I have an abstract base class with a virtual method. In the derived class, this method is implemented. However, I want the function in the derived class as a static method in order to be able to call the function without instantiating an object of that class.
class Base
{
virtual double Foo(double rParam) const;
};
class Derived1 : public Base
{
static double Foo(double rParam);
};
class Derived2 : public Base
{
static double Foo(double rParam);
};
Essentially, Derived1 and Derived2 provide different implementations of a static function (that do not depend on object data), but I want those functions to be virtual in order to be able to call those functions in other functions of the base class. The only solution I see right now is to implement two member functions in the derived class, one being the virtual function from the base class, and the other one (with a different name) being a static function. In order to prevent doubling the source code, the virtual function could then directly call the static function (could be inlined). Any other solutions?
class Derived : public Base
{
double Foo(double rParam)const
{
return FooStatic(rParam);
}
inline static double FooStatic(double rParam);
};
I do think that the FooStatic approach you have is the best way to go about it, in terms of adhering to good OO practice. Essentially, your non-static Foo overrides work like function pointers into your static versions.
(Emphasis on "work like": you can't actually use function pointers directly, since your override for Foo has to throw away the reference to this before invoking FooStatic.)
It would not make sense to make a virtual method static. Virtual means that you choose which implementation to run based on the type of an object. In a static context you have no object.
EDIT: While the first example will compile, it will not do what you expect it to do. The static methods will not override the base implementation, but rather hide it, meaning that the static methods will never be called instead of that of the base class. The second example is fine though, I cannot see any problem with it.
Related
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).
I'd like to add the extra functionality without changing the existing class.
Say,
class base{
public:
int i;
base(){i = 1;}
virtual void do_work(){ /*Do some work*/ }
};
If I want to add serialization member function to it, I will simply create a derived class
class derived : public base{
public:
void serialize();
};
void derived::serialize(){
cout << "derived class" << endl;
}
And I do need to handle existing base objects,e.g.
int main(){
base a;
derived & b = static_cast<derived &>(a);
b.serialize();
}
This example runs without problems. But I do know the downcast through static_cast is something to be avoided in general.
But I'd like to know if the downcast for this particular use case can be considered safe since the derived class only has one extra member function. Will it has some potential undefined behavior for accessing vtable?
The way you're extending Base you're not making use of the vtable because you have no virtual methods. It may be easier to think of it as Derived has A Base; That you created a new class that contains a Base member variable.
My Suggestion.
Template Function
I personally would go with a template function. You can keep all the work in your original question, and avoid the need of adding virtual calls to your class.
template<typename T>
void serialize_object(T& t)
{
t.serialize()
}
And then based on your example.
Derivied d;
serialize_object(d);
The big benefit is that you're not adding runtime cast here. The compiler will inform you if you pass an object that doesn't have a method serialize.
Go Virtual
If you really want to handle this through a virtual interface do so.
struct Serializable{
virtual void serialize()=0;
virtual ~Serializable(){}
}
class Derived : public Serializable {
public:
void serialize() override;
}
void Derivied::serialize()
{
std::cout << "Yah\n";
}
Then in your code.
Derivied d;
Serializable& s = dynamic_cast<Serializable&>(d);
However, the big concern here is who is the owner of your base class? Did they provide a virtual dtor? If not, then making use of std::unique_ptr or std::shared_ptr could cause you to not deal directly with the interface.
If you can write the serialize function in a derived class without touching private or protected members then you should simply make it a free function. That solves everything.
You can't just cast a base class object to an inherited class. Any members in the inherited class will not have been created and initialized. You need to start with an object of the inherited class.
You can go the other way, of course, and call base class functions from a function of a class inherited from the base, because the base class object will always be there as part of the inherited class object, by definition.
If you have a pointer to the base class, and you build with RTTI, you can use dynamic_cast to cast to an inherited class, but that cast will fail and return NULL if the object is not of the class you're trying to cast to. But usually it's better to call a virtual function than to use dynamic_cast.
The code segment below works fine:
#include <iostream>
#include <string>
using namespace std;
class HelpInterface {
public:
void getHelp();
};
class Applicaiton : public HelpInterface {
public:
void getHelp() {
cout << "General help";
}
};
int main(void) {
Applicaiton applicaiton;
applicaiton.getHelp();
}
Make the getHelp function virtual in HelpInterface class and I will get a linker error:
class HelpInterface {
public:
virtual void getHelp();
};
If I make an empty implemenation of getHelp as below things will work again:
class HelpInterface {
public:
virtual void getHelp() {};
};
Can someone please help me understand why virtual throws a linker error unless I have an implementation for getHelp in the base class and why a non-virtual function with no implementation works just fine? In this example, the base function never gets called.
Here is a screenshot of VS2013 with the linker error:
If you want your base class method to be virtual and provide no implementation, you must set it equal to zero like this:
class HelpInterface {
public:
virtual void getHelp() = 0;
};
This is known as a pure virtual method. It has the effect of making your class abstract and forcing all of its derived classes to provide an implementation of the method. Consequently, take note that you will no longer be able to create an instance of the base class because it is abstract.
When a method is in the base class but isn't virtual the linker will not actually reference the implementation of the method if you don't have an explicit call to that method on either a pointer/reference to the base type (On an instance of the derived class) or an instance of the base type.
Virtual functions can be implemented a number of different ways, and are implementation specific. One of the most common is by using a virtual table. (Also known by virtual method table, virtual function table, virtual call table, dispatch table, vtable, or vftable), and I'm pretty sure your compiler (VS2013) uses this method of implementing virtual functions.
A virtual table is a table of function pointers for the virtual member functions. The class instance would contain a pointer to the table that the class belongs to.
When you make a function virtual the linker tries to put it into the virtual table for that type. It doesn't matter whether you call it or not or instantiate a base class (also an implementation specific detail).
As qexyn has already answered, to get around that you declare the method as pure virtual by adding = 0 after the virtual function declaration. This tells the linker to put a null pointer in the virtual function table for the class. You can also declare the virtual function pure virtual and also provide an implementation. This forces the derived classes to implement that function, but allows them to choose to use the default method.
The reason for this is that your base class definition is using indirection to get the actual function for Application.
This is often implemented with a function pointer, but in any case there could be a derived class that doesn't override the base class implementation.
Although this isn't usually reproducible that is simply because member functions are implicitly declared inline and when the implementation is visible an optimizing compiler will do exactly that; inline the function. The proof is in the optimization.
What you want, is to make sure that every derived class from your base implements getHelp(). This is a common idiom and is a core language feature. You want a "pure virtual function". This will make your base class an "abstract base class" which is actually a synonym for "interface" in object-oriented jibber-jabber.
The way to do this in C++ is with a special member-function specific syntax after all of the member function qualifiers (I'm using trailing return types here cause I think they're pretty):
class base{
public:
virtual auto func() -> void = 0;
};
This specifies that base is an abstract base class with a pure virtual function func() that all classes derived from it will implement.
In your case you would write:
class HelperInterface{
public:
virtual void getHelp() = 0; // probably want const qualifier
};
And just leave Application as it is.
You learn something every day, huh?
Suppose I have a Base class and its derived class Derived as follows:
class Base{
private:
_privateVar;
protected:
protectedVar;
public:
publicVar;
void publicMethod(someValue, anotherValue)
{
protectedVar = someValue;
publicVar = anotherValue;
}
};
class Dervied: public Base{
protected:
protectedVar:
};
int main(void)
{
Dervied d;
d.publicMethod(valueA, valueB);
}
My question-
When I call d.publicMethod(...), does the protectedVar in Derived get set or the one in Base class?
Thanks
--A
It is of Base class. Base class cannot access derived class members.
When I call d.publicMethod(...), does the protectedVar in Derived get set or the one in Base class?
The method is a member of the Base class and hence it can access only the members of the Base class.
If the method belonged to your Derived class, then it would access Derived class member.
Because Derived class data members always hide Base class data members when accessed inside their own member functions.
You cannot override a member variable, you can create another different variable in a different level in the hierarchy that has the same name, but they will be two unrelated variables. Within the context of the use of the variable, lookup will find one or the other and that is the one that will be picked up and used.
Polymorphism only applies to virtual member functions, not to non-virtual functions, not to member variables either.
Since publicMethod is a methoc of the Base class, the protectedVar of the base class is set.
I don't know if this is what you want or expect, but even if this is what you want, I think it is not advised to do it like this. Tools like PC-LINT will probably also warn you about such a construction.
The member in the base class is modified. Data members never behave as if they are virtual.
When the compiler reaches the definition of Base::publicMethod, it has to statically resolve all the names it uses. At that point, the only possible resolution of those names are the members of Base, so the generated code will access those members.
Subclassing Base later on does not, and could not possibly, go back and change the code generated for Base::publicMethod. The derived class could be in a different translation unit, or in a dynamically-loaded library.
If you want dynamic lookup, access protectedVar via a protected virtual accessor function: that would allow a derived class to interpose its own storage for the variable.
If Base::protectedVar and Derived::protectedVar have the same type anyway, it's hard to see what you expect to gain, but it would look like:
class Base
{
Type protectedVarImpl;
protected:
virtual Type & protectedVar();
virtual Type const & protectedVar() const;
public:
void publicMethod(someValue, anotherValue)
{
protectedVar() = someValue; // Note the method call
publicVar = anotherValue;
}
};
Type& Base::protectedVar() { return protectedVarImpl; }
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.