I am trying to understand the internals of https://github.com/vshymanskyy/TinyGSM/tree/master/src and am confused with how the classes are constructed.
In particular I see that in TinyGsmClientBG96.h they define a class that inherits from multiple templated parent classes.
class TinyGsmBG96 : public TinyGsmModem<TinyGsmBG96>,
public TinyGsmGPRS<TinyGsmBG96>,
public TinyGsmTCP<TinyGsmBG96, TINY_GSM_MUX_COUNT>,
public TinyGsmCalling<TinyGsmBG96>,
public TinyGsmSMS<TinyGsmBG96>,
public TinyGsmTime<TinyGsmBG96>,
public TinyGsmGPS<TinyGsmBG96>,
public TinyGsmBattery<TinyGsmBG96>,
public TinyGsmTemperature<TinyGsmBG96>
Fair enough. If I look at one of these, for example TinyGsmTemperature, I find some confusing code.
It looks like the static cast is in place so the we can call the hardware agnostic interface getTemperature() and use the implementation defined in TinyGsmBG96.
Why not use function overriding in this case?
What is the thinking behind this implementation?
Is this a common pattern in c++?
template <class modemType>
class TinyGsmTemperature
{
public:
/*
* Temperature functions
*/
float getTemperature()
{
return thisModem().getTemperatureImpl();
}
/*
* CRTP Helper
*/
protected:
inline const modemType &thisModem() const
{
return static_cast<const modemType &>(*this);
}
inline modemType &thisModem()
{
return static_cast<modemType &>(*this);
}
float getTemperatureImpl() TINY_GSM_ATTR_NOT_IMPLEMENTED;
};
Is this a common pattern in c++?
Yes, it is called CRTP - curiously recurring template pattern.
Why not use function overriding in this case?
override relies on virtual tables, causing extra runtime overhead.
What is the thinking behind this implementation?
Say, we want a class hierarchy with overridable methods. The classic OOP approach is virtual functions. However, they aren't zero-cost: when you have
void foo(Animal& pet) { pet.make_noise(); }
you don't statically know (in general) which implementation has been passed to foo() because you've erased its type from Dog (or Cat? or something else?) to Animal. So, the OOP approach uses virtual tables to find the right function at runtime.
How do we avoid this? We can instead remember the derived type statically:
template<typename Derived /* here's where we keep the type */> struct Animal {
void make_noise() {
// we statically know we're a Derived - no runtime dispatch!
static_cast<Derived&>(*this).make_noise();
}
};
struct Dog: public Animal<Dog /* here's how we "remember" the type */> {
void make_noise() { std::cout << "Woof!"; }
};
Now, let's rewrite foo() in a zero-cost manner:
template<typename Derived> void foo(Animal<Derived>& pet) { pet.make_noise(); }
Unlike the first attempt, we haven't erased the type from ??? to Animal: we know Animal<Derived> is actually a Derived, which is a templated - therefore, fully known to the compiler - type. This turns the virtual function call into a direct one (so, even allows inlining).
Assuming I had two classes, the first one for writing primitive types (bool, int, float, etc.) and the second one extending the first to also write complex types:
struct Writer {
virtual void Write(int value) = 0;
};
struct ComplexWriter : public Writer {
template <typename TValue> void Write(const TValue &value) {
boost::any any(value);
Write(any);
}
//virtual void Write(int value) = 0; // see question below
virtual void Write(const boost::any &any) = 0;
};
The idea is that if someone calls myWriter.Write(someIntValue);, the int overload will receive priority over the templated method.
Instead, my compiler (Visual C++ 11.0 RC) always picks the template method. The following code snippet, for example, will print Wrote any to the console:
struct ComplexWriterImpl : public ComplexWriter {
virtual void Write(int value) { std::cout << "Wrote an int"; }
virtual void Write(const boost::any &any) { std::cout << "Wrote any"; }
};
void TestWriter(ComplexWriter &writer) {
int x = 0;
writer.Write(x);
}
int main() {
ComplexWriterImpl writer;
TestWriter(writer);
}
The behavior suddenly changes when I declare the Write(int) method in the ComplexWriter class as well (see commented out line in the first snippet). It then prints Wrote an int to the console.
Is this how my compiler ought to behave? Does the C++ standard explicitly say that only overloads defined in the same class (and not a base class) shall be prioritized over a templated method?
The problem is that at the point you're calling writer.Write(x) the compiler sees a ComplexWriter not a ComplexWriterImpl, so it is only aware of the functions defined in ComplexWriter - the template function and the boost::any function.
ComplexWriter does not contain any virtual functions that accept an int, so it has no way to call through to the int overload defined in ComplexWriterImpl
When you add in the virtual overload to the ComplexWriter class, then the compiler becomes aware that there is an integer overload in the ComplexWriter class and therefore calls through to it's implementation in ComplexWriterImpl
EDIT: Now that you've edited in the inheritance between ComplexWriter & Writer, I've got a more complete explanation for you:
When you create a subclass and define a function in it then all of the functions of that name in the base class will be hidden, regardless of their argument types.
You can get around this with the using keyword I believe:
struct ComplexWriter : public Writer {
template <typename TValue> void Write(const TValue &value) {
boost::any any(value);
Write(any);
}
using Writer::Write;
virtual void Write(const boost::any &any) = 0;
};
For more details see this FAQ entry: http://www.parashift.com/c++-faq-lite/strange-inheritance.html#faq-23.9
EDIT 2: Just to confirm that this does indeed solve your problem: http://ideone.com/LRb5a
When you access the object via the ComplexWriter "interface", the compiler will try to resolve the function call to Write(int) using the definitions in that class. If it does not able to do so, it will consider base classes.
In this case, you have two candidates: Write(any) and the templated version. Since there is no explicit Write(int) available at this point, it will have to choose between these two options. Write(any) requires an implicit conversion, while the templated version does not, so the templated version is called (which in turn calls Write(any)).
To make the Write(int) from Writer available, import the Writer::Write functions:
class ComplexWriter : public Writer
{
using Writer::Write;
// rest is as before
};
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.
I have a base class which implements the following:
struct Consumer
{
template <typename T>
void callback(T msg) { /*null implementation */ }
};
I then have a class implement this:
struct Client : public Consumer
{
void callback(Msg1 msg);
void callback(Msg2 msg);
void callback(Msg3 msg);
};
The issue is I have a container of Client objects treated as Consumer* and I can't think of a way to get these Consumer objects to call the derived functions. My intended functionality is to have multiple Clients each of which implement an overloaded function for each Msg class that means something to them and the rest of the calls simply call the null implementation in the base class
Any thoughts how I can get the derived class to be called? Right now I need to implement every overloaded function in Consumer and mark them as virtual.
Cheers,
Graeme
If you really don't want to use virtual functions (this seems to be a perfect use case for them actually, but I don't know about your message classes), you can use the CRTP:
template <typename U>
struct Consumer
{
template <typename T>
void callback(T msg)
{ static_cast<U*>(this)->callback(msg); }
};
struct Client : Consumer<Client>
{
void callback(Msg1 msg);
void callback(Msg2 msg);
void callback(Msg3 msg);
};
The problem, of course, is that you cannot store Consumer objects in a container any more. Since everything is compile time, the actual type of the client must be stored alongside the consumer object for the compiler to call the right callback function. Virtual functions allow you to wait until runtime for this...
Is there a reason not to have Msg classes polymorphic and use standard virtual functions (other than "I have to rewrite all the code and I cannot") ?
EDIT If your concern is about message classes, why not use something like that, assuming message classes implement a DoSomething member function: (this technique is known as Type Erasure)
struct AnyMsg
{
template <typename Msg>
AnyMsg(Msg x) : impl(newImpl(x)) {}
void DoSomething() { impl->DoSomething(); }
private:
struct Impl
{
virtual ~Impl() {}
virtual void DoSomething() = 0;
};
// Probably better is std::unique_ptr if you have
// C++0x. Or `boost::scoped_ptr`, but you have to
// provide copy constructors yourself.
boost::shared_ptr<Impl> impl;
template <typename Msg>
Impl* newImpl(Msg m)
{
class C : public Impl
{
void DoSomething() { x.DoSomething(); }
Msg x;
public:
C(Msg x) : x(x) {}
};
return new C(m);
}
};
You can customize the behavior of newImpl to get what you want (eg. default actions if there is no DoSomething member function in the message class, specialization for some message classes or anything else). This way, you implement Msg classes like you would have done with your template solution, and you have a unique facade that you can pass to the virtual functions in your client classes.
If the Message classes are going to be very different, and client classes may react differently to them, and you are going to have a lot of message classes, this begins to smell. Or perhaps you have a candidate for the ugly and scary Visitor pattern.
Since you don't want to use virtual methods, the compiler would have to know statically (i.e. at compile time) which function to call. If you have different client objects in your container, there is now way the compiler could possibly know this. So I think there's no solution to your problem without using virtual methods (which are btw. exactly designed for this kind of situations...).
Of course you could alternatively using some switch statements for manually deriving the concrete type, but this is neither elegant nor efficient (and you would have to hardcode all possible client types ...)
EDIT
Personally, I'd implement some base message class containing a type code and implement a switch statement in the client class to handle different message types like:
struct MsgBase {
int type;
};
struct Consumer {
virtual void callback(MsgBase msg) { };
};
struct Client : public Consumer {
void callback(MsgBase msg) {
switch (msg.type) {
case MSGTYPE1:
callback((Msg1)msg);
break;
case MSGTYPE2:
callback((Msg2)msg);
break;
// ...
}
}
void callback(Msg1 msg) { /* ... */ }
void callback(Msg2 msg) { /* ... */ }
};
You could also make MsgBase polymorphic (e.g. virtual destructor) and use typeid to differentiate (more elegant but slightly less efficient ...)
struct Client : public Consumer {
void callback(MsgBase* msg) {
if (typeid(*msg) == typeof(Msg1))
callback(static_cast<Msg1*>(msg));
else if (typeid(*msg) == typeof(Msg2))
callback(static_cast<Msg2*>(msg));
}
// ...
};
This is always a difficult situation to make totally extensible, as is the case usually with the Visitor pattern.
You end up needing up to V*T implementations where V is the number of "visitors" and T is the number of types being visited and will probably end up having to use a mixture of visitor and class factory pattern.
visitors here would be your consumers
class factory would be used for the message types.
and your best way to make it totally extensible is to create new function "objects" for message/consumer pairs and a bit of double-dispatch to ensure the right one gets called.
In your case you have different messages come in and then you give them to your consumers who might handle them? So each message should have an identifiable "type" and your consumer
should look up this type in a table to create a handler for it.
You can have one handler per type per consumer class.
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;
}