Access base class variable from child static function - c++

class A
{
public:
void init();
void method1();
void method2();
private:
bool _var1 = false;
bool _var2 = false;
};
class B : public A
{
public:
static void method3();
};
B::method3()
{
_var1 = false
}
If I call method3, I get the expected error "invalid use of member 'A::_var1' in static member function".
I have many classes which need to be static (have static functions), and many of the common variables and properties I need to inherit from a base class for code tidiness. What's the best solution for this problem?
Point me in the right direction, thank you.

a static function is not associated with an object hence cannot access class members.
see reference for more details.
you may need to pass an object instance as a function argument if that static function requires to use one e.g.
B::method3(B& b)
{
// do something with b
if b.isConnected() ) {}
}
bool B::isConnected(){return _isconnected;}

Related

Override virtual method with static method

Is there any specific reason why I cannot override virtual method from base class with static one?
Anyone knows why it would be bad idea?
Example:
#include <cstdio>
class Foo
{
public:
virtual void SomeMethod() = 0;
};
class Bar : public Foo
{
public:
static void SomeMethod() override
{
printf("SomeMethod");
}
};
void SomeFunctionWithFoo( Foo *p )
{
p->SomeMethod();
}
int main()
{
Bar o;
o.SomeMethod();
SomeFunctionWithFoo( &o );
Bar::SomeMethod();
o.StaticSomeMethod();
}
Instead I have to do this:
#include <cstdio>
class Foo
{
public:
virtual void SomeMethod() = 0;
};
class Bar : public Foo
{
public:
void SomeMethod() override
{
StaticSomeMethod();
}
static void StaticSomeMethod()
{
printf("SomeMethod");
}
};
void SomeFunctionWithFoo( Foo *p )
{
p->SomeMethod();
}
int main()
{
Bar o;
o.SomeMethod();
SomeFunctionWithFoo( &o );
Bar::StaticSomeMethod();
o.StaticSomeMethod();
}
I think as long as you don't need to access member variables, your function can be static, so that it can serve behaviour without object. In the same time such static function can serve behaviour when using interface. But maybe I am wrong and I am missing something?
With one method and two classes, it is not problem, but I have case of 10 such methods inside class, and many classes that inherit.
In real world scenario, such possibility would make my code simpler.
Summary: member functions have an invisible first parameter that your static method doesn't have.
Details: Member functions (effectively) are effectively all static methods that have an "invisible" first parameter, which is the Bar* this parameter, which tells the method which instance of the class to use. So the signature of virtual void SomeMethod() is, under the covers, actually static void SomeMethod(Foo*), but static StaticSomeMethod() doesn't have the same number of parameters.
C++ is mostly able to pretend this parameter doesn't exist, but overrides are one case where it pops up. You also see it occur when trying to bind a member function to a std::function, where you have to explicitly pass the this as the first pointer.

Dont allow access to member variable directly within same class

I am not sure is my question is right or not? But let me still try to ask once.
I have a Class with have few member variables defined. As per OO concepts, every member function can access , all member variables of its class.
But I want these member variable to be accessed via specific methods (Lets say Getters) , even within same class member functions.
It there any way to do it?
class A {
public:
void func1();
void func2();
B getB();
private:
B b;
}
void A::func1() {
b.functionFromB(); // function uses member variable b directly
}
void A::func2() {
B b1=getB(); // function ask for B from a function and then uses it. // I need something like this... And ensure each function uses same way otherwise there should be warning...
b1.functionFromB();
}
Thanks,
Kailas
No, there is not. You can do it via encapsulation and inheritance like:
class shape
{
private:
int angles;
protected:
shape(int angles_):angles(angles_){};
int getAngles() const;
}
class square : private shape
{
public:
square():shape(4){}
void doSth()
{
\\ you can access angles only via getAngles();
}
}
Any private members of the class can be accessed from within the class, but not by users of the class. So it looks like you need private members and public methods that allow access to them.
class A
{
private:
int a;
public:
int getA() {return a;}
};
int main()
{
A inst;
int t;
inst.a =5; // error member a is private
t = inst.getA(); //OK
}
The concept extends fine to nested class declarations in case you only want to allow instance of a class to be created from another class; details here
As others have said - you have to add an additional layer.
If you want to give access to specific methods then you can use the friend keyword. E.g.
// Public.h
#pragma once
class Public
{
public:
Public();
int GetI() const;
float GetF() const;
private:
std::unique_ptr<Private> p_;
};
//Public.cpp
#include "Public.h"
Public::Public()
: p_(new Private)
{
}
int Public::GetI() const
{
return p_->i_;
}
float Public::GetF() const
{
return p_->f_;
}
// Private.h
#pragma once
class Private
{
friend int Public::GetI() const;
friend float Public::GetF() const;
int i_;
float f_;
};
Keep in mind that every friend method can access ALL private members.
If you really really want to limit which methods can access which members then you can wrap each member in a separate class/struct and make only the getter/setter of that member a friend of that class/struct but I would not recommend this approach.

How to make a member function in an inheritance hierarchy return always the same value?

I have an inheritance hierarchy and I want to make each class in this hierarchy have a set of attributes which are particular for that class and which do not change during the run of the program. For example:
class Base
{
public:
const std::string getName() const;
bool getAttribute1() const;
int getAttribute2() const;
};
Now I want these functions to return the same result all the time. Furthermore, when another class inherits Base this class should have its own set of attributes and any instance of this derived class should have the same attributes. Also the name should be unique for each class.
I want to know a way to make this as transparent and elegant as possible. Sofar I have considered 2 ideas that I can use:
Make some lock system.
That is provide setters for these attributes, but make them throw a runtime exception when they are called more than once.
Make the getters pure virtual.
In this case, the result of the functions would not be stored inside the object itself. This would make it vaguely clear that the result depends on the dynamic type.
Both ideas sound incredibly lousy, so I need your help.
I am new to C++, but I know there are a lot of idioms and patterns to solve general problems like this one. Do you know any?
I have an inheritance hierarchy and I want to make each class in this hierarchy have a set of attributes which are particular for that class and which do not change during the run of the program
Well, then just provide the corresponding values as arguments to a class constructor, and do not expose any setter method on the public interface. This will make sure the values remain constant throughout the life-time of the object.
To protect against possible errors that would alter the value of those data members from member functions of your class (which of course can access the private data), make those data members const. Notice, that this will force you to initialize those members in the constructor's initializer list.
class Base
{
public:
// Forwarding constructor (requires C++11)
Base() : Base("base", true, 42) { }
const std::string getName() const { return _s; }
bool getAttribute1() const { return _a1; }
int getAttribute2() const { return _a2; }
protected:
// Constructor that can be called by derived classes
Base(std::string s, bool a1, int a2)
: _s(s), _a1(a1), _a2(a2) { }
private:
const std::string _s;
const bool _a1;
const bool _a2;
};
Derived classes would then just construct the base subobject with the appropriate arguments:
class Derived : public Base
{
public:
// Provide the values for the constant data members to the base constructor
Derived() : Base("derived", false, 1729) { }
};
This way you would not incur in the overhead of a virtual function call, and you won't have to rewrite similar virtual functions for each of these members in derived classes.
Make them virtual and hard-code the result which the functions should return:
class Base
{
public:
virtual const std::string getName() const { return "BaseName"; }
virtual bool getAttribute1() const { return whatEverAttributeValueYouWant; }
virtual int getAttribute2() const { return attributeValueHere; }
};
class Derived : public Base {
public:
virtual const std::string getName() const { return "DerivedName"; }
virtual bool getAttribute1() const { return whatEverOtherAttributeValueYouWant; }
virtual int getAttribute2() const { return otherAttributeValueHere; }
};
If you want to describe classes rather than objects, use (kind-of) traits:
template<class T> struct AttributeValues;
template<> struct AttributeValues<Base> {
static const std::string name () { return "BaseName"; }
};
template<> struct AttributeValues<Derived> {
static const std::string name () { return "DerivedName"; }
};
//...
auto nameBase = AttributeValues<Base>::name ();
auto nameDerived = AttributeValues<Derived>::name ();

Confusion over extending abstract classes in C++

In my header file for Class1 I have:
class Class1 : Class2::Callback {
public:
//Class2::Callback method
virtual bool class2Method(int i);
}
in Class1.cxx I have:
bool Class1::class2Method(int i) {
if (i == 1) return true;
return false;
}
In another place I have:
IWantAClass2Callback((Class2::Callback)instanceOfClass1);
When I try and compile I get the following error:
MyApp.cxx In constructor 'MyApp()':
MyApp.cxx:55:55: error: 'Class2::Callback is an inaccessible base of Class1'
MyApp.cxx:55:55: error: cannot allocate an object of abstract type 'Class2::Callback'
Class2.h:16:10: note: because the following virtual functions are pure within 'Class2::Callback'
Class2.h:19:18: note: virtual bool Class2::Callback::class2Method(int)
What am I doing wrong?
Here's the definition of Class2::Callback as defined in Class2.h :
class Class2
{
public:
struct Callback {
virtual bool class2Method(int i) = 0;
};
}
You have used private inheritance here. The declaration of Class1 should look like this:
class Class1 : public Class2::Callback { // note the keyword "public" here
public:
//Class2::Callback method
virtual bool class2Method(int i);
};
In c++ the default access specifier for classes is private:
class Class1 : public Class2::Callback
Should do the trick. That or make it a struct whose default access specifier is public.
There're two problems in your code:
As many pointed out, you used private inheritance instead of public
one.
You use slicing when converting to Class2::Callback: (Class2::Callback)instanceOfClass1. To use polimorphism you need a pointer or a reference. Assuming that you IWantAClass2Callback has following signature:
void IWantAClass2Callback(Class2::Callback&);
(return type doesn't matter), correct call would be:
IWantAClass2Callback(instanceOfClass1);
The Callback must reside in public: area of Class2, is that the case? I mean:
class Class2 {
public:
class Callback {
virtual bool class2Method(int) = 0
};
};
Do you really have
IWantAClass2Callback((Class2::Callback)instanceOfClass1);
in there? what do you expect it to do? you can cast pointers and references, but no instances.
try changing the signature to
void IWantAClass2Callback(Class2::Callback &instance)
and calling
IWantAClass2Callback((Class2::Callback&)instance)

C++ static virtual members?

Is it possible in C++ to have a member function that is both static and virtual? Apparently, there isn't a straightforward way to do it (static virtual member(); is a compile error), but is there at least a way to achieve the same effect?
I.E:
struct Object
{
struct TypeInformation;
static virtual const TypeInformation &GetTypeInformation() const;
};
struct SomeObject : public Object
{
static virtual const TypeInformation &GetTypeInformation() const;
};
It makes sense to use GetTypeInformation() both on an instance (object->GetTypeInformation()) and on a class (SomeObject::GetTypeInformation()), which can be useful for comparisons and vital for templates.
The only ways I can think of involves writing two functions / a function and a constant, per class, or use macros.
Any other solutions?
No, there's no way to do it, since what would happen when you called Object::GetTypeInformation()? It can't know which derived class version to call since there's no object associated with it.
You'll have to make it a non-static virtual function to work properly; if you also want to be able to call a specific derived class's version non-virtually without an object instance, you'll have to provide a second redunduant static non-virtual version as well.
Many say it is not possible, I would go one step further and say it is not meaningfull.
A static member is something that does not relate to any instance, only to the class.
A virtual member is something that does not relate directly to any class, only to an instance.
So a static virtual member would be something that does not relate to any instance or any class.
I ran into this problem the other day: I had some classes full of static methods but I wanted to use inheritance and virtual methods and reduce code repetition. My solution was:
Instead of using static methods, use a singleton with virtual methods.
In other words, each class should contain a static method that you call to get a pointer to a single, shared instance of the class. You can make the true constructors private or protected so that outside code can't misuse it by creating additional instances.
In practice, using a singleton is a lot like using static methods except that you can take advantage of inheritance and virtual methods.
While Alsk has already given a pretty detailed answer, I'd like to add an alternative, since I think his enhanced implementation is overcomplicated.
We start with an abstract base class, that provides the interface for all the object types:
class Object
{
public:
virtual char* GetClassName() = 0;
};
Now we need an actual implementation. But to avoid having to write both the static and the virtual methods, we will have our actual object classes inherit the virtual methods. This does obviously only work, if the base class knows how to access the static member function. So we need to use a template and pass the actual objects class name to it:
template<class ObjectType>
class ObjectImpl : public Object
{
public:
virtual char* GetClassName()
{
return ObjectType::GetClassNameStatic();
}
};
Finally we need to implement our real object(s). Here we only need to implement the static member function, the virtual member functions will be inherited from the ObjectImpl template class, instantiated with the name of the derived class, so it will access it's static members.
class MyObject : public ObjectImpl<MyObject>
{
public:
static char* GetClassNameStatic()
{
return "MyObject";
}
};
class YourObject : public ObjectImpl<YourObject>
{
public:
static char* GetClassNameStatic()
{
return "YourObject";
}
};
Let's add some code to test:
char* GetObjectClassName(Object* object)
{
return object->GetClassName();
}
int main()
{
MyObject myObject;
YourObject yourObject;
printf("%s\n", MyObject::GetClassNameStatic());
printf("%s\n", myObject.GetClassName());
printf("%s\n", GetObjectClassName(&myObject));
printf("%s\n", YourObject::GetClassNameStatic());
printf("%s\n", yourObject.GetClassName());
printf("%s\n", GetObjectClassName(&yourObject));
return 0;
}
Addendum (Jan 12th 2019):
Instead of using the GetClassNameStatic() function, you can also define the the class name as a static member, even "inline", which IIRC works since C++11 (don't get scared by all the modifiers :)):
class MyObject : public ObjectImpl<MyObject>
{
public:
// Access this from the template class as `ObjectType::s_ClassName`
static inline const char* const s_ClassName = "MyObject";
// ...
};
It is possible!
But what exactly is possible, let's narrow down. People often want some kind of "static virtual function" because of duplication of code needed for being able to call the same function through static call "SomeDerivedClass::myfunction()" and polymorphic call "base_class_pointer->myfunction()". "Legal" method for allowing such functionality is duplication of function definitions:
class Object
{
public:
static string getTypeInformationStatic() { return "base class";}
virtual string getTypeInformation() { return getTypeInformationStatic(); }
};
class Foo: public Object
{
public:
static string getTypeInformationStatic() { return "derived class";}
virtual string getTypeInformation() { return getTypeInformationStatic(); }
};
What if base class has a great number of static functions and derived class has to override every of them and one forgot to provide a duplicating definition for virtual function. Right, we'll get some strange error during runtime which is hard to track down. Cause duplication of code is a bad thing. The following tries to resolve this problem (and I want to tell beforehand that it is completely type-safe and doesn't contain any black magic like typeid's or dynamic_cast's :)
So, we want to provide only one definition of getTypeInformation() per derived class and it is obvious that it has to be a definition of static function because it is not possible to call "SomeDerivedClass::getTypeInformation()" if getTypeInformation() is virtual. How can we call static function of derived class through pointer to base class? It is not possible with vtable because vtable stores pointers only to virtual functions and since we decided not to use virtual functions, we cannot modify vtable for our benefit. Then, to be able to access static function for derived class through pointer to base class we have to store somehow the type of an object within its base class. One approach is to make base class templatized using "curiously recurring template pattern" but it is not appropriate here and we'll use a technique called "type erasure":
class TypeKeeper
{
public:
virtual string getTypeInformation() = 0;
};
template<class T>
class TypeKeeperImpl: public TypeKeeper
{
public:
virtual string getTypeInformation() { return T::getTypeInformationStatic(); }
};
Now we can store the type of an object within base class "Object" with a variable "keeper":
class Object
{
public:
Object(){}
boost::scoped_ptr<TypeKeeper> keeper;
//not virtual
string getTypeInformation() const
{ return keeper? keeper->getTypeInformation(): string("base class"); }
};
In a derived class keeper must be initialized during construction:
class Foo: public Object
{
public:
Foo() { keeper.reset(new TypeKeeperImpl<Foo>()); }
//note the name of the function
static string getTypeInformationStatic()
{ return "class for proving static virtual functions concept"; }
};
Let's add syntactic sugar:
template<class T>
void override_static_functions(T* t)
{ t->keeper.reset(new TypeKeeperImpl<T>()); }
#define OVERRIDE_STATIC_FUNCTIONS override_static_functions(this)
Now declarations of descendants look like:
class Foo: public Object
{
public:
Foo() { OVERRIDE_STATIC_FUNCTIONS; }
static string getTypeInformationStatic()
{ return "class for proving static virtual functions concept"; }
};
class Bar: public Foo
{
public:
Bar() { OVERRIDE_STATIC_FUNCTIONS; }
static string getTypeInformationStatic()
{ return "another class for the same reason"; }
};
usage:
Object* obj = new Foo();
cout << obj->getTypeInformation() << endl; //calls Foo::getTypeInformationStatic()
obj = new Bar();
cout << obj->getTypeInformation() << endl; //calls Bar::getTypeInformationStatic()
Foo* foo = new Bar();
cout << foo->getTypeInformation() << endl; //calls Bar::getTypeInformationStatic()
Foo::getTypeInformation(); //compile-time error
Foo::getTypeInformationStatic(); //calls Foo::getTypeInformationStatic()
Bar::getTypeInformationStatic(); //calls Bar::getTypeInformationStatic()
Advantages:
less duplication of code (but we
have to call
OVERRIDE_STATIC_FUNCTIONS in every
constructor)
Disadvantages:
OVERRIDE_STATIC_FUNCTIONS in every
constructor
memory and performance
overhead
increased complexity
Open issues:
1) there are different names for static and virtual functions
how to solve ambiguity here?
class Foo
{
public:
static void f(bool f=true) { cout << "static";}
virtual void f() { cout << "virtual";}
};
//somewhere
Foo::f(); //calls static f(), no ambiguity
ptr_to_foo->f(); //ambiguity
2) how to implicitly call OVERRIDE_STATIC_FUNCTIONS inside every constructor?
It is possible. Make two functions: static and virtual
struct Object{
struct TypeInformation;
static const TypeInformation &GetTypeInformationStatic() const
{
return GetTypeInformationMain1();
}
virtual const TypeInformation &GetTypeInformation() const
{
return GetTypeInformationMain1();
}
protected:
static const TypeInformation &GetTypeInformationMain1(); // Main function
};
struct SomeObject : public Object {
static const TypeInformation &GetTypeInformationStatic() const
{
return GetTypeInformationMain2();
}
virtual const TypeInformation &GetTypeInformation() const
{
return GetTypeInformationMain2();
}
protected:
static const TypeInformation &GetTypeInformationMain2(); // Main function
};
No, this is not possible, because static member functions lack a this pointer. And static members (both functions and variables) are not really class members per-se. They just happen to be invoked by ClassName::member, and adhere to the class access specifiers. Their storage is defined somewhere outside the class; storage is not created each time you instantiated an object of the class. Pointers to class members are special in semantics and syntax. A pointer to a static member is a normal pointer in all regards.
virtual functions in a class needs the this pointer, and is very coupled to the class, hence they can't be static.
It's not possible, but that's just because an omission. It isn't something that "doesn't make sense" as a lot of people seem to claim. To be clear, I'm talking about something like this:
struct Base {
static virtual void sayMyName() {
cout << "Base\n";
}
};
struct Derived : public Base {
static void sayMyName() override {
cout << "Derived\n";
}
};
void foo(Base *b) {
b->sayMyName();
Derived::sayMyName(); // Also would work.
}
This is 100% something that could be implemented (it just hasn't), and I'd argue something that is useful.
Consider how normal virtual functions work. Remove the statics and add in some other stuff and we have:
struct Base {
virtual void sayMyName() {
cout << "Base\n";
}
virtual void foo() {
}
int somedata;
};
struct Derived : public Base {
void sayMyName() override {
cout << "Derived\n";
}
};
void foo(Base *b) {
b->sayMyName();
}
This works fine and basically what happens is the compiler makes two tables, called VTables, and assigns indices to the virtual functions like this
enum Base_Virtual_Functions {
sayMyName = 0;
foo = 1;
};
using VTable = void*[];
const VTable Base_VTable = {
&Base::sayMyName,
&Base::foo
};
const VTable Derived_VTable = {
&Derived::sayMyName,
&Base::foo
};
Next each class with virtual functions is augmented with another field that points to its VTable, so the compiler basically changes them to be like this:
struct Base {
VTable* vtable;
virtual void sayMyName() {
cout << "Base\n";
}
virtual void foo() {
}
int somedata;
};
struct Derived : public Base {
VTable* vtable;
void sayMyName() override {
cout << "Derived\n";
}
};
Then what actually happens when you call b->sayMyName()? Basically this:
b->vtable[Base_Virtual_Functions::sayMyName](b);
(The first parameter becomes this.)
Ok fine, so how would it work with static virtual functions? Well what's the difference between static and non-static member functions? The only difference is that the latter get a this pointer.
We can do exactly the same with static virtual functions - just remove the this pointer.
b->vtable[Base_Virtual_Functions::sayMyName]();
This could then support both syntaxes:
b->sayMyName(); // Prints "Base" or "Derived"...
Base::sayMyName(); // Always prints "Base".
So ignore all the naysayers. It does make sense. Why isn't it supported then? I think it's because it has very little benefit and could even be a little confusing.
The only technical advantage over a normal virtual function is that you don't need to pass this to the function but I don't think that would make any measurable difference to performance.
It does mean you don't have a separate static and non-static function for cases when you have an instance, and when you don't have an instance, but also it might be confusing that it's only really "virtual" when you use the instance call.
Well , quite a late answer but it is possible using the curiously recurring template pattern. This wikipedia article has the info you need and also the example under static polymorphism is what you are asked for.
This question is over a decade old, but it looks like it gets a good amount of traffic, so I wanted to post an alternative using modern C++ features that I haven't seen anywhere else.
This solution uses CRTP and SFINAE to perform static dispatching. That, in itself, is nothing new, but all such implementations I've found lack strict signature checking for "overrides." This implementation requires that the "overriding" method signature exactly matches that of the "overridden" method. This behavior more closely resembles that of virtual functions, while also allowing us to effectively overload and "override" a static method.
Note that I put override in quotes because, strictly speaking, we're not technically overriding anything. Instead, we're calling a dispatch method X with signature Y that forwards all of its arguments to T::X, where T is to the first type among a list of types such that T::X exists with signature Y. This list of types considered for dispatching can be anything, but generally would include a default implementation class and the derived class.
Implementation
#include <experimental/type_traits>
template <template <class...> class Op, class... Types>
struct dispatcher;
template <template <class...> class Op, class T>
struct dispatcher<Op, T> : std::experimental::detected_t<Op, T> {};
template <template <class...> class Op, class T, class... Types>
struct dispatcher<Op, T, Types...>
: std::experimental::detected_or_t<
typename dispatcher<Op, Types...>::type, Op, T> {};
// Helper to convert a signature to a function pointer
template <class Signature> struct function_ptr;
template <class R, class... Args> struct function_ptr<R(Args...)> {
using type = R (*)(Args...);
};
// Macro to simplify creation of the dispatcher
// NOTE: This macro isn't smart enough to handle creating an overloaded
// dispatcher because both dispatchers will try to use the same
// integral_constant type alias name. If you want to overload, do it
// manually or make a smarter macro that can somehow put the signature in
// the integral_constant type alias name.
#define virtual_static_method(name, signature, ...) \
template <class VSM_T> \
using vsm_##name##_type = std::integral_constant< \
function_ptr<signature>::type, &VSM_T::name>; \
\
template <class... VSM_Args> \
static auto name(VSM_Args&&... args) \
{ \
return dispatcher<vsm_##name##_type, __VA_ARGS__>::value( \
std::forward<VSM_Args>(args)...); \
}
Example Usage
#include <iostream>
template <class T>
struct Base {
// Define the default implementations
struct defaults {
static std::string alpha() { return "Base::alpha"; };
static std::string bravo(int) { return "Base::bravo"; }
};
// Create the dispatchers
virtual_static_method(alpha, std::string(void), T, defaults);
virtual_static_method(bravo, std::string(int), T, defaults);
static void where_are_the_turtles() {
std::cout << alpha() << std::endl; // Derived::alpha
std::cout << bravo(1) << std::endl; // Base::bravo
}
};
struct Derived : Base<Derived> {
// Overrides Base::alpha
static std::string alpha(){ return "Derived::alpha"; }
// Does not override Base::bravo because signatures differ (even though
// int is implicitly convertible to bool)
static std::string bravo(bool){ return "Derived::bravo"; }
};
int main() {
Derived::where_are_the_turtles();
}
I think what you're trying to do can be done through templates. I'm trying to read between the lines here. What you're trying to do is to call a method from some code, where it calls a derived version but the caller doesn't specify which class. Example:
class Foo {
public:
void M() {...}
};
class Bar : public Foo {
public:
void M() {...}
};
void Try()
{
xxx::M();
}
int main()
{
Try();
}
You want Try() to call the Bar version of M without specifying Bar. The way you do that for statics is to use a template. So change it like so:
class Foo {
public:
void M() {...}
};
class Bar : public Foo {
public:
void M() {...}
};
template <class T>
void Try()
{
T::M();
}
int main()
{
Try<Bar>();
}
No, Static member function can't be virtual .since virtual concept is resolved at run time with the help of vptr, and vptr is non static member of a class.due to that static member function can't acess vptr so static member can't be virtual.
No, its not possible, since static members are bound at compile time, while virtual members are bound at runtime.
If your desired use for a virtual static is to be able to define an interface over the static section of a class then there is a solution to your problem using C++20 concept's.
class ExBase { //object properties
public: virtual int do(int) = 0;
};
template <typename T> //type properties
concept ExReq = std::derived_from<T, ExBase> && requires(int i) { //~constexpr bool
{
T::do_static(i) //checks that this compiles
} -> std::same_as<int> //checks the expression type is int
};
class ExImpl : virtual public ExBase { //satisfies ExReq
public: int do(int i) override {return i;} //overrides do in ExBase
public: static int do_static(int i) {return i;} //satisfies ExReq
};
//...
void some_func(ExReq auto o) {o.do(0); decltype(o)::do_static(0);}
(this works the same way on members aswell!)
For more on how concepts work: https://en.cppreference.com/w/cpp/language/constraints
For the standard concepts added in C++20: https://en.cppreference.com/w/cpp/concepts
First, the replies are correct that what the OP is requesting is a contradiction in terms: virtual methods depend on the run-time type of an instance; static functions specifically don't depend on an instance -- just on a type. That said, it makes sense to have static functions return something specific to a type. For example, I had a family of MouseTool classes for the State pattern and I started having each one have a static function returning the keyboard modifier that went with it; I used those static functions in the factory function that made the correct MouseTool instance. That function checked the mouse state against MouseToolA::keyboardModifier(), MouseToolB::keyboardModifier(), etc. and then instantiated the appropriate one. Of course later I wanted to check if the state was right so I wanted write something like "if (keyboardModifier == dynamic_type(*state)::keyboardModifier())" (not real C++ syntax), which is what this question is asking.
So, if you find yourself wanting this, you may want to rething your solution. Still, I understand the desire to have static methods and then call them dynamically based on the dynamic type of an instance. I think the Visitor Pattern can give you what you want. It gives you what you want. It's a bit of extra code, but it could be useful for other visitors.
See: http://en.wikipedia.org/wiki/Visitor_pattern for background.
struct ObjectVisitor;
struct Object
{
struct TypeInformation;
static TypeInformation GetTypeInformation();
virtual void accept(ObjectVisitor& v);
};
struct SomeObject : public Object
{
static TypeInformation GetTypeInformation();
virtual void accept(ObjectVisitor& v) const;
};
struct AnotherObject : public Object
{
static TypeInformation GetTypeInformation();
virtual void accept(ObjectVisitor& v) const;
};
Then for each concrete Object:
void SomeObject::accept(ObjectVisitor& v) const {
v.visit(*this); // The compiler statically picks the visit method based on *this being a const SomeObject&.
}
void AnotherObject::accept(ObjectVisitor& v) const {
v.visit(*this); // Here *this is a const AnotherObject& at compile time.
}
and then define the base visitor:
struct ObjectVisitor {
virtual ~ObjectVisitor() {}
virtual void visit(const SomeObject& o) {} // Or = 0, depending what you feel like.
virtual void visit(const AnotherObject& o) {} // Or = 0, depending what you feel like.
// More virtual void visit() methods for each Object class.
};
Then the concrete visitor that selects the appropriate static function:
struct ObjectVisitorGetTypeInfo {
Object::TypeInformation result;
virtual void visit(const SomeObject& o) {
result = SomeObject::GetTypeInformation();
}
virtual void visit(const AnotherObject& o) {
result = AnotherObject::GetTypeInformation();
}
// Again, an implementation for each concrete Object.
};
finally, use it:
void printInfo(Object& o) {
ObjectVisitorGetTypeInfo getTypeInfo;
Object::TypeInformation info = o.accept(getTypeInfo).result;
std::cout << info << std::endl;
}
Notes:
Constness left as an exercise.
You returned a reference from a static. Unless you have a singleton, that's questionable.
If you want to avoid copy-paste errors where one of your visit methods calls the wrong static function, you could use a templated helper function (which can't itself be virtual) t your visitor with a template like this:
struct ObjectVisitorGetTypeInfo {
Object::TypeInformation result;
virtual void visit(const SomeObject& o) { doVisit(o); }
virtual void visit(const AnotherObject& o) { doVisit(o); }
// Again, an implementation for each concrete Object.
private:
template <typename T>
void doVisit(const T& o) {
result = T::GetTypeInformation();
}
};
With c++ you can use static inheritance with the crt method. For the example, it is used widely on window template atl & wtl.
See https://en.wikipedia.org/wiki/Curiously_recurring_template_pattern
To be simple, you have a class that is templated from itself like class myclass : public myancestor. From this point the myancestor class can now call your static T::YourImpl function.
I had a browse through the other answers and none of them seem to mention virtual function tables (vtable), which explains why this is not possible.
A static function inside a C++ class compiles to something which is effectively the same as any other function in a regular namespace.
In other words, when you declare a function static you are using the class name as a namespace rather than an object (which has an instance, with some associated data).
Let's quickly look at this...
// This example is the same as the example below
class ExampleClass
{
static void exampleFunction();
int someData;
};
// This example is the same as the example above
namespace ExampleClass
{
void exampleFunction();
// Doesn't work quite the same. Each instance of a class
// has independent data. Here the data is global.
int someData;
}
With that out of the way, and an understanding of what a static member function really is, we can now consider vtables.
If you declare any virtual function in a class, then the compiler creates a block of data which (usually) precedes other data members. This block of data contains runtime information which tells the program at runtime where in memory it needs to jump to in order to execute the correct (virtual) function for each instance of a class which might be created during runtime.
The important point here is "block of data". In order for that block of data to exist, it has to be stored as part of an instance of an object (class). If your function is static, then we already said it uses the name of the class as a namespace. There is no object associated with that function call.
To add slightly more detail: A static function does not have an implicit this pointer, which points to the memory where the object lives. Because it doesn't have that, you can't jump to a place in memory and find the vtable for that object. So you can't do virtual function dispatch.
I'm not an expert in compiler engineering by any means, but understanding things at least to this level of detail is helpful, and (hopefully?) makes it easy to understand why (at least in C++) static virtual does not make sense, and cannot be translated into something sensible by the compiler.
Maybe you can try my solution below:
class Base {
public:
Base(void);
virtual ~Base(void);
public:
virtual void MyVirtualFun(void) = 0;
static void MyStaticFun(void) { assert( mSelf != NULL); mSelf->MyVirtualFun(); }
private:
static Base* mSelf;
};
Base::mSelf = NULL;
Base::Base(void) {
mSelf = this;
}
Base::~Base(void) {
// please never delete mSelf or reset the Value of mSelf in any deconstructors
}
class DerivedClass : public Base {
public:
DerivedClass(void) : Base() {}
~DerivedClass(void){}
public:
virtual void MyVirtualFun(void) { cout<<"Hello, it is DerivedClass!"<<endl; }
};
int main() {
DerivedClass testCls;
testCls.MyStaticFun(); //correct way to invoke this kind of static fun
DerivedClass::MyStaticFun(); //wrong way
return 0;
}
Like others have said, there are 2 important pieces of information:
there is no this pointer when making a static function call and
the this pointer points to the structure where the virtual table, or thunk, are used to look up which runtime method to call.
A static function is determined at compile time.
I showed this code example in C++ static members in class; it shows that you can call a static method given a null pointer:
struct Foo
{
static int boo() { return 2; }
};
int _tmain(int argc, _TCHAR* argv[])
{
Foo* pFoo = NULL;
int b = pFoo->boo(); // b will now have the value 2
return 0;
}