I'm trying to figure out how the template method generation works inside a class.
I've found code that describes the functionality which I would like to use but cannot understand how are the methods being generated.
Desired functionality is the automatic creation of methods that are templated based on the std::tuple arguments.
Code snippet was taken from:
https://commschamp.github.io/comms_protocols_cpp/
Here's the code:
template <typename TCommon, typename TAll>
class GenericHandler;
template <typename TCommon, typename TFirst, typename... TRest>
class GenericHandler<TCommon, std::tuple<TFirst, TRest...> > :
public GenericHandler<TCommon, std::tuple<TRest...> >
{
using Base = GenericHandler<TCommon, std::tuple<TRest...> >;
public:
using Base::handle; // Don't hide all handle() functions from base classes
virtual void handle(TFirst& msg)
{
// By default call handle(TCommon&)
this->handle(static_cast<TCommon&>(msg));
}
};
template <typename TCommon>
class GenericHandler<TCommon, std::tuple<> >
{
public:
virtual ~GenericHandler() {}
virtual void handle(TCommon&)
{
// Nothing to do
}
};
Example of usage:
class Message
{
public:
Message() = default;
virtual ~Message() {};
};
class X : public Message
{
public:
X() = default;
virtual ~X() {}
int x;
};
class Y : public Message
{
public:
Y() = default;
virtual ~Y() {}
int y;
};
using AllMessages = std::tuple<X,Y>;
class Handler : public GenericHandler<Message, AllMessages> {};
int main()
{
X x;
Handler h;
h.handle(x);
std::cout<<"Hello World";
return 0;
}
As I understand the template <typename TCommon, typename TFirst, typename... TRest> class GenericHandler<TCommon, std::tuple<TFirst, TRest...> > generates the handle() method and when the std::tuple is empty then the template <typename TCommon> class GenericHandler<TCommon, std::tuple<> > is used.
How does using Base::handle; work in this example?
How does GenericHandler inheritance work in this case template <typename TCommon, typename TFirst, typename... TRest> class GenericHandler<TCommon, std::tuple<TFirst, TRest...> > : public GenericHandler<TCommon, std::tuple<TRest...> >?
What is the process of generating the template methods?
Edit: I'm bound to C++ 11, but I'm interested in what can be done on the newer versions of C++.
How does using Base::handle; work in this example?
It's explained by its comment:
// Don't hide all handle() functions from base classes
The template class inherits from a parent template class, which also has its implementation of handle(). In the template's handle():
this->handle(static_cast<TCommon&>(msg));
Since this has a handle() (which just happens to be where this function call is made), normally overload resolution would find this handle() and only this handle().
However this ends up calling the parent class's handle(), thanks to the using declaration which has the effect of pulling in the parent class's handle(), which then gets picked by overload resolution. This is one of the fundamental rules of overload resolution, if you have:
pointer->method();
and the class has a method(), overload resolution stops, and will not include any method()s from the parent class. Here the intent is to call the parent class's method. The using declaration has the effect of adding the parent class's method to the subclass's namespace, and include it in any overload resolution.
The same thing can probably be accomplished without the using declaration, and by calling
this->Base::handle(static_cast<TCommon&>(msg));
explicitly.
Edit: I'm bound to the C++ 11, but I'm interested what can be done on the newer versions of C++.
The newer versions of C++ still work the same way.
Related
I was wondering how we can declare an interface in C++ without using virtual functions. After some internet searching I put together this solution:
#include <type_traits>
using namespace std;
// Definition of a type trait to check if a class defines a member function "bool foo(bool)"
template<typename T, typename = void>
struct has_foo : false_type { };
template<typename T>
struct has_foo<T, typename enable_if<is_same<bool, decltype(std::declval<T>().foo(bool()))>::value, void>::type> : true_type { };
// Definition of a type trait to check if a class defines a member function "void bar()"
template<typename T, typename = void>
struct has_bar : false_type { };
template<typename T>
struct has_bar<T, typename enable_if<is_same<void, decltype(std::declval<T>().bar())>::value, void>::type> : true_type { };
// Class defining the interface
template <typename T>
class Interface{
public:
Interface(){
static_assert(has_foo<T>::value == true, "member function foo not implemented");
static_assert(has_bar<T>::value == true, "member function bar not implemented");
}
};
// Interface implementation
class Implementation:Interface<Implementation>{
public:
// If the following member functions are not declared a compilation error is returned by the compiler
bool foo(bool in){return !in;}
void bar(){}
};
int main(){}
I'm planning to use this design strategy in a project where I will use static polymorphism only.
The C++ standard I will use in the project is C++11.
What do you think are the pros and cons of this approach?
What improvements can be made on the code I proposed?
EDIT 1:
I just realised that inheriting from Interface is not needed. This code could also be used:
class Implementation{
Interface<Implementation> unused;
public:
bool foo(bool in){return !in;}
void bar(){}
};
EDIT 2-3:
One major difference between the static_assert solution (with or without CRTP) and the standard CRTP is that the CRTP does not guarantee that the derived class implements all the interface members. E.g., the following code compiles correctly:
#include <type_traits>
using namespace std;
template< typename T>
class Interface{
public:
bool foo(bool in){
return static_cast<T*>(this)->foo(in);
}
void bar(){
static_cast<T*>(this)->bar();
}
};
class Implementation: public Interface<Implementation>{
public:
// bool foo(bool in){return !in;}
// void bar(){}
};
int main(){}
An error about a missing member function will be returned by the compiler only when the functions foo or bar will be required.
The way I see it, the static_assert solution feels more like an interface declaration than CRTP alone.
An common way to implement static polymorphism is to use CRTP.
With this pattern, you define an templated interface class, whose methods forward to the template:
// Interface
template <typename T>
struct base {
void foo(int arg) {
static_cast<T*>(this)->do_foo(arg);
}
};
You implementation the inherits from the base class and implements the methods:
// Implementation
struct derived : base<derived> {
void do_foo(int arg) {
std::cout << arg << '\n'
}
};
This pattern has the advantage that it looks "feels" a lot like regular runtime polymorphism, and the error messages are generally quite sane. Because all the code is visible to the compiler, everything can be inlined so there's no overhead.
It appears that you want to implement concepts (lite). You may want to read the article before attempting an implementation.
Absent compiler support, you can partially implement this idea. Your static_assert idea is a known way to express interface requirements.
Consider the Sortable example from the link. You can create a class template Sortable, use static_assert to assert all kind of thinks about the template parameter. You explain to your users that they need to implement a certain cet of methods, and to enforce that set is implemented, they need to make use of Sortable<TheirClass> one way or another.
In order to express, right in a function declaration. the idea that your function requires a Sortable, you will have to resort to something like this:
template <typename Container>
auto doSomethingWithSortable (Container&) -> std::enable_if<Implements<Container, Sortable>>::type;
So I have run into a problem in which I need a method to be defined for any types that inherit from a base class that I have created, yet I need that method to be static, is there anyway I can force it to be created?
The reason I need this I will have people be extending my base class, but I need to be able to guarantee a call to a function like so derivedType derivedType::createFromSerialized(std::string) will work to create a new instance from a serialization.
Edit: I am trying to follow Richard J. Ross III's advice and use static_assert however I am running into some problems with that and I have a feeling its due to it being called from a templated class but I can't figure out how to fix it.
template <typename indType> class population {
static_assert(std::is_function<indType::createFromSerialized>::value, "message");
....
};
However that is giving me an error of to refer to a type member of a template parameter, use ‘typename indType:: createFromSerialized’ and no type named 'createFromSerialized' in 'class test'
The reason I am trying to use static_assert is to get a nicer error message that will give information about the proper function signature for createFromSerialized instead of just one that says its not defined.
This can be done by combining static_assert and SFINAE detection technique.
template<typename T, typename V = bool>
struct has_deserialize : std::false_type { };
template<typename T>
struct has_deserialize<T,
typename std::enable_if<
std::is_same<decltype(&T::createFromSerialized),
T* (*)(const std::string&)>::value,
bool
>::type
> : std::true_type { };
template <typename T>
class test
{
static_assert(has_deserialize<T>::value, "No suitable createFromSerialized");
};
What you are trying to accomplish is possible through a factory pattern very easily. Not sure whether you can accomplish it using templates.
Here's a skeleton of how I would go about implementing the serialization functionality.
SerializationFunctor.h
class Base;
class SerializationFunctor
{
virtual Base* operator(FILE* in) const = 0;
};
Base.h
class Base
{
bool registerSerializationFunction(std::string const& identifier,
SerializationFunctor* functor);
};
ConcreteA.h
class ConcreteA
{
};
ASerializationFunctor.cc
class ASerializationFunctor : public SerializationFunctor
{
virtual Base* operator(FILE* in)
{
// Restore ConcreteA and return a pointer.
}
};
bool dummy = registerSerializationFunction("ConcreteA", new ASerializationFunctor());
ConcreteB.h
class ConcreteB
{
};
BSerializationFunctor.cc
class BSerializationFunctor : public SerializationFunctor
{
virtual Base* operator(FILE* in)
{
// Restore ConcreteB and return a pointer.
}
};
bool dummy = registerSerializationFunction("ConcreteB", new BSerializationFunctor());
In the CRTP pattern, we run into problems if we want to keep the implementation function in the derived class as protected. We must either declare the base class as a friend of the derived class or use something like this (I have not tried the method on the linked article). Is there some other (simple) way that allows keeping the implementation function in the derived class as protected?
Edit: Here is a simple code example:
template<class D>
class C {
public:
void base_foo()
{
static_cast<D*>(this)->foo();
}
};
class D: public C<D> {
protected: //ERROR!
void foo() {
}
};
int main() {
D d;
d.base_foo();
return 0;
}
The above code gives error: ‘void D::foo()’ is protected with g++ 4.5.1 but compiles if protected is replaced by public.
It's not a problem at all and is solved with one line in derived class:
friend class Base< Derived >;
#include <iostream>
template< typename PDerived >
class TBase
{
public:
void Foo( void )
{
static_cast< PDerived* > ( this )->Bar();
}
};
class TDerived : public TBase< TDerived >
{
friend class TBase< TDerived > ;
protected:
void Bar( void )
{
std::cout << "in Bar" << std::endl;
}
};
int main( void )
{
TDerived lD;
lD.Foo();
return ( 0 );
}
As lapk recommended, problem can be solved with simple friend class declaration:
class D: public C<D> {
friend class C<D>; // friend class declaration
protected:
void foo() {
}
};
However, that exposes all protected/private members of derived class and requires custom code for each derived class declaration.
The following solution is based on the linked article:
template<class D>
class C {
public:
void base_foo() { Accessor::base_foo(derived()); }
int base_bar() { return Accessor::base_bar(derived()); }
private:
D& derived() { return *(D*)this; }
// accessor functions for protected functions in derived class
struct Accessor : D
{
static void base_foo(D& derived) {
void (D::*fn)() = &Accessor::foo;
(derived.*fn)();
}
static int base_bar(D& derived) {
int (D::*fn)() = &Accessor::bar;
return (derived.*fn)();
}
};
};
class D : public C<D> {
protected: // Success!
void foo() {}
int bar() { return 42; }
};
int main(int argc, char *argv[])
{
D d;
d.base_foo();
int n = d.base_bar();
return 0;
}
PS: If you don't trust your compiler to optimize away the references, you can replace the derived() function with the following #define (resulted in 20% fewer lines of disassembly code using MSVC 2013):
int base_bar() { return Accessor::base_bar(_instance_ref); }
private:
#define _instance_ref *static_cast<D*>(this) //D& derived() { return *(D*)this; }
After some I came with a solution that works event for private members of templated derived classes. It does not solves the problem of not exposing all the members of the derived class to the base, since it uses a friend declaration on the whole class. On the other hand, for the simple case, this does not requires repeating the base name, nor it's template parameters and will always work.
First the simple case when the derived is non-template. The base takes an additional void template parameter just to show that everything still works in the case of extra template parameters of the base. The only needed one, as per the CRTP, is the typename Derived.
//Templated variadic base
template <typename Derived, typename...>
struct Interface
{
using CRTP = Interface; //Magic!
void f() { static_cast<Derived*>(this)->f(); }
};
//Simple usage of the base with extra types
//This can only be used when the derived is NON templated
class A : public Interface<A, void>
{
friend CRTP;
void f() {}
};
The only thing needed for this to work is the using CRTP = Interface; declaration in the base and the friend CRTP; declaration in the derived.
For the case when the derived is itself templated the situation is trickier. It took me some time to come to the solution, and I'm sure it's still not perfect.
Most of the magic happens inside these templates:
namespace CRTP
{
template <template <typename, typename...> class _Base, typename _Derived, typename... _BaseArgs>
struct Friend { using Base = _Base<_Derived, _BaseArgs...>; };
template <template <typename, typename...> class _Base, typename ..._BaseArgs>
struct Base
{
template <template <typename...> class _Derived, typename... _DerivedArgs>
struct Derived : public _Base<_Derived<_DerivedArgs...>, _BaseArgs...> {};
};
}
Their usage is more or less straightforward. Two use the above templates several steps are needed.
First, when inheriting in the derived class the inherited-from base class, and it's optional parameters, needs to be given. This is done using CRTP::Base<MyBase, BaseOptional....>, where MyBase is the name of the class used for CRTP, and the BaseOptional... are template parameters that are passed to the base class as-is, directly after passing our derived class that is supplied in the next step. When the base class does not accepts any additional template parameters they can be omitted completely: CRTP::Base<MyBase>.
The next step is to introduce the derived class (the whole point of CRTP). This is done by following the above CRTP::Base<...> with a ::Derived<ThisDerived, DerivedOptional...>. Where ThisDerived is the class this is defined in, and DerivedOptional... are all the template parameters declared in this class'es template declaration. The optional parameters much be specified exactly as they appear in the class template declaration.
The last step is declaring the base class as a friend. This is done by declaring friend typename CRTP::Friend<MyBase, ThisDerived, BaseOptional...>::Base somewhere in the class. The BaseOptional... template perameters must be repeated exactly as they appear in the CRTP::Base<MyBase, BaseOptional...> that is inherited from.
Follows is an example of using a templated derived when the base does not depends on the templated types (but it still can take other template parameters, void in this example).
//Templated derived with extra, non-dependant types, passed to the base
//The arguments passed to CRTP::Base::Derived<, ARGS> must exactly match
// the template
template <typename T, typename... Args>
class B : public CRTP::Base<Interface, void>::Derived<B, T, Args...>
{
friend typename CRTP::Friend<Interface, B, void>::Base;
void f() {}
};
Next is an example for when the base depends on template parameters of the derived. The only difference from the previous example is the template keyword. An experiment shows that if the keyword is specified for the previous, non dependant, case the code also complies cleanly.
//Templated derived with extra dependant types passed to the base
//Notice the addition of the "template" keyword
template <typename... Args>
class C : public CRTP::Base<Interface, Args...>::template Derived<C, Args...>
{
friend typename CRTP::Friend<Interface, C, Args...>::Base;
void f() {}
};
Please note that these templates do not work for non-templated derived classes. I will update this answer when I find the solution, so a unified syntax could be used for all cases. The closest thing that can be done is just using some fake template parameter. Note that it still must be named and passed to the CRTP machinery. For example:
template <typename Fake = void>
class D : public CRTP::Base<Interface>::Derived<D, Fake>
{
friend typename CRTP::Friend<Interface, D>::Base;
void f() {}
};
Note that A, B, C & D are declared as class. That is, all their members are private.
Follows is some code that uses the above classes.
template <typename... Args>
void invoke(Interface<Args...> & base)
{
base.f();
}
int main(int, char *[])
{
{
A derived;
//Direct invocation through cast to base (derived.f() is private)
static_cast<A::CRTP &>(derived).f();
//Invocation through template function accepting the base
invoke(derived);
}
{
B<int> derived;
static_cast<B<int>::CRTP &>(derived).f();
invoke(derived);
}
{
C<void> derived;
static_cast<C<void>::CRTP &>(derived).f();
invoke(derived);
}
{
D<void> derived;
static_cast<D<>::CRTP &>(derived).f();
invoke(derived);
}
return 0;
}
The invoke free-standing templated function works for any class derived from the base.
Also shown is how to cast the derived to the base without the need to actually specify the name of the base.
Surprisingly, this does not depend on any system headers.
The full code is available here: https://gist.github.com/equilibr/b27524468a0519aad37abc060cb8bc2b
Comments and corrections are welcome.
I want to do:
template <class Derived=BattleData>
class BattleData : public BattleCommandManager<Derived> {
};
But obviously BattleData isn't declared, so I tried a forward declaration:
template <class T> class BattleData;
template <class Derived=BattleData>
class BattleData : public BattleCommandManager<Derived> {
};
But then I get
error: "wrong number of template parameter on the second line, with
BattleData.
I really fail to see a solution to this!
Edit:
The reason I'm doing this is because I want to be able to use BattleData directly as a class, but I also want to be able to subclass it in which case I have to specify the derived class as the second template parameter.
For example let's say the corpus of my BattleData class is :
template <class Derived> class BattleData: public BaseClass<Derived> {
void foo1(){};
void foo2(){};
void foo3(){};
}
And I have a subclass
template class SubBattleData: public BattleData<SubBattleData> {
void foo1(){};
}
I would still want, in some cases, to be able to write code like this:
BattleData *x = new BattleData(...);
I can't even do the following without being able to use default arguments:
BattleData<BattleData> *x = new BattleData<BattleData>(...);
On one side, the reason functions aren't virtualized in the BattleData class is the benefit of having no virtual function. The other reason it doesn't work for me is that one of the parent CRTP classes invokes functions only if they're present in the derived type (using decltype(Derived::function) and enable-if like structures), and fall back to default behavior otherwise. Since there can be a great deal of those functions with a particular design pattern (like a CRTP that reads a protocol with many different cases and processes a case a particular way only if the derived class specify the corresponding function, otherwise just transfer it without processing).
So those functions can be present in SubBattleData and not BattleData, but both classes would work fine if instantiated, yet it's impossible to instantiate BattleData.
You should be able to accomplish your original design goals more naturally than the above. You can't use the actual Derived typename as the default clearly because what you're really trying to write is the following:
template <class Derived=BattleData <BattleData <BattleData <...>>>
class BattleData : public BattleCommandManager<Derived> {
};
You get the idea. Instead, just use a placeholder like void:
template <typename T = void>
class BattleData : public BattleCommandManager <
typename std::conditional <
std::is_same <T, void>::value,
BattleData <void>,
T
>::type>
{
};
Disclaimer: I did not compile the above.
Can't you use an Empty class for the second template parameter?
template <class T=DataContainer, class Derived=BattleData<T, Empty> >
class BattleData : public BattleCommandManager<Derived> {
};
I don't see what you are trying to do. What is wrong with
template <class T=DataContainer>
class BattleData : public BattleCommandManager< BattleData<T> > {
};
If you specify Derived to be something else than the actual derived class static polymorphism is not going to work and CRTP becomes somewhat useless anyway.
Edit: From what I have gathered this is what you want to in abstract terms:
template <class Derived>
struct Base {
void interface() {
static_cast<Derived*>(this)->implementation();
}
};
template<typename T>
struct Derived : Base<Derived> {
// dummy so we get you example
T t;
void implementation() {
std::cout << "derived" << std::endl;
}
};
struct Derived2 : public Derived<int> {
// hide implementation in Derived
// but still have Base::interface make the right call statically
void implementation() {
std::cout << "derived2" << std::endl;
}
};
There is no way I know of that you can make this work. Another
approach would be to use policy classes instead of CRTP. They are
compatible with inheritance and you can achieve similar behaviour.
template<typename Policy>
struct BattleCmdManager : public Policy {
using Policy::foo;
};
template<typename T>
struct BattleData {
// ...
protected:
void foo();
};
struct BattleData2 : public BattleData<int {
// ...
protected:
void foo();
};
Here is how I solved it:
template <class Derived> class BattleDataInh: public BaseClass<Derived> {
void foo1(){};
void foo2(){};
void foo3(){};
};
template class SubBattleData: public BattleDataInh<SubBattleData> {
void foo1(){};
};
class BattleData : public BattleDataInh<BattleData> {
};
And that way, I can add any other template parameters too. The solution was in front of my eyes the whole time but I didn't see it...
I'm wanting to invoke a specialized templated function by using a pointer to it's base type. I'm not sure if this possible so I'm open to suggestions and/or alternatives. Here is an example of my situation:
class CBase {};
class CDerivedClass : public CBase {};
template<class T>
int func<T>(T &x) { ... };
template<>
int func<CDerivedClass>(CDerivedClass &x) { ... };
I have another function that manages a list of CBase pointers and then calls the func() function.
void doStuff()
{
CBase *foo[10] = { ...... };
for (int i = 0; i < 10; ++i)
func(*foo[i]);
}
Is there a way to get the derived type, so that func(CDerivedClass &) is called?
What about Template Subclassing? This idiom allows you to use compile-time polymorphism in C++. The cost of it is higher verbosity (such as specifying the whole class hierarchy up to the current class). In your case:
template <typename TSpec> class Klass {};
template <typename TSpec> struct SpecTag {};
template <typename TSpec> class Klass<SpecTag<TSpec> > {};
template <typename TSpec>
int func(Klass<TSpec> &x) { ... };
template <typename TSpec>
int func(Klass<SpecTag<TSpec> > &x) { ... };
The "Visitor" pattern comes to the rescue in this case. It enables polymorphic behavior in an algorithm implemented outside the class. Some support code is required inside the class, but new algorithms can later be added, existing algorithms modified, etc., without affecting the class.
Alternative solution : from your example, it's obvious that you just should to use a virtual method in CBase, so you just have to define a virtual function in CBase and an overriding function in the derived class.