I have a class template that derives from a parent class. I am storing the children in a vector as Parent*. I can later use typeid to determine if two objects are the same type. What I want to do is compare a property of two objects that are the same type. For simplicity, I have omitted storing the objects in a vector, but the concept is demonstrated below:
#include <iostream>
#include <typeinfo>
#include <vector>
class Parent{ public: virtual ~Parent(){}};
template<typename T>
class TypedChild : public Parent
{
public:
virtual ~TypedChild(){}
T getValue() {return mValue;}
private:
T mValue;
};
int main()
{
Parent* child1 = new TypedChild<int>();
Parent* child2 = new TypedChild<float>();
std::vector<Parent*> objects;
objects.push_back(child1);
objects.push_back(child2);
if(typeid(*(objects[0])) == typeid(*(objects[1])))
if(objects[0]->getValue() == objects[1]->getValue()) // compiler error: Parent has no member named getValue
std::cout << "Success";
return 0;
}
Of course in this example I could dynamic_cast to TypedChild<int> before calling getValue(), but in the real case where the objects are in a vector I don't know their types, I just know that they are the same type so their getValue() functions should return the same type and can hence be compared.
Is there any way to do this comparison?
With your use case, being able to avoid a dynamic_cast will be hard, if not impossible. If you want to get the value of just one object, you'll need to use dynamic_cast, such as:
Parent* child = ...;
auto typedChild = dynamic_cast<TypedChild*>(child):
if ( typedChild )
{
int val = typedChild->getValue();
}
If you want to compare two objects for equality, the best case scenario is to have a virtual operator==() function.
class Parent
{
public:
virtual ~Parent(){}
virtual bool operator==(Parent const& rhs) const = 0;
};
template<typename T>
class TypedChild : public Parent
{
public:
virtual ~TypedChild(){}
T getValue() {return mValue;}
virtual bool operator==(Parent const& rhs) const
{
auto derivedRHS = dynamic_cast<TypedChild<T> const*>(&rhs);
if ( !derivedRHS )
{
return false;
}
return (this->mValue == derivedRHS->mValue);
}
private:
T mValue;
};
Related
I have a base class B with derived classes X, Y and Z (in fact, more than 20 derived classes). Each class has a tag() function that identifies which (derived) class it is. My program stores instances of the derived classes as pointers in a vector defined as vector<B*>. Each derived class may appear in this vector 0..n times.
I would like to have a function that looks through the vector for instances of a derived type and returns a new vector with the type of the derived class, eg
#include <vector>
using namespace std;
class B {
public:
// ...
virtual int tag() {return 0xFF;};
};
class X : public B {
// ...
int tag() {return 1;};
vector<X*> find_derived(vector<B*> base_vec) {
vector<X*> derived_vec;
for (auto p : base_vec) {
if (p->tag() == tag()) {
derived_vec.push_back((X*) p);
}
}
return derived_vec;
}
};
Obviously I don't want to have to define find_derived in each derived class but I don't see how to do this as a virtual function. Currently I am doing it using a macro but, since I am learning C++, I woudl prefer a method that used language constructs rather than those in the pre-processor. Is there another way?
One possibility:
template <typename D>
class FindDerivedMixin {
public:
vector<D*> find_derived(const vector<B*>& base_vec) {
int my_tag = static_cast<D*>(this)->tag();
vector<D*> derived_vec;
for (auto p : base_vec) {
if (p->tag() == my_tag) derived_vec.push_back(static_cast<D*>(p));
}
return derived_vec;
}
};
class X : public B, public FindDerivedMixin<X> {};
Like the previous answer, what you need is some template programming.
This is an example without mixin though:
#include <vector>
#include <iostream>
#include <type_traits>
#include <string>
//-----------------------------------------------------------------------------
// Base class
class Base
{
public:
virtual ~Base() = default;
// pure virtual method to be implemented by derived classes
virtual void Hello() const = 0;
protected:
// example of a constuctor with parameters
// it is protected since no instances of Base
// should be made by accident.
explicit Base(const std::string& message) :
m_message(message)
{
}
// getter for private member variable
const std::string& message() const
{
return m_message;
}
private:
std::string m_message;
};
//-----------------------------------------------------------------------------
// Class which contains a collection of derived classes of base
class Collection
{
public:
Collection() = default;
virtual ~Collection() = default;
// Add derived classes to the collection.
// Forward any arguments to the constructor of the derived class
template<typename type_t, typename... args_t>
void Add(args_t&&... args)
{
// compile time check if user adds a class that's derived from base.
static_assert(std::is_base_of_v<Base, type_t>,"You must add a class derived from Base");
// for polymorphism to work (casting) we need pointers to derived classes.
// use unique pointers to ensure it is the collection that will be the owner of the
// instances
m_collection.push_back(std::make_unique<type_t>(std::forward<args_t>(args)...));
}
// Getter function to get derived objects of type_t
template<typename type_t>
std::vector<type_t*> get_objects()
{
static_assert(std::is_base_of_v<Base, type_t>, "You must add a class derived from Base");
// return non-owning pointers to the derived classes
std::vector<type_t*> retval;
// loop over all objects in the collection of type std::unique_ptr<Base>
for (auto& ptr : m_collection)
{
// try to cast to a pointer to derived class of type_t
type_t* derived_ptr = dynamic_cast<type_t*>(ptr.get());
// if cast was succesful we have a pointer to the derived type
if (derived_ptr != nullptr)
{
// add the non-owning pointer to the vector that's going to be returned
retval.push_back(derived_ptr);
}
}
return retval;
}
private:
std::vector<std::unique_ptr<Base>> m_collection;
};
//-----------------------------------------------------------------------------
// some derived classes for testing.
class Derived1 :
public Base
{
public:
explicit Derived1(const std::string& message) :
Base(message)
{
}
virtual ~Derived1() = default;
void Hello() const override
{
std::cout << "Derived1 : " << message() << "\n";
}
};
//-----------------------------------------------------------------------------
class Derived2 :
public Base
{
public:
explicit Derived2(const std::string& message) :
Base(message)
{
}
virtual ~Derived2() = default;
void Hello() const override
{
std::cout << "Derived2 : " << message() << "\n";
}
};
//-----------------------------------------------------------------------------
int main()
{
Collection collection;
collection.Add<Derived1>("Instance 1");
collection.Add<Derived1>("Instance 2");
collection.Add<Derived2>("Instance 1");
collection.Add<Derived2>("Instance 2");
collection.Add<Derived1>("Instance 3");
// This is where template programming really helps
// the lines above where just to get the collection filled
auto objects = collection.get_objects<Derived1>();
for (auto& derived : objects)
{
derived->Hello();
}
return 0;
}
How do I ensure my derived class implements at least one of two chosen methods in the base class?
class base {
public:
virtual int sample()=0;
virtual Eigen::VectorXf sample()=0;
};
class Derived : Base {
int sample() override {return 1;}
}
This code returns an error, as the sample method is not implemented with the VectorXf return type. However, my intention is that only one of these need to be implemented. The only reason they are seperate in the base class is that they have different return type. How can I do this in C++?
Overloading by return type is not possible. You may use std::variant instead:
#include <variant>
class Base {
public:
virtual std::variant<int, Eigen::VectorXf> sample()=0;
};
class Derived : public Base {
std::variant<int, Eigen::VectorXf> sample() override {return 1;}
};
If one is restricted to C++11, then there are many alternatives.
Implement and use something like variant: a class that has a enumerator selecting between two active types, and a union to contain these types.
Use Boost variant.
std::pair
Implement a hierarchy of classes (a simplification of std::any), and return on the right pointer to object:
class AbstractBase {
public:
virtual ~AbstractBase() = 0;
template <class T>
const T* get() const;
};
template <class T>
class ValueWrapper : public AbstractBase {
public:
ValueWrapper(const T& value) : m_value(value) {}
const T & getValue() const { return m_value; }
private:
T m_value;
};
template <class T>
inline const T * AbstractBase::get() const {
auto child = dynamic_cast<ValueWrapper<T> const*>(this);
return child ? &child->getValue() : nullptr;
}
class Base {
public:
virtual std::unique_ptr<AbstractBase> sample()=0;
};
The question is, why would you need this?
I did a small exemple to try to explain you with my poor english what I want to do :).
I have a main class who is my engine. This is my parent class of several children.
this is the parent class :
#include <string>
#include <iostream>
#include <vector>
template <typename Type>
class A
{
public:
A(std::string const &str)
: m_str(str)
{
}
void run(void) const {
unsigned int i;
for(i = 0; ACTIONS[i].f != nullptr; i++) {
if(m_str == ACTIONS[i].key) {
return ((*(this).*ACTIONS[i].f)(m_str));
}
}
}
protected:
typedef struct s_action {
std::string key;
void (Type::*f)(std::string const &);
} t_action;
static t_action const ACTIONS[];
std::string m_str;
};
class B : public A<B>
{
public:
B(std::string const &str);
protected:
static t_action const ACTIONS[];
void error(std::string const &str);
void success(std::string const &str);
};
I would like to call children method with table pointer of member function in this parent class A::run as you can see above
This code does not compile.
I know it's not possible to have a static variable virtual, but it's
exactly that I need to do have for A::ACTIONS. I absolutely need to initialise B::ACTIONS to A::run works.
In first Is it possible? Have you got a small exemple of this case?
This is the end of my small code :
#include "Class.hpp"
B::t_action const B::ACTIONS[] = {
{"ERROR", &B::error},
{"SUCCESS", &B::success},
{"", nullptr}
};
B::B(std::string const &str)
: A<B>(str)
{
}
void B::error(std::string const &str) {
std::cerr << str << std::endl;
}
void B::success(std::string const &str) {
std::cout << str <<std::endl;
}
And the main:
#include "Class.hpp"
int main() {
B b("SUCCESS");
b.run();
return (0);
}
I didn't try, normally this code should Display SUCCESS on stdout
Thank you for your help
void run(void) const
{
unsigned int i;
for(i = 0; ACTIONS[i].f != nullptr; i++)
if (m_str == ACTIONS[i].key)
return ((*(this).*ACTIONS[i].f)(m_str));
}
There are multiple reasons why this fails to compile. Not one, but several reasons. This entire dispatching mechanism must be completely redesigned.
The first order of business is that this is a
void run(void) const
A const class method.
The method pointer in question is:
void (Type::*f)(std::string const &);
The method pointer is not const, but mutable. From an existing const class method, you can only invoke other const methods. You cannot invoke non-const methods, either directly or indirectly via a method pointer, from a const class methods.
So the first order of business is to change this to
void (Type::*f)(std::string const &) const;
This also means that all your methods, in the child class, error() and success(), must also be const class methods too.
If it's necessary to use this dispatch mechanism with non-const methods, the run() method cannot be a const class method itself. But this is not the only problem here, so I'll continue with the const method, at hand.
return ((*(this).*ACTIONS[i].f)(m_str));
The this here, is a A<Type>. This is a method of that class. That's what this is here.
The method pointer, f is pointer to a method of Type, not A<Type>. Type is a subclass of A<Type>, and you cannot convert a pointer or a reference to a base class to a pointer or a reference to a subclass, any more than you can take a pointer to A, and convert to a pointer to B when B inherits from A. C++ does not work this way.
The solution is simple, and requires only a few small tweaks. This run() should take a reference to const Type &, and invoke the method via the passed-in reference, then a replacement abstract run() method invokes it, passing *this as a parameter:
public:
virtual void run()=0;
protected:
void run_me(const Type &me) const
{
unsigned int i;
for(i = 0; ACTIONS[i].f != nullptr; i++)
if (m_str == ACTIONS[i].key)
return ((me.*ACTIONS[i].f)(m_str));
}
Then, each subclass that inherits this template only needs to implement a simple facade:
class B : public A<B>
{
public:
void run() const override
{
run_me(*this);
}
EDIT: This addresses the compilation error, but additional work is needed to deal with the fact that static class members cannot be overridden. The solution is also pretty simple: also leverage virtual class methods in order to implement this.
Remove the declaration of ACTIONS from the template base class, and replace it with an abstract function:
virtual const t_action *get_actions() const=0;
And use it in run_me():
const t_action *ACTIONS=this->get_actions();
The rest of run_me() remains as is, and then implement get_actions() in the child class:
const t_action *get_actions() const override
{
return ACTIONS;
}
Pretty much everything else remains the same.
The problem is that A will always use is own defined set of actions, not B's.
You don't need to create A at all, as you want to use B methods and list of methods.
Let's say that you create first a run call function:
template<typename T>
void run(T* obj, const std::string method)
{
const auto& available_methods = obj->get_methods();
auto iter = available_methods.find(method);
if(iter == available_methods.end())
{
// Handle this case
}
std::invoke(iter->second, obj); //C++17, or (obj->*(iter->second))();
}
Now for the class B, you need something very simple:
class B
{
public:
typedef std::unordered_map<std::string, void(B::*)()> MethodMap;
void foo();
static MethodMap& get_methods()
{
static MethodMap map{{"foo", &B::foo}};
return map;
}
};
Populate the map with get_methods() in the static function, and then call run through:
int main()
{
B b;
run(&b, "foo");
}
If you are going to use CRTP, IMO you need to google for CRTP first.
By the way here's a quick direct ans 2 your q:
template<typename crtp_child>
class crtp_base{
using crtp_target=crtp_child;
auto crtp_this(){
return static_cast<crtp_target*>(this);
};
auto crtp_this() const {
return static_cast<crtp_target const*>(this);
};
public:
void run(){
auto range=crtp_this()->actions.equal_range(m_str);
for(auto entry:range)
(crtp_this()->*(entry.second))(m_str);
};
protected:
crtp_base(std::string str):
m_str(str)
{};
std::string m_str;
//...
};
struct crtp_user:
crtp_base<crtp_user>
{
using crtp_base::crtp_base;//ctor fwding
protected:
friend class crtp_base<crtp_user>;
std::unordered_multimap<std::string, void (crtp_user::*)(std::string)> actions;
//...
};
I'm trying to make an abstract class, and one of the methods that its children must override should return an instance of the child class.
class JsonSerializable {
public:
virtual <child_of_JsonSerializable> fromJson(string jsonStr) const = 0;
};
class ConcreteSerializable : public JsonSerializable {
public:
ConcreteSerializable fromJson(string jsonStr) const {
return ConcreteSerializable();
}
};
I tried using templates following this answer, but I get an error that templates may not be virtual.
Is there a way to do what I'm looking for without using raw pointers as the return type?
You cannot create an object of an abstract type. And because you cannot create such an object, you also cannot return it. This is the reason why all examples returning a base/derived object, always return a pointer or some reference to the base class
struct B {
virtual B *fromJson(const std::string &jsonStr) const = 0;
};
struct D : public B {
D(const std::string &jsonStr);
D *fromJson(const std::string &jsonStr) const;
};
D *D::fromJson(const std::string &jsonStr) const
{
return new D(jsonStr);
}
Are you trying to implement something that historically was done using CRTP?
struct Interface {
virtual Interface *inflate(std::string const &json) = 0;
virtual ~Interface() {}
};
template<typename Child> struct Base: public Interface {
Interface *inflate(std::string const &json) { return new Child(json); }
};
struct Child: public Base<Child> {
Child(std::string const &json);
};
I have got this problem and don't know how to solve it.
Suppose I have these base classes:
class ValueBase
{
private:
int base_value;
public:
int GetValue();
void SetValue(int val);
virtual ValueBase* Meet(ValueBase* const a, ValueBase* const b) = 0;
}
class NodeBase
{
private:
ValueBase* base_nodeValue;
public:
bool AddValue(int val);
}
and derived class:
class Value : public ValueBase
{
public:
Value* Meet(ValueBase* a, ValueBase* b) override;
}
Is there a way to create instance of class Value in method AddValue in class NodeBase? I know that I should probably make AddValue pure virtual and implement it in derived class of NodeBase, but is there possibility to do it without this option? Can I use e.g. template method or maybe a callback to method in Value that would construct that object? Or is it just too evil to do it?
EDITED:
I don't have access to derived class Value in class NodeBase
Add a creation member function:
class ValueBase
{
public:
virtual ValueBase * create() = 0;
// ...
};
Then in NodeBase you can use base_nodeValue->create().
Derived classes implement it:
class Value : public ValueBase
{
Value * create() override { return new Value; }
};
The more common form of this pattern is a clone function, though, which does not produce a default-constructed object of the same type, but a copy:
Derived * clone() override { return new Derived(*this); }
I see no approach without changing class definitions. However, there are many approaches that involve changing class definitions, depending on what you are "allowed" to use.
A. Make AddValue() templated on the object type it should create:
class NodeBase
{
private:
ValueBase* base_nodeValue;
public:
template<class ValueType>
bool AddValue(int val) { base_nodeValue = new ValueType; }
}
...
// other code that has access to Value
node.AddValue<Value>(10);
B. Create function that creates Value (forwarding any arguments to constructor if needed) and pass it as an argument toAddValue`:
// might need to adapt syntax
class NodeBase
{
private:
ValueBase* base_nodeValue;
public:
bool AddValue(int val, ValueBase* (*creator)()) { base_nodeValue = (*creator)(); }
}
...
// other code that has access to Value
ValueBase* valueCreator() { return new Value; }
...
node.AddValue(10, valueCreator);
(May use a functor or a lambda here as well)
C. You can create a function in ValueBase that returns Value*.
class ValueBase
{
public:
static ValueBase* createValue();
};
class NodeBase
{
private:
ValueBase* base_nodeValue;
public:
bool AddValue(int val) { base_nodeValue = ValueBase::createValue(); }
};
// in a separate cpp
ValueBase* ValueBase::createValue() { return new Value; }
This is in fact similar to a factory approach: you can have createValue() accept a parameter and create different ValueBases depending on it. Having ValueBase store some pointer to creator function, you can have it not know about Value at all, just initialize that pointer at some other place, like you can register a subclass in a factory table in base class.