I'm making a class which has a method that launches some threads of member functions in the same class. I'm quite new to threads in c++, especially when classes are involved but this is what iv'e come up with.
class A
{
public:
void StartThreads()
{
std::thread fooThread(&A::FooFunc, this);
fooThread.join();
}
protected:
virtual void FooFunc()
{
while (true)
std::cout << "hello\n";
}
};
My question is, if i can get the name of the current object, because now if i create a class B which inherits from A but overwrites FooFunc, FooFunc from class A will be called when i do:
B b;
b.StartThreads();
So i'm looking for a way to replace std::thread fooThread(&A::FooFunc, this) with something like std::thread fooThread(&this->GetClass()::FooFunc, this). I could just make StartThreads virtual and overwrite it in derived classes, but It would be better just to write it once and being done with it. Is there a way to do this or something that results in the same thing?
In case of that your this is known at compile-time then static metaprogramming to the rescue.
C++, Swift and Rust (and now Scala also) are static languages that has a lot of compile time tricks to do for problems like that.
How? In your case templates could help you.
Also, you don't need it to be a member function, it can be a friend function (so that you can easily use templates).
class A
{
public:
template<typename T>
friend void StartThreads(const T& obj);
protected:
virtual void FooFunc()
{
while (true)
std::cout << "hello\n";
}
};
template<typename T>
void StartThreads(const T& obj) {
std::thread fooThread(&T::FooFunc, obj);
fooThread.join();
}
WARNING: This ONLY works if the class is known at compile time, i.e.
class B: public A {
};
...
B b;
A &a = b;
StartThreads(a); // Will call it AS IF IT IS A, NOT B
Another solution:
Functional programming to the rescue, you can use lambdas (or functors using structs if you are on C++ prior to C++11)
C++11:
void StartThreads()
{
std::thread fooThread([=](){ this->FooFunc(); });
fooThread.join();
}
C++98:
// Forward declaration
class A;
// The functor class (the functor is an object that is callable (i.e. has the operator (), which is the call operator overloaded))
struct ThreadContainer {
private:
A &f;
public:
ThreadContainer(A &f): f(f) {}
void operator() ();
};
class A
{
public:
// To allow access of the protected FooFunc function
friend void ThreadContainer::operator() ();
void StartThreads()
{
// Create the functor
ThreadContainer a(*this);
// Start the thread with the "call" operator, the implementation of the constructor tries to "call" the operand, which here is a
std::thread fooThread(a);
fooThread.join();
}
protected:
virtual void FooFunc()
{
while (true)
std::cout << "hello\n";
}
};
class B: public A {
protected:
virtual void FooFunc() {
while(true)
std::cout << "overridden\n";
}
};
void ThreadContainer::operator() () {
f.FooFunc();
}
You've looked at using a virtual FooFunc() directly, and somehow surmised that it doesn't work. (I won't address the accuracy of that here, as that is being brought up in the question's comments.) You don't like the idea of moving the virtual function earlier in the process. So why not move it later? There is a somewhat-common paradigm out there that uses non-virtual wrappers to virtual functions. (Usually the wrapper is public while the virtual function is protected or private.) So, something like:
class A
{
public:
void StartThreads()
{
std::thread fooThread(&A::FooFuncCaller, this); // <-- call the new function
fooThread.join();
}
protected:
void FooFuncCaller() // <-- new function layer
{
FooFunc();
}
virtual void FooFunc()
{
while (true)
std::cout << "hello\n";
}
};
Of course, if the direct call to the virtual Foofunc works, might as well use that. Still, this is simpler than using templates or custom functor classes. A lambda is a reasonable alternative, with the benefit of not changing your class' interface (header file).
Thanks for all of your answers, it turned out that my question was unrelated and that i messed up some other members in the class.
Thanks for your answers giving me some insight into other ways you can do the same thing using different methods. (https://stackoverflow.com/users/9335240/user9335240)
Related
class MyObj{
public:
void myFunc(){
//ToBeExecutedJustOnce
}
};
I have a function that I want to be executable only once for the lifetime of MyObj. There may be many instances of MyObj, and each should be able to execute that function once. So if I have:
MyObj first;
MyObj second;
MyObj third:
first.myFunc(); // Should execute
second.myFunc(); // Should execute
third.myFunc(); // Should execute
first.myFunc(); // Should not execute
second.myFunc(); // Should not execute
third.myFunc(); // Should not execute
Options:
member variable: If I use a member variable, then other functions within MyObj can access it and change it.
global static variable: Can't work because first,second and third will all be checking the same variable.
local static: Same problem as #2.
The only solution I have found, is to have MyObj inherit from another class
MyOtherObj{
private:
bool _isInit = false;
public:
bool isInit(){
bool ret = true;
if (!_isInit){
ret = false;
_isInit = true;
}
return ret;
}
};
class MyObj : public MyOtherObj {
public:
void MyFunc(){
if (!isInit()){
//Do stuff...
}
}
};
Any better suggestion ?
EDIT: I don't care about thread safety!
EDIT: I do not want to execute the method in the constructor, simply because the method may need to be executed much later in the lifetime of the object....
Use std::once_flag. It is not resettable from other methods (then again, if you cannot trust other methods of the same class, your development process is highly questionable), easy to use, and it is even thread-safe if you ever do care about that. It can be a bit less efficient in a single-threaded program.
#include <mutex>
class MyObj {
public:
void MyFunc() {
std::call_once(initFlag, [=] {
//Do stuff...
});
}
private:
std::once_flag initFlag;
};
I don't see what is wrong with Option 1. If a class has so many responsibilities that another function may accidentally mess with the is_init member variable then the class should probably be made smaller.
However, if you want to encapsulate into another class that is less error prone, rather than using inheritance, I suggest you use composition:
class FirstTime {
bool first_time = true;
public:
bool operator()(){
if (!first_time)
return false;
first_time = false;
return true;
}
};
class MyObj {
FirstTime first_time;
public:
void myFunc(){
if (first_time()){
std::cout << "First time!\n";
}
}
};
Live demo.
As with Option 1, you should think about what copy/move behavior do you want. e.g Should a copy of an initialized MyObj be considered initialized?
I see three reasonable options:
Just use your option #1, a bool member variable.
Create a little class for an init flag, that can be set, but not be unset.
Use the public non-virtual interface (NVI) idiom, if you really want to be sure, that no-one messes with your flag.
A bool member variable
This would be my first choice. Make it private, of course. If your class has so many other data fields, that adding this new member appears painful, then this could be a sign of bad design of the entire class in the first place.
Often init() methods can be avoided completely by splitting up a class into two: A class A that contains the constructed data before the call to init() and a class B that is initialized upon construction. That way you can see if an object is initialized only by its type.
An init flag that can be set, but not reset
This class could look somewhat like this:
class InitFlag
{
public:
void set()
{
isSet_ = true;
}
operator bool() const
{
return isSet_;
}
private:
bool isSet_ = false;
};
This way, member functions cannot mess up your flag as easily. As an author of a class, you should be able to trust your member functions enough, that they don't set this flag, unless they are called init().
The non-virtual interface idiom
You create a base class with an init() function that is public and non-virtual. This function checks, if init() has been called before, calls a private purely virtual doInit() function which is supposed to do the actual initialization and sets the init flag after that. It looks like this:
class InitializeBase
{
public:
virtual ~InitializeBase() = default;
bool isInit() const
{
return isInit_;
}
void init()
{
assert( !isInit() );
doInit();
isInit_ = true;
}
private:
virtual void doInit() = 0;
bool isInit_ = false;
};
This has several security advantages:
Derived classes cannot modify isInit_.
Derived classes cannot call doInit(), as long as they don't make it public or protected (which would be very nasty). However, they can and must implement this function.
Hence doInit() function is statically guaranteed not to be called more than once, unless an assert() will trigger.
If you don't want the init() function to be public, then you can derive with the protected or the private attribute from InitializeBase.
The obvious drawback is that the design is more complicated and you get an additional virtual function call. For this reason the NVI idiom has become somewhat controversial.
Here's a variant that wraps a function in a class.
Once the function is called, it's replaced with one that does nothing.
const std::function<void()> nop = [](){};
class Once
{
public:
Once(std::function<void()> f) : m_function(f) {}
void operator()()
{
m_function();
m_function = nop;
}
private:
std::function<void()> m_function;
};
class Foo
{
public:
Foo(int x)
: m_function([this](){m_x += 1;}),
m_x(x) {}
int get() const { return m_x; }
void dostuff() { m_function(); }
private:
int m_x;
Once m_function;
};
int main()
{
Foo f(0);
cout << f.get() << endl; // 0
f.dostuff();
cout << f.get() << endl; // 1
f.dostuff();
cout << f.get() << endl; // 1
}
molbdnilo's answer is pretty good and was along the same lines I was thinking. I've changed a few things which I personally think makes it more idiomatic.
#include <iostream>
#include <functional>
class Once
{
bool isDone = false;
public:
void exec(std::function<void()> function)
{
if (!isDone)
{
function();
isDone = true;
}
}
};
class MyObj {
Once once = Once();
public:
void myFunc()
{
once.exec( []{
std::cout << "Hello, world!";
// do some stuff
});
}
};
int main()
{
MyObj foo = MyObj();
foo.myFunc();
foo.myFunc();
foo.myFunc();
}
The solution at the top is very good, but this might be a better solution for an interesting special case.
I assume that the method shall only be executed once because it modifies the state of the class. For the special case that the method initializes some parts of the class, I think it is best to use an optional, either boost::optional or std::optional or std::experimental::optional, depending on what is available to you:
#include <boost/optional.hpp>
class X
{
public:
void init()
{
if( ! _x )
{
_x.reset( 5 );
}
}
private:
boost::optional<int> _x;
};
I was wondering whether there's a way to override a function for a specific instance only. For ex,
class A
{
public:
...
void update();
...
}
int main()
{
...
A *first_instance = new A();
// I want this to have a specific update() function.
// ex. void update() { functionA(); functionB(); ... }
A *second_instance = new A();
// I want this to have a different update() function than the above one.
// ex. void update() { functionZ(); functionY(); ...}
A *third_instance = new A();
// ....so on.
...
}
Is there a way to achieve this?
I think virtual function is just what you want, with virtual function, different instances of the same type can have different functions, but you need to inherit the base class. for example
class A
{
public:
...
virtual void update()
{
std::cout << "Class A\n";
}
...
};
class B: public A
{
public:
virtual void update()
{
std::cout << "Class B\n";
}
};
class C: public A
{
public:
virtual void update()
{
std::cout << "Class C\n";
}
};
int main()
{
...
A *first_instance = new A();
// I want this to have a specific update() function.
// ex. void update() { functionA(); functionB(); ... }
A *second_instance = new B();
// I want this to have a different update() function than the above one.
// ex. void update() { functionZ(); functionY(); ...}
A *third_instance = new C();
// ....so on.
...
}
each instance in the above code will bind different update functions.
Besides, you can also use function pointer to implement your requirement, but it is not recommended. For example
class A
{
public:
A(void(*u)())
{
this->update = u;
}
...
void (*update)();
};
void a_update()
{
std::cout << "update A\n";
}
void b_update()
{
std::cout << "update B\n";
}
void c_update()
{
std::cout << "update C\n";
}
int main()
{
...
A first_instance(a_update);
// I want this to have a specific update() function.
// ex. void update() { functionA(); functionB(); ... }
A second_instance(b_update);
// I want this to have a different update() function than the above one.
// ex. void update() { functionZ(); functionY(); ...}
A third_instance(c_update);
// ....so on.
...
}
Hope helps!
Hold a function in the class.
#include <iostream>
#include <functional>
using namespace std;
class Foo
{
public:
Foo(const function<void ()>& f) : func(f)
{
}
void callFunc()
{
func();
}
private:
function<void ()> func;
};
void printFoo() { cout<<"foo"<<endl; }
void printBar() { cout<<"bar"<<endl; }
int main()
{
Foo a(printFoo);
Foo b(printBar);
a.callFunc();
b.callFunc();
}
You may have noticed that the end brace of a class is often followed by a semicolon, whereas the end braces of functions, while loops etc don't. There's a reason for this, which relates to a feature of struct in C. Because a class is almost identical to a struct, this feature exists for C++ classes too.
Basically, a struct in C may declare a named instance instead of (or as well as) a named "type" (scare quotes because a struct type in C isn't a valid type name in itself). A C++ class can therefore do the same thing, though AFAIK there may be severe limitations on what else that class can do.
I'm not in a position to check at the moment, and it's certainly not something I remember using, but that may mean you can declare a named class instance inheriting from a base class without giving it a class name. There will still be a derived type, but it will be anonymous.
If valid at all, it should look something like...
class : public baseclass // note - no derived class name
{
public:
virtual funcname ()
{
...
}
} instancename;
Personally, even if this is valid, I'd avoid using it for a number of reasons. For example, the lack of a class name means that it's not possible to define member functions separately. That means that the whole class declaration and definition must go where you want the instance declared - a lot of clutter to drop in the middle of a function, or even in a list of global variables.
With no class name, there's presumably no way to declare a constructor or destructor. And if you have non-default constructors from the base class, AFAIK there's no way to specify constructor parameters with this.
And as I said, I haven't checked this - that syntax may well be illegal as well as ugly.
Some more practical approaches to varying behaviour per-instance include...
Using dependency injection - e.g. providing a function pointer or class instance (or lambda) for some part of the behavior as a constructor parameter.
Using a template class - effectively compile-time dependency injection, with the dependency provided as a function parameter to the template.
I think it will be the best if you'll tell us why do you need to override a function for a specific instance.
But here's another approach: Strategy pattern.
Your class need a member that represent some behaviour. So you're creating some abstract class that will be an interface for different behaviours, then you'll implement different behaviours in subclasses of that abstract class. So you can choose those behaviours for any object at any time.
class A;//forward declaration
class Updater
{
public:
virtual ~Updater() {};//don't forget about virtual destructor, though it's not needed in this case of class containing only one function
virtual void update(A&) = 0;
}
class SomeUpdater
{
public:
virtual void update(A & a);//concrete realisation of an update() method
}
class A
{
private:
Updater mUpdater;
public:
explicit A(Updater updater);//constructor takes an updater, let's pretend we want to choose a behaviour once for a lifetime of an object - at creation
void update()
{
mUpdater.update(this);
}
}
You can use local classes, yet, personally, I consider the "hold function in the class" approach mentioned in the other answer better. I'd recommend the following approach only if doFunc must access internals of your base class, which is not possible from a function held in a member variable:
class ABase {
public:
void Func () { this->doFunc (); }
private:
virtual void doFunc () = 0;
public:
virtual ~ABase () { }
};
ABase* makeFirstA () {
class MyA : public ABase {
virtual void doFunc () { std::cout << "First A"; }
};
return new MyA;
}
ABase* makeSecondA () {
class MyA : public ABase {
virtual void doFunc () { std::cout << "Second A"; }
};
return new MyA;
}
int main () {
std::shared_ptr<ABase> first (makeFirstA ());
std::shared_ptr<ABase> second (makeSecondA ());
first->Func ();
second->Func ();
}
From a design patterns point of view, the "local classes" approach implements the template method pattern, while the "hold a function(al) in a member variable" approach reflects the strategy pattern. Which one is more appropriate depends on what you need to achieve.
Compare the following two variants (that should do the same thing)
class Foo
{
public:
void void doStuff()
{
//...
doStuffImpl();
//...
}
virtual void doStuffImpl()=0;
void affectStateInFoo()
{}
};
class Bar:public Foo
{
public:
void doStuffImpl()
{
affectStateInFoo();
}
};
And
class Foo;
class Callback
{
public:
virtual void doStuff(Foo& foo)=0;
};
class Foo
{
public:
Foo(Callback& o):obj(o){}
void void doStuff()
{
//...
obj.doStuff(*this);
//...
}
void affectStateInFoo()
{}
Callback& obj;
};
class Bar:public Callback
{
public:
void doStuff(Foo& foo)
{
foo.affectStateInFoo();
}
};
When is one of the two variants to prefer?
Your first method requires Bar to inherit from Foo, which closely couples these classes. For callbacks this is not always what you want to do. Your second method doesn't require this.
I would use the first method if you actually extending a class, but for notifications I would use the second approach, or as Igor R. mentioned in the comments a function pointer like object.
I would prefer the second one because it is more unit-testable with mocks. But that's just my opinion.
Overdoing stuff per inheritance is a common failure to newcomers of object orientation.
Using delegation, you called it callback, is in common more flexible.
Less numbers of classes
possible reuse of "callback" class
Exchangeable at runtime instead of compiletime
Is there some library that allows me to easily and conveniently create Object-Oriented callbacks in c++?
the language Eiffel for example has the concept of "agents" which more or less work like this:
class Foo{
public:
Bar* bar;
Foo(){
bar = new Bar();
bar->publisher.extend(agent say(?,"Hi from Foo!", ?));
bar->invokeCallback();
}
say(string strA, string strB, int number){
print(strA + " " + strB + " " + number.out);
}
}
class Bar{
public:
ActionSequence<string, int> publisher;
Bar(){}
invokeCallback(){
publisher.call("Hi from Bar!", 3);
}
}
output will be:
Hi from Bar! 3 Hi from Foo!
So - the agent allows to to capsule a memberfunction into an object, give it along some predefined calling parameters (Hi from Foo), specify the open parameters (?), and pass it to some other object which can then invoke it later.
Since c++ doesn't allow to create function pointers on non-static member functions, it seems not that trivial to implement something as easy to use in c++. i found some articles with google on object oriented callbacks in c++, however, actually i'm looking for some library or header files i simply can import which allow me to use some similarily elegant syntax.
Anyone has some tips for me?
Thanks!
The most OO way to use Callbacks in C++ is to call a function of an interface and then pass an implementation of that interface.
#include <iostream>
class Interface
{
public:
virtual void callback() = 0;
};
class Impl : public Interface
{
public:
virtual void callback() { std::cout << "Hi from Impl\n"; }
};
class User
{
public:
User(Interface& newCallback) : myCallback(newCallback) { }
void DoSomething() { myCallback.callback(); }
private:
Interface& myCallback;
};
int main()
{
Impl cb;
User user(cb);
user.DoSomething();
}
People typically use one of several patterns:
Inheritance. That is, you define an abstract class which contains the callback. Then you take a pointer/reference to it. That means that anyone can inherit and provide this callback.
class Foo {
virtual void MyCallback(...) = 0;
virtual ~Foo();
};
class Base {
std::auto_ptr<Foo> ptr;
void something(...) {
ptr->MyCallback(...);
}
Base& SetCallback(Foo* newfoo) { ptr = newfoo; return *this; }
Foo* GetCallback() { return ptr; }
};
Inheritance again. That is, your root class is abstract, and the user inherits from it and defines the callbacks, rather than having a concrete class and dedicated callback objects.
class Foo {
virtual void MyCallback(...) = 0;
...
};
class RealFoo : Foo {
virtual void MyCallback(...) { ... }
};
Even more inheritance- static. This way, you can use templates to change the behaviour of an object. It's similar to the second option but works at compile time instead of at run time, which can yield various benefits and downsides, depending on the context.
template<typename T> class Foo {
void MyCallback(...) {
T::MyCallback(...);
}
};
class RealFoo : Foo<RealFoo> {
void MyCallback(...) {
...
}
};
You can take and use member function pointers or regular function pointers
class Foo {
void (*callback)(...);
void something(...) { callback(...); }
Foo& SetCallback( void(*newcallback)(...) ) { callback = newcallback; return *this; }
void (*)(...) GetCallback() { return callback; }
};
There are function objects- they overload operator(). You will want to use or write a functional wrapper- currently provided in std::/boost:: function, but I'll also demonstrate a simple one here. It's similar to the first concept, but hides the implementation and accepts a vast array of other solutions. I personally normally use this as my callback method of choice.
class Foo {
virtual ... Call(...) = 0;
virtual ~Foo();
};
class Base {
std::auto_ptr<Foo> callback;
template<typename T> Base& SetCallback(T t) {
struct NewFoo : Foo {
T t;
NewFoo(T newt) : t(newt) {}
... Call(...) { return t(...); }
};
callback = new NewFoo<T>(t);
return this;
}
Foo* GetCallback() { return callback; }
void dosomething() { callback->Call(...); }
};
The right solution mainly depends on the context. If you need to expose a C-style API then function pointers is the only way to go (remember void* for user arguments). If you need to vary at runtime (for example, exposing code in a precompiled library) then static inheritance can't be used here.
Just a quick note: I hand whipped up that code, so it won't be perfect (like access modifiers for functions, etc) and may have a couple of bugs in. It's an example.
C++ allows function pointers on member objects.
See here for more details.
You can also use boost.signals or boost.signals2 (depanding if your program is multithreaded or not).
There are various libraries that let you do that. Check out boost::function.
Or try your own simple implementation:
template <typename ClassType, typename Result>
class Functor
{
typedef typename Result (ClassType::*FunctionType)();
ClassType* obj;
FunctionType fn;
public:
Functor(ClassType& object, FunctionType method): obj(&object), fn(method) {}
Result Invoke()
{
return (*obj.*fn)();
}
Result operator()()
{
return Invoke();
}
};
Usage:
class A
{
int value;
public:
A(int v): value(v) {}
int getValue() { return value; }
};
int main()
{
A a(2);
Functor<A, int> fn(a, &A::getValue);
cout << fn();
}
Joining the idea of functors - use std::tr1::function and boost::bind to build the arguments into it before registering it.
There are many possibilities in C++, the issue generally being one of syntax.
You can use pointer to functions when you don't require state, but the syntax is really horrid. This can be combined with boost::bind for an even more... interesting... syntax (*)
I correct your false assumption, it is indeed feasible to have pointer to a member function, the syntax is just so awkward you'll run away (*)
You can use Functor objects, basically a Functor is an object which overloads the () operator, for example void Functor::operator()(int a) const;, because it's an object it has state and may derive from a common interface
You can simply create your own hierarchy, with a nicer name for the callback function if you don't want to go the operator overloading road
Finally, you can take advantage of C++0x facilities: std::function + the lambda functions are truly awesome when it comes to expressiveness.
I would appreciate a review on lambda syntax ;)
Foo foo;
std::function<void(std::string const&,int)> func =
[&foo](std::string const& s, int i) {
return foo.say(s,"Hi from Foo",i);
};
func("Hi from Bar", 2);
func("Hi from FooBar", 3);
Of course, func is only viable while foo is viable (scope issue), you could copy foo using [=foo] to indicate pass by value instead of pass by reference.
(*) Mandatory Tutorial on Function Pointers
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;
}