual. Functions that provide fixed functionality should not.
Why declare something virtual until you are really overriding it? I believe it's not a question of being sure or not. Follow the facts: is it overriden somewhere? No? Then it must not be virtual.