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;
}
Related
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.
I wasn't sure about a proper title. It does seem confusing. Sorry for that.
I am not a programmer. So please, bear with me. I am surely using some terminology wrongly.
Here's what I want to achieve:
Have a base class A_base with some nested class (sub-class?) defined in it (B_base). A_base will have all kinds of common functions that do not require specialization. The nested class is providing some interface for communication with another application (there are callbacks in it, which I define according to my needs).
class A_base
{
public:
void a_function()
{
std::cout << "This is a base a_function\n" ;
}
void a_b_function();
class B_base
{
public:
void b_function()
{
std::cout << "This is a base b_function\n" ;
}
void b_a_function();
virtual void b_c_function1(){return;};
virtual void b_c_function2(){return;};
B_base(const std::string& name_in, A_base* ptr_A_in):
name(name_in), ptr_A(ptr_A_in)
{}
protected:
const std::string name;
A_base* ptr_A;
};
B_base* b_object;
A_base(const std::string& name_in):
b_object(new B_base(name_in, this)), name(name_in)
{}
protected:
const std::string name;
};
void A_base::a_b_function()
{
b_object->b_function();
}
void A_base::B_base::b_a_function()
{
ptr_A->a_function();
}
Derive a templated class out of A_base (A_derived) with some additional specialized functionality and another nested class in it (C_class, which is also templated). I chose this to optimize the latency of my program. It doesn't have to be always lightning fast, but there's one particular path in the logic that must be as fast as possible. Therefore, I am trying to eliminate unnecessary branching by using templates.
enum Side{BB=0,SS=1};
template<bool B>
class A_derived : public A_base
{
public:
void a_function()
{
std::cout << "This is a derived a_function\n" ;
}
template<Side S>
class C_class
{
public:
void c_function();
C_class(const std::string& name_in, A_derived* ptr_A_in):
name(name_in), ptr_A(ptr_A_in)
{}
protected:
const std::string name;
A_derived* ptr_A;
};
C_class<Side::BB> c_object_bb;
C_class<Side::SS> c_object_ss;
A_derived(const std::string& name_in):
A_base(name_in), c_object_bb(name_in, this), c_object_ss(name_in, this)
{}
};
Here I am defining the specialized functions (the ones that are supposed to be fast):
template<> template<>
void A_derived<false>::C_class<Side::BB>::c_function()
{
std::cout << "This is a false-BB c_function\n" ;
}
template<> template<>
void A_derived<false>::C_class<Side::SS>::c_function()
{
std::cout << "This is a false-SS c_function\n" ;
}
template<> template<>
void A_derived<true>::C_class<Side::BB>::c_function()
{
std::cout << "This is a true-BB c_function\n" ;
}
template<> template<>
void A_derived<true>::C_class<Side::SS>::c_function()
{
std::cout << "This is a true-SS c_function\n" ;
}
And on top of that there's a need to be able to call some of the specialized functions out of a method of a nested class B_base. The specialized functions belong to a C_class. Hence, B_base knows nothing about it. But I am trying to circumvent this by casting the A_base pointer into A_derived pointer. I know for sure what kind of A_derived will be used in each case.
The problem is, the C_class function has to be called depending on the version of A_derived that is used. But B_base itself is not a templated class. So, clearly the following is not working the way I want it to work:
void A_derived<false>::B_base::b_c_function1()
{
static_cast<A_derived<false>*>(ptr_A)->c_object_bb.c_function();
}
void A_derived<false>::B_base::b_c_function2()
{
static_cast<A_derived<false>*>(ptr_A)->c_object_ss.c_function();
}
void A_derived<true>::B_base::b_c_function1()
{
static_cast<A_derived<true>*>(ptr_A)->c_object_bb.c_function();
}
void A_derived<true>::B_base::b_c_function2()
{
static_cast<A_derived<true>*>(ptr_A)->c_object_ss.c_function();
}
It just complains about redefining functions. And if I add template<> syntaxis it complains that there is not template out there. And it's kinda true. Can't blame the compiler for this.
Can you please suggest me a reasonable solution? How can I call a templated function of a nested class of a derived class from a method of a nested class of a base class?
Thank you.
Added: So, one person suggested in the comments I should use virtual functions instead. After some thinking I decided to clarify. B_base has its functions as virtual, actually. And this is not my code. I just have to deal with it. The reason virtual functions are of little help is that I cannot simply derive my A_base off B_base (due to some reasons). And for the most of B_base's functions I am fine with some common definitions. However, there are 10% of B_base's functions that I want to be specialized, depending on A_derived that is being instantiated.
My back up solution is to not use B_base at all but rather have some derived B_derived that'd be fully defined inside my A_derived. Unfortunately that would mean writing an infinite amount of useless code. Because most of B_base functions do not depend on type of A_derived.
Templated virtual member functions are not supported in C++ but I have a scenario where it would be ideal. Im wondering if someone has ideas for ways to accomplish this.
#include <iostream>
class Foo {
public:
virtual void bar(int ){}
// make a clone of my existing data, but with a different policy
virtual Foo* cloneforDB() = 0;
};
struct DiskStorage {
static void store(int x) { std::cout << "DiskStorage:" << x << "\n"; }
};
struct DBStorage {
static void store(int x) { std::cout << "DBStorage:" << x << "\n"; }
};
template<typename Storage>
class FooImpl : public Foo {
public:
FooImpl():m_value(0) {}
template<typename DiffStorage>
FooImpl(const FooImpl<DiffStorage>& copyfrom) {
m_value = copyfrom.m_value;
}
virtual void bar(int x) {
Storage::store(m_value);
std::cout << "FooImpl::bar new value:" << x << "\n";
m_value = x;
}
virtual Foo* cloneforDB() {
FooImpl<DBStorage> * newfoo = new FooImpl<DBStorage>(*this);
return newfoo;
}
int m_value;
};
int main()
{
Foo* foo1 = new FooImpl<DiskStorage>();
foo1->bar(5);
Foo* foo2 = foo1->cloneforDB();
foo2->bar(21);
}
Now if I want to clone the Foo implmemetation, but with a different Storagepolicy, I have to explicitly spell out each such implementation:
cloneforDB()
cloneforDisk()
A template parameter would have simplified that.
Can anyone think of a cleaner way to do this?
Please focus on the idea and not the example, since its obviously a contrived example.
Usually if you want to use a virtual template method, it means that something is wrong in the design of your class hierarchy. The high level reason for that follows.
Template parameters must be known at compile-time, that's their semantics. They are used to guarantee soundness properties of your code.
Virtual functions are used for polymorphism, ie. dynamic dispatching at runtime.
So you cannot mix static properties with runtime dispatching, it does not make sense if you look at the big picture.
Here, the fact that you store something somewhere should not be part of the type of your method, since it's just a behavioral trait, it could change at runtime. So it's wrong to include that information in the type of the method.
That's why C++ does not allow that: you have to rely on polymorphism to achieve such a behavior.
One easy way to go would be to pass a pointer to a Storage object as an argument (a singleton if you just want one object for each class), and work with that pointer in the virtual function.
That way, your type signature does not depend on the specific behavior of your method. And you can change your storage (in this example) policy at runtime, which is really what you should ask for as a good practice.
Sometimes, behavior can be dictated by template parameters (Alexandrescu's policy template parameters for example), but it is at type-level, not method level.
Just use templates all the way:
class Foo {
public:
virtual void bar(int ){}
template <class TargetType>
Foo* clonefor() const;
};
class FooImpl { ... };
template
inline <class TargetType>
Foo* Foo::clonefor() const
{
return new FooImpl<TargetType>(*this);
}
Now call it:
int main()
{
Foo* foo1 = new FooImpl<DiskStorage>();
foo1->bar(5);
Foo* foo2 = foo1->clonefor<DBStorage>();
foo2->bar(21);
}
A trick I have sometimes used to get around this issue is this:
template<typename T>
using retval = std::vector<T const*>;
struct Bob {};
// template type interface in Base:
struct Base {
template<typename T>
retval<T> DoStuff();
virtual ~Base() {};
// Virtual dispatch so children can implement it:
protected:
virtual retval<int> DoIntStuff() = 0;
virtual retval<double> DoDoubleStuff() = 0;
virtual retval<char> DoCharStuff() = 0;
virtual retval<Bob> DoBobStuff() = 0;
};
// forward template interface through the virtual dispatch functions:
template<> retval<int> Base::DoStuff<int>() { return DoIntStuff(); }
template<> retval<double> Base::DoStuff<double>() { return DoDoubleStuff(); }
template<> retval<char> Base::DoStuff<char>() { return DoCharStuff(); }
template<> retval<Bob> Base::DoStuff<Bob>() { return DoBobStuff(); }
// CRTP helper so the virtual functions are implemented in a template:
template<typename Child>
struct BaseHelper: public Base {
private:
// In a real project, ensuring that Child is a child type of Base should be done
// at compile time:
Child* self() { return static_cast<Child*>(this); }
Child const* self() const { return static_cast<Child const*>(this); }
public:
virtual retval<int> DoIntStuff() override final { self()->DoStuff<int>(); }
virtual retval<double> DoDoubleStuff() override final { self()->DoStuff<double>(); }
virtual retval<char> DoCharStuff() override final { self()->DoStuff<char>(); }
virtual retval<Bob> DoBobStuff() override final { self()->DoStuff<Bob>(); }
};
// Warning: if the T in BaseHelper<T> doesn't have a DoStuff, infinite
// recursion results. Code and be written to catch this at compile time,
// and I would if this where a real project.
struct FinalBase: BaseHelper<FinalBase> {
template<typename T>
retval<T> DoStuff() {
retval<T> ret;
return ret;
}
};
where I go from template-based dispatch, to virtual function dispatch, back to template based dispatch.
The interface is templated on the type I want to dispatch on. A finite set of such types are forwarded through a virtual dispatch system, then redispatched at compile time to a single method in the implementation.
I will admit this is annoying, and being able to say "I want this template to be virtual, but only with the following types" would be nice.
The reason why this is useful is that it lets you write type-agnostic template glue code that operates on these methods uniformly without having to do stuff like pass through pointers to methods or the like, or write up type-trait bundles that extract which method to call.
I'm finding it difficult to describe this problem very concisely, so I've attached the code for a demonstration program.
The general idea is that we want a set of Derived classes that are forced to implement some abstract Foo() function from a Base class. Each of the derived Foo() calls must accept a different parameter as input, but all of the parameters should also be derived from a BaseInput class.
We see two possible solutions so far, neither we're very happy with:
Remove the Foo() function from the base class and reimplement it with the correct input types in each Derived class. This, however, removes the enforcement that it be implemented in the same manner in each derived class.
Do some kind of dynamic cast inside the receiving function to verify that the type received is correct. However, this does not prevent the programmer from making an error and passing the incorrect input data type. We would like the type to be passed to the Foo() function to be compile-time correct.
Is there some sort of pattern that could enforce this kind of behaviour? Is this whole idea breaking some sort of fundamental idea underlying OOP? We'd really like to hear your input on possible solutions outside of what we've come up with.
Thanks so much!
#include <iostream>
// these inputs will be sent to our Foo function below
class BaseInput {};
class Derived1Input : public BaseInput { public: int d1Custom; };
class Derived2Input : public BaseInput { public: float d2Custom; };
class Base
{
public:
virtual void Foo(BaseInput& i) = 0;
};
class Derived1 : public Base
{
public:
// we don't know what type the input is -- do we have to try to cast to what we want
// and see if it works?
virtual void Foo(BaseInput& i) { std::cout << "I don't want to cast this..." << std::endl; }
// prefer something like this, but then it's not overriding the Base implementation
//virtual void Foo(Derived1Input& i) { std::cout << "Derived1 did something with Derived1Input..." << std::endl; }
};
class Derived2 : public Base
{
public:
// we don't know what type the input is -- do we have to try to cast to what we want
// and see if it works?
virtual void Foo(BaseInput& i) { std::cout << "I don't want to cast this..." << std::endl; }
// prefer something like this, but then it's not overriding the Base implementation
//virtual void Foo(Derived2Input& i) { std::cout << "Derived2 did something with Derived2Input..." << std::endl; }
};
int main()
{
Derived1 d1; Derived1Input d1i;
Derived2 d2; Derived2Input d2i;
// set up some dummy data
d1i.d1Custom = 1;
d2i.d2Custom = 1.f;
d1.Foo(d2i); // this compiles, but is a mistake! how can we avoid this?
// Derived1::Foo() should only accept Derived1Input, but then
// we can't declare Foo() in the Base class.
return 0;
}
Since your Derived class is-a Base class, it should never tighten the base contract preconditions: if it has to behave like a Base, it should accept BaseInput allright. This is known as the Liskov Substitution Principle.
Although you can do runtime checking of your argument, you can never achieve a fully type-safe way of doing this: your compiler may be able to match the DerivedInput when it sees a Derived object (static type), but it can not know what subtype is going to be behind a Base object...
The requirements
DerivedX should take a DerivedXInput
DerivedX::Foo should be interface-equal to DerivedY::Foo
contradict: either the Foo methods are implemented in terms of the BaseInput, and thus have identical interfaces in all derived classes, or the DerivedXInput types differ, and they cannot have the same interface.
That's, in my opinion, the problem.
This problem occured to me, too, when writing tightly coupled classes that are handled in a type-unaware framework:
class Fruit {};
class FruitTree {
virtual Fruit* pick() = 0;
};
class FruitEater {
virtual void eat( Fruit* ) = 0;
};
class Banana : public Fruit {};
class BananaTree {
virtual Banana* pick() { return new Banana; }
};
class BananaEater : public FruitEater {
void eat( Fruit* f ){
assert( dynamic_cast<Banana*>(f)!=0 );
delete f;
}
};
And a framework:
struct FruitPipeLine {
FruitTree* tree;
FruitEater* eater;
void cycle(){
eater->eat( tree->pick() );
}
};
Now this proves a design that's too easily broken: there's no part in the design that aligns the trees with the eaters:
FruitPipeLine pipe = { new BananaTree, new LemonEater }; // compiles fine
pipe.cycle(); // crash, probably.
You may improve the cohesion of the design, and remove the need for virtual dispatching, by making it a template:
template<class F> class Tree {
F* pick(); // no implementation
};
template<class F> class Eater {
void eat( F* f ){ delete f; } // default implementation is possible
};
template<class F> PipeLine {
Tree<F> tree;
Eater<F> eater;
void cycle(){ eater.eat( tree.pick() ); }
};
The implementations are really template specializations:
template<> class Tree<Banana> {
Banana* pick(){ return new Banana; }
};
...
PipeLine<Banana> pipe; // can't be wrong
pipe.cycle(); // no typechecking needed.
You might be able to use a variation of the curiously recurring template pattern.
class Base {
public:
// Stuff that don't depend on the input type.
};
template <typename Input>
class Middle : public Base {
public:
virtual void Foo(Input &i) = 0;
};
class Derived1 : public Middle<Derived1Input> {
public:
virtual void Foo(Derived1Input &i) { ... }
};
class Derived2 : public Middle<Derived2Input> {
public:
virtual void Foo(Derived2Input &i) { ... }
};
This is untested, just a shot from the hip!
If you don't mind the dynamic cast, how about this:
Class BaseInput;
class Base
{
public:
void foo(BaseInput & x) { foo_dispatch(x); };
private:
virtual void foo_dispatch(BaseInput &) = 0;
};
template <typename TInput = BaseInput> // default value to enforce nothing
class FooDistpatch : public Base
{
virtual void foo_dispatch(BaseInput & x)
{
foo_impl(dynamic_cast<TInput &>(x));
}
virtual void foo_impl(TInput &) = 0;
};
class Derived1 : public FooDispatch<Der1Input>
{
virtual void foo_impl(Der1Input & x) { /* your implementation here */ }
};
That way, you've built the dynamic type checking into the intermediate class, and your clients only ever derive from FooDispatch<DerivedInput>.
What you are talking about are covariant argument types, and that is quite an uncommon feature in a language, as it breaks your contract: You promised to accept a base_input object because you inherit from base, but you want the compiler to reject all but a small subset of base_inputs...
It is much more common for programming languages to offer the opposite: contra-variant argument types, as the derived type will not only accept everything that it is bound to accept by the contract, but also other types.
At any rate, C++ does not offer contravariance in argument types either, only covariance in the return type.
C++ has a lot of dark areas, so it's hard to say any specific thing is undoable, but going from the dark areas I do know, without a cast, this cannot be done. The virtual function specified in the base class requires the argument type to remain the same in all the children.
I am sure a cast can be used in a non-painful way though, perhaps by giving the base class an Enum 'type' member that is uniquely set by the constructor of each possible child that might possibly inherit it. Foo() can then check that 'type' and determine which type it is before doing anything, and throwing an assertion if it is surprised by something unexpected. It isn't compile time, but it's the closest a compromise I can think of, while still having the benefits of requiring a Foo() be defined.
It's certainly restricted, but you can use/simulate coviarance in constructors parameters.
Im following Lars Haendel's Functor tutorial on newty.de to setup a callback system. I am a bit confused however and I am hoping someone can assist me.
Here is my Functor template
#include <igameevents.h>
// Abstract Base Class (Functor)
class TBaseCallback
{
public:
// two possible functions to call member function. virtual cause derived
// classes will use a pointer to an object and a pointer to a member function
// to make the function call
virtual void operator()(IGameEvent *pEvent){}; // call using operator
virtual void Call(IGameEvent *pEvent) {}; // call using function
};
// Derived Template Class
template <class TClass> class TEventCallback : public TBaseCallback
{
private:
void (TClass::*funcPtr)(IGameEvent*); // pointer to member function
TClass* thisPtr; // pointer to object
public:
// constructor - takes pointer to an object and pointer to a member and stores them in two private variables
TEventCallback(TClass* _thisPtr, void(TClass::*_funcPtr)(const char*))
{ thisPtr = _thisPtr; funcPtr=_funcPtr; };
// override operator "()"
virtual void operator()(IGameEvent *pEvent)
{ (*thisPtr.*funcPtr)(pEvent); }; // execute member function
// override function "Call"
virtual void Call(IGameEvent *pEvent)
{ (*thisPtr.*funcPtr)(pEvent); }; // execute member function
};
What I want to do is basically allow other .dlls to use my HookGameEvent() function, and when a Game Event is called, I can run through a vector||list of my hooks, check if the event name matches, then execute the callbacks as needed. What is confusing me though is how I can store the callback in my HookEvent struct which looks like this.
std::vector<EventHook*> m_EventHooks;
struct EventHook
{
char *name;
EventHookMode mode;
//TEventCallback<IGameEvent*> pEventCallback;
};
I have it commented out for now, but im sure its obvious what im confused on and where I am screwing up. If anyone can provide any assistance it would be much appreciated.
Most people don't understand inheritance. Generally, derived classes are implementation details. The only time you utter their names are to construct them. Furthermore, virtual functions in a base should be private and pure, and should be completely inaccessible in derived classes, it's a design bug in C++ that this isn't enforced.
struct TBaseCallback
void operator()(IGameEvent *pEvent) { _Call(pEvent); };
void Exec(IGameEvent *pEvent) { _Call(PEvent); }
private:
virtual void _Call(IGameEvent *pEvent)=0;
};
struct EventHook
{
char *name;
EventHookMode mode;
TBaseCallback *p;
void dispatch(char *msg; IGameEvent *e) const {
if(strcmp(msg,name)==0) p->Exec(e);
}
};
With this design, it doesn't make any difference what is in classes derived from TBaseCallback, and nor should it. Only the abstraction should ever be publically visible. In normal code this is hard to enforce .. when you're using DLLs to get the derived classes it is absolutely mandatory because the set of derived classes is open/arbitrary/infinite/indeterminate (take your pick).
BTW: when you push this to more complex abstractions you will soon discover why Object Orientation is a broken concept. With DLL loaded derived classes, you simply cannot cheat with dynamic_cast switches (because they're closed/specific/finite/determinate).
The class that is going to do the callbacks should hold a list of Functor objects to be called. These would be your
std::vector<EventHook*> m_EventHooks;
Now the EventHook should have a virtual function:
struct EventHook
{
...
virtual void notifyMe();
}
Then everyone that is interested in getting notified will create his own implementation of the hook:
struct MyEventHook : public EventHook
{
virtual void notifyMe() { ... whatever I want to do in that case ... }
}
Through the wonders of polymorphism, when you then iterate over all elements of your m_EventHooks container and call notifyMe() for those, the correct class' version will be called.
The problem I see (and there could very well be others) is that in pEventCallback's type, the template parameter should be a class type but is actually a pointer type. One fix (without limiting what types the callback wraps) is to use the base type:
struct EventHook
{
char *name;
EventHookMode mode;
TBaseCallback* pCallback;
};
If there's more to TEventCallback's API, and it needs to be accessible through an EventHook, you should move the code in TEventCallback that deals with an object and its method into a separate subclass.
// Example EventCallback that takes other args
class EventCallback : public TBaseCallback {
public:
EventCallback();
EventCallback(const EventArgs& evtArgs);
// EventCallback specific methods ...
virtual EventArgs& args();
virtual const EventArgs& args() const;
}
/* TReturn allows for calling methods with a non-void return. Return value is ignored.
*/
template <class TClass, typename TReturn = void>
class TMethodCallback : public EventCallback
{
private:
typedef TReturn (TClass::*TMeth)(IGameEvent*);
TMeth funcPtr; // pointer to member function
TClass* thisPtr; // pointer to object
public:
// constructor - takes pointer to an object and pointer to a member and stores them in two private variables
TMethodCallback(TClass* _thisPtr, TMeth _funcPtr)
{ thisPtr = _thisPtr; funcPtr=_funcPtr; };
// override operator "()"
virtual void operator()(IGameEvent *pEvent)
{ (*thisPtr.*funcPtr)(pEvent); }; // execute member function
// override function "Call"
virtual void Call(IGameEvent *pEvent)
{ (*thisPtr.*funcPtr)(pEvent); }; // execute member function
};
Off-Topic
You might as well make the default implementation of TBaseCallback::Call call TBaseCallback::operator().
void TBaseCallback::Call(IGameEvent *pEvent) { this->operator()(pEvent); };
I think you will be getting a complicated compiler error because you are using T* instead of T in your template instantiation.
Try this:
struct EventHook
{
char *name;
EventHookMode mode;
TEventCallback<IGameEvent> pEventCallback;
};
should compile, if that what you want.