For example, I have 2 classes (in reality, it's more, that's why I'm asking this question) with the same methods:
class class1{
public:
void init(){
//something
}
void dostuff(){
//something
}
//
};
class class2{
public:
void init(){
//something
}
void dostuff(){
//something
}
//
};
And now a third one in which I want to deal with the two classes in the same manner:
class upclass{
public:
upclass(class12* argclass){
myclass=argclass;
myclass->init();
}
void domorestuff(){
myclass->dostuff();
}
private:
class12* myclass; //pointer to class 1 OR class 2
};
My question is now, do I need multiple constructors and multiple declarations to make it work or is there a way around it? Is it even possible to make "class12" a spacekeeper for these types without preprocessor-directives?
I am sorry to say, this is a wide field and there are really many many possible solution.
But I guess that we are talking about object- oriented programming, derivation and plymorphic functions. What you describe, will be typically solved with a class hierachy.
You have one base class with virtual (polymorphic) functions.
Then you derive other classes from this base class and override the virtual functions from the base class.
In a 3rd step, you create some instances of the derived classes dynamically, during runtime and you store the newly created classes (their address) in a pointer to the base class.
Later, you can call any of the virtual overriden function through the base class pointer. And mechanism behind the scenes will call the correct function for you.
Additionally. You defined some function init. Such a function name suggests the usage of a class-constructor. This will be called automatically in the correct sequence. First the base class constructor and then the derived class constructor.
Please see the below example:
#include <iostream>
#include <string>
class Base {
std::string baseName{};
public:
Base() { // Do initialization stuff
baseName = "Base";
std::cout << "\nConstructor Base\n";
}
virtual void doStuff() { // virtual function
std::cout << baseName << '\n';
}
};
class Derived1 : public Base {
std::string derivedName{};
public:
Derived1() : Base() { // Do initialization stuff
derivedName = "Derived1";
std::cout << "Constructor Derived1\n";
}
void doStuff() override { // Override virtaul function
std::cout << derivedName << '\n';
}
};
class Derived2 : public Base {
std::string derivedName{};
public:
Derived2() : Base() { // Do initialization stuff
derivedName = "Derived2";
std::cout << "Constructor Derived2\n\n";
}
void doStuff() override { // Override virtaul function
std::cout << derivedName << '\n';
}
};
int main() {
Base* base = new Base();
Base* derived1 = new Derived1(); // Store in base class pointer
Base* derived2 = new Derived2(); // Store in base class pointer
base->doStuff();
derived1->doStuff(); // Magic of polymorphism
derived2->doStuff(); // Magic of polymorphism
}
The Base class pointer will accept all classes derived from Base.
Please note. In reality you ould not use raw pointers and also to the constructor differently. This is just fotr demo.
But, you need to read several books about it to get the complete understanding.
You can explicitly write "store one of these" via std::variant and obtain the actual type (when needed) through std::visit:
#include <variant>
using class12 = std::variant<class1*, class2*>;
class upclass {
public:
upclass(class12 argclass): myclass{argclass} {
visit([](auto classn) { classn->init(); }, myclass);
}
void domorestuff() {
visit([](auto classn) { classn->dostuff(); }, myclass);
}
private:
class12 myclass;
};
If those visits get too repetitive, you might consider writing a pretty API to hide them:
class prettyclass12: public std::variant<class1*, class2*> {
private: // both g++ and clang want variant_size<>, a quick hack:
auto& upcast() { return static_cast<std::variant<class1*, class2*>&>(*this); }
public:
using std::variant<class1*, class2*>::variant;
void init() { visit([](auto classn) { classn->init(); }, upcast()); }
void dostuff() { visit([](auto classn) { classn->dostuff(); }, upcast()); }
};
class prettyupclass {
public:
prettyupclass(prettyclass12 argclass): myclass{argclass} { myclass.init(); }
void domorestuff() { myclass.dostuff(); }
private:
prettyclass12 myclass;
};
Suppose I have code like that one, and I want to get access to the myClassB members. How can I do that? I need to use the functionality of functionA.
I can't change it because it is from the 3rd party library. And I need to use functionA to create it, and get values created by it. In this case "Test_1" string
class myClassA {
public:
myClassA(){}
~myClassA(){}
};
class myClassB : public myClassA
{
public:
myClassB(){}
void setString1(std::string newString)
std::string getS1()
private:
std::string privateMember;
};
std::shared_ptr<myClassA> functionA()
{
std::shared_ptr<myClassB> temporary(new myClassB());
temporary->setString1("TEST_1");
return std::move(temporary);
}
int main()
{
std::shared_ptr<myClassA> myPtr; // = functionA(); ??
}
Theoretically, you could use a dynamic_cast (or in this case specifically, std::dynamic_pointer_cast to get the derived pointer type. Like so:
std::shared_ptr<MyClassA> a_ptr = functionA();
std::shared_ptr<MyClassB> b_ptr = std::dynamic_pointer_cast<MyClassB>(a_ptr);
if(b_ptr) {//Check to verify the cast was successful
b_ptr->setString("Test1");
}
There is, however, a major caveat to this. In order for dynamic_cast (and therefore std::dynamic_pointer_cast) to work, your object hierarchy must have a virtual table defined. That means at least one of the methods defined by MyClassA must be declared virtual. The simplest solution is to declare the destructor virtual, since that's good practice whenever you're defining polymorphic objects (since you need it to ensure that any derived classes clean up their resources correctly).
class MyClassA {
public:
MyClassA() = default;
virtual ~MyClassA() = default;
};
Agree with dynamic_cast but without a virtual function table in ClassA, something like this would have to do:
Test This Code
#include <string>
#include <memory>
#include <iostream>
#include <set>
class myClassA {
public:
myClassA(){}
~myClassA(){}
};
class myClassB;
class ClassB_Registry{
private:
ClassB_Registry(){
}
~ClassB_Registry(){
}
public:
static ClassB_Registry* Get(){ static ClassB_Registry obj; return &obj; }
static void Register(myClassB* ptr){
Get()->mPointers.insert(ptr);
}
static void UnRegister(myClassB* ptr){
Get()->mPointers.erase(ptr);
}
static myClassB* Cast(myClassA* ptr){
if(Get()->mPointers.count((myClassB*)ptr) > 0) return (myClassB*)ptr;
return nullptr;
}
private:
std::set<myClassB*> mPointers;
};
class myClassB : public myClassA
{
public:
myClassB(){ ClassB_Registry::Register(this); }
~myClassB(){ ClassB_Registry::UnRegister(this); }
void setString1(std::string newString){privateMember = newString;}
std::string getS1() { return privateMember; }
private:
std::string privateMember;
};
std::shared_ptr<myClassA> functionA()
{
std::shared_ptr<myClassB> temporary(new myClassB());
temporary->setString1("TEST_1");
return std::move(temporary);
}
int main()
{
std::shared_ptr<myClassA> myPtr = functionA(); //??
std::shared_ptr<myClassA> myPtr_a(new myClassA()); //??
myClassB* pDerrived = ClassB_Registry::Cast(myPtr.get()); // bridge the RTTI gap
if(pDerrived)
std::cout << pDerrived->getS1();
pDerrived = ClassB_Registry::Cast(myPtr_a.get()); // works on A pointers to return null
if(pDerrived)
std::cout << pDerrived->getS1() << " \n";
else
std::cout << "Not A Pointer of Type B" << " \n";
}
It's not pretty, but if myClassB had a virtual table as mentioned previously, and all future derived classes used myClassB as the base, then you could bridge the gap for RTTI.
One of the main benefits of virtual in C++ is being able to use the base class (pointer or reference) to call derived methods.
I'm reading up on using CRTP to implement static polymorphism, but I can't understand how to achieve what I've mentioned above using this technique, because I can't declare a function as taking type Base when this requires a template.
It seems to me that what is described in the article could be achieved by simply using function overloading, so I'm sure that there must be more to this technique.
(PS: this exact problem is alluded to in a comment to an answer to this question, but unfortunately no one had replied to it: "What vtables truly provide is using the base class (pointer or reference) to call derived methods. You should show how it is done with CRTP here.")
Here is my minimal code, which gives the error "missing template arguments before ‘&’ token
void Print(Base& Object)".
#include <cstring>
#include <iostream>
template <typename Derived>
struct Base
{
std::string ToStringInterface() { return static_cast<Derived*>(this)->ToString(); }
std::string ToString() { return "This is Base."; }
};
struct Derived : Base<Derived>
{
std::string ToString() { return "This is Derived."; }
};
void Print(Base& Object)
{
std::cout << Object->ToStringInterface() << std::endl;
}
int main()
{
Derived MyDerived;
// This works, but could have been achieved with a function overload.
std::cout << MyDerived.ToStringInterface() << std::endl;
// This does not work.
Print(MyDerived);
}
Thanks to the comments and answers received, I'm posting my implementation, in case it may come in useful to anyone else.
#include <cstring>
#include <iostream>
template <typename Derived>
class Base
{
public:
std::string ToStringInterface()
{
return static_cast<Derived*>(this)->ToString();
}
};
template<>
class Base<void> : public Base<Base<void> >
{
public:
std::string ToString()
{
return "This is Base (default implementation).";
}
};
class Derived : public Base<Derived>
{
public:
std::string ToString()
{
return "This is Derived.";
}
};
template <typename T>
void Print(Base<T>& Object)
{
std::cout << Object.ToStringInterface() << std::endl;
}
int main()
{
int Decision;
std::cout << "Do you want to create an object of type Base (input 0) or Derived (input 1)? ";
std::cin >> Decision;
if (Decision == 0)
{
Base<void> MyBase;
Print(MyBase);
}
else
{
Derived MyDerived;
Print(MyDerived);
}
}
Well, you need to declare print a template function :
template<class T>
void Print(Base<T>& Object)
{
std::cout << Object.ToStringInterface() << std::endl;
}
Sorry, but it CRTP indeed doesn't work that way. The idea is typically to inject some code into the dependency hierarchy, in a way that is very specific to C++. In your example, you could have e.g. an interface that requires the ToStringInterface() function and use CRTP to bind it to an existing class hierarchy's ToString():
class IStringable
{
virtual string ToStringInterface() = 0;
};
class Unchangeable
{
virtual string ToString();
};
template<class Derived>
class UnchangeableToIStringableMixin
{
virtual string ToStringInterface()
{
return static_cast<Derived*>(this)->ToString();
}
};
class StringableUnchangeable:
public Unchangeable, UnchangeableToIStringableMixin<StringableUnchangeable>
{
};
However, if Unchangeable can actually be changed, you wouldn't do something like that. Don't forget to consider the possibility that CRTP just isn't the right tool for what you are doing.
Scenario: I have the following defined classes.
class Baseclass { };
class DerivedTypeA : public Baseclass { };
class DerivedTypeB : public Baseclass { };
// ... and so on ...
class Container
{
list<Baseclass*> stuff;
list<DerivedTypeA*> specific_stuff;
// ... initializing constructors and so on ...
public:
void add(Baseclass * b)
{
stuff.add(b);
}
void add(DerivedTypeA * a)
{
stuff.add(a);
specific_stuff.add(a);
}
};
class ContainerOperator
{
Container c;
// ... initializing constructors and so on ...
public:
void operateOnStuff(Baseclass * b)
{
// This will always use "void add(Baseclass * b)" no matter what object b really is.
c.add(b);
}
};
// ...
containerOperator.operateOnStuff(new DerivedTypeA());
So, what I want to do is to handle a certain derived class in some special way in Container.
Problem: void add(DerivedTypeA * a) is never called. I'm obviously doing something wrong. What is the correct way of doing what I am trying to achieve here?
Overload resolution in C++ happens at compile-time, not run-time. The "usual" way to solve problems like this is to use Visitor pattern.
You can reduce the amount of boilerplate copy-paste by implementing Visitor with CRTP.
If you use CRTP for Base::accept, you don't need to define it any more in derived classes.
Here is a similar program to yours, but a little simpler:
#include <iostream>
class Base; class Derived;
struct Operation {
void add(Base *b) {
std::cout << "Base\n";
}
void add(Derived *b) {
std::cout << "Derived\n";
}
void visit(Base *b); // need to define this after Base class
};
struct Base {
virtual ~Base() {}
virtual void accept(Operation &o)
{
o.add(this);
}
};
void Operation::visit(Base *b) {
b->accept(*this);
}
struct Derived : public Base {
void accept(Operation &o)
{
o.add(this);
}
};
int main() {
Operation o;
Base b;
Derived d;
Base *ptrb = &b;
Base *ptrd = &d;
o.add(ptrb); // These two print "Base"
o.add(ptrd);
o.visit(ptrb); // "Base"
o.visit(ptrd); // "Derived"
}
You can use RTTI to determine whether the provided object is of the derived type, and if so, call the second add() function.
void add(Baseclass * b)
{
stuff.add(b);
DerivedTypeA * a = dynamic_cast<DerivedTypeA *>(b);
if ( a != 0 )
specific_stuff.add(a);
}
Unlike the visitor pattern this solution violates the Open-Closed Principle, but it's a lot simpler and easier to understand when the number of derived classes do not change or change slowly over time.
I was thinking along the lines of using typeid() but I don't know how to ask if that type is a subclass of another class (which, by the way, is abstract)
class Base
{
public: virtual ~Base() {}
};
class D1: public Base {};
class D2: public Base {};
int main(int argc,char* argv[]);
{
D1 d1;
D2 d2;
Base* x = (argc > 2)?&d1:&d2;
if (dynamic_cast<D2*>(x) == nullptr)
{
std::cout << "NOT A D2" << std::endl;
}
if (dynamic_cast<D1*>(x) == nullptr)
{
std::cout << "NOT A D1" << std::endl;
}
}
You really shouldn't. If your program needs to know what class an object is, that usually indicates a design flaw. See if you can get the behavior you want using virtual functions. Also, more information about what you are trying to do would help.
I am assuming you have a situation like this:
class Base;
class A : public Base {...};
class B : public Base {...};
void foo(Base *p)
{
if(/* p is A */) /* do X */
else /* do Y */
}
If this is what you have, then try to do something like this:
class Base
{
virtual void bar() = 0;
};
class A : public Base
{
void bar() {/* do X */}
};
class B : public Base
{
void bar() {/* do Y */}
};
void foo(Base *p)
{
p->bar();
}
Edit: Since the debate about this answer still goes on after so many years, I thought I should throw in some references. If you have a pointer or reference to a base class, and your code needs to know the derived class of the object, then it violates Liskov substitution principle. Uncle Bob calls this an "anathema to Object Oriented Design".
You can do it with dynamic_cast (at least for polymorphic types).
Actually, on second thought--you can't tell if it is SPECIFICALLY a particular type with dynamic_cast--but you can tell if it is that type or any subclass thereof.
template <class DstType, class SrcType>
bool IsType(const SrcType* src)
{
return dynamic_cast<const DstType*>(src) != nullptr;
}
The code below demonstrates 3 different ways of doing it:
virtual function
typeid
dynamic_cast
#include <iostream>
#include <typeinfo>
#include <typeindex>
enum class Type {Base, A, B};
class Base {
public:
virtual ~Base() = default;
virtual Type type() const {
return Type::Base;
}
};
class A : public Base {
Type type() const override {
return Type::A;
}
};
class B : public Base {
Type type() const override {
return Type::B;
}
};
int main()
{
const char *typemsg;
A a;
B b;
Base *base = &a; // = &b; !!!!!!!!!!!!!!!!!
Base &bbb = *base;
// below you can replace base with &bbb and get the same results
// USING virtual function
// ======================
// classes need to be in your control
switch(base->type()) {
case Type::A:
typemsg = "type A";
break;
case Type::B:
typemsg = "type B";
break;
default:
typemsg = "unknown";
}
std::cout << typemsg << std::endl;
// USING typeid
// ======================
// needs RTTI. under gcc, avoid -fno-rtti
std::type_index ti(typeid(*base));
if (ti == std::type_index(typeid(A))) {
typemsg = "type A";
} else if (ti == std::type_index(typeid(B))) {
typemsg = "type B";
} else {
typemsg = "unknown";
}
std::cout << typemsg << std::endl;
// USING dynamic_cast
// ======================
// needs RTTI. under gcc, avoid -fno-rtti
if (dynamic_cast</*const*/ A*>(base)) {
typemsg = "type A";
} else if (dynamic_cast</*const*/ B*>(base)) {
typemsg = "type B";
} else {
typemsg = "unknown";
}
std::cout << typemsg << std::endl;
}
The program above prints this:
type A
type A
type A
dynamic_cast can determine if the type contains the target type anywhere in the inheritance hierarchy (yes, it's a little-known feature that if B inherits from A and C, it can turn an A* directly into a C*). typeid() can determine the exact type of the object. However, these should both be used extremely sparingly. As has been mentioned already, you should always be avoiding dynamic type identification, because it indicates a design flaw. (also, if you know the object is for sure of the target type, you can do a downcast with a static_cast. Boost offers a polymorphic_downcast that will do a downcast with dynamic_cast and assert in debug mode, and in release mode it will just use a static_cast).
I don't know if I understand your problem correctly, so let me restate it in my own words...
Problem: Given classes B and D, determine if D is a subclass of B (or vice-versa?)
Solution: Use some template magic! Okay, seriously you need to take a look at LOKI, an excellent template meta-programming library produced by the fabled C++ author Andrei Alexandrescu.
More specifically, download LOKI and include header TypeManip.h from it in your source code then use the SuperSubclass class template as follows:
if(SuperSubClass<B,D>::value)
{
...
}
According to documentation, SuperSubClass<B,D>::value will be true if B is a public base of D, or if B and D are aliases of the same type.
i.e. either D is a subclass of B or D is the same as B.
I hope this helps.
edit:
Please note the evaluation of SuperSubClass<B,D>::value happens at compile time unlike some methods which use dynamic_cast, hence there is no penalty for using this system at runtime.
I disagree that you should never want to check an object's type in C++. If you can avoid it, I agree that you should. Saying you should NEVER do this under any circumstance is going too far though. You can do this in a great many languages, and it can make your life a lot easier. Howard Pinsley, for instance, showed us how in his post on C#.
I do a lot of work with the Qt Framework. In general, I model what I do after the way they do things (at least when working in their framework). The QObject class is the base class of all Qt objects. That class has the functions isWidgetType() and isWindowType() as a quick subclass check. So why not be able to check your own derived classes, which is comparable in it's nature? Here is a QObject spin off of some of these other posts:
class MyQObject : public QObject
{
public:
MyQObject( QObject *parent = 0 ) : QObject( parent ){}
~MyQObject(){}
static bool isThisType( const QObject *qObj )
{ return ( dynamic_cast<const MyQObject*>(qObj) != NULL ); }
};
And then when you are passing around a pointer to a QObject, you can check if it points to your derived class by calling the static member function:
if( MyQObject::isThisType( qObjPtr ) ) qDebug() << "This is a MyQObject!";
#include <stdio.h>
#include <iostream.h>
class Base
{
public: virtual ~Base() {}
template<typename T>
bool isA() {
return (dynamic_cast<T*>(this) != NULL);
}
};
class D1: public Base {};
class D2: public Base {};
class D22: public D2 {};
int main(int argc,char* argv[]);
{
D1* d1 = new D1();
D2* d2 = new D2();
D22* d22 = new D22();
Base* x = d22;
if( x->isA<D22>() )
{
std::cout << "IS A D22" << std::endl;
}
if( x->isA<D2>() )
{
std::cout << "IS A D2" << std::endl;
}
if( x->isA<D1>() )
{
std::cout << "IS A D1" << std::endl;
}
if(x->isA<Base>() )
{
std::cout << "IS A Base" << std::endl;
}
}
Result:
IS A D22
IS A D2
IS A Base
I was thinking along the lines of using typeid()...
Well, yes, it could be done by comparing: typeid().name(). If we take the already described situation, where:
class Base;
class A : public Base {...};
class B : public Base {...};
void foo(Base *p)
{
if(/* p is A */) /* do X */
else /* do Y */
}
A possible implementation of foo(Base *p) would be:
#include <typeinfo>
void foo(Base *p)
{
if(typeid(*p) == typeid(A))
{
// the pointer is pointing to the derived class A
}
else if (typeid(*p).name() == typeid(B).name())
{
// the pointer is pointing to the derived class B
}
}
I see some good answers here and I see some dumb response.
"Trying to query the type of an object is a design flaw". Which means that Java instanceof and C# is keywords are design flaws. These are response of people that dont rate polymorphism. If you have an interface, that interface is derived by another interface that impelments more features. If you need these extra features you must first check that you have such an interface. Even microsoft COM API makes use of this design.
Then in terms of how to deduce if an object is a instanceof a class, many good answers have already been given
typeid
having a virtual type function
dynamic cast
is_base_of has nothing to do with polymorphism.
And having each virtual function define its own type method is unnecessary as it is redundant. Each virtual class already has a pointer to its virtual table.
class Base
{
void *p_virtual_table = BASE_VIRTUAL_TABLE;
}
class Derived : Base
{
void *p_virtual_table = DERIVED_VIRTUAL_TABLE;
}
void *BASE_VIRTUAL_TABLE[n];
void *DERIVED_VIRTUAL_TABLE[n];
The point here is that the address of the virtual tables are fixed and a simple comparrison will decide if a virtual object is an instanceof a virtual class.
Since cpp doesnt give us a standard way of accessing the virtual tables, it would be hard to do these comparrisons manually. But the cpp abstract machine has absolutely no problems deducing the exact instance of a virtual object.
You can only do it at compile time using templates, unless you use RTTI.
It lets you use the typeid function which will yield a pointer to a type_info structure which contains information about the type.
Read up on it at Wikipedia
You can do it with templates (or SFINAE (Substitution Failure Is Not An Error)). Example:
#include <iostream>
class base
{
public:
virtual ~base() = default;
};
template <
class type,
class = decltype(
static_cast<base*>(static_cast<type*>(0))
)
>
bool check(type)
{
return true;
}
bool check(...)
{
return false;
}
class child : public base
{
public:
virtual ~child() = default;
};
class grandchild : public child {};
int main()
{
std::cout << std::boolalpha;
std::cout << "base: " << check(base()) << '\n';
std::cout << "child: " << check(child()) << '\n';
std::cout << "grandchild: " << check(grandchild()) << '\n';
std::cout << "int: " << check(int()) << '\n';
std::cout << std::flush;
}
Output:
base: true
child: true
grandchild: true
int: false
As a spin off of multiple other answers (including one I previously posted myself!), here's a macro to help:
#define isInstance( ptr, clazz ) (dynamic_cast<const clazz*>(ptr) != NULL)
In c# you can simply say:
if (myObj is Car) {
}