I'm having a little problem with inheritance in C++ that I can't quite figure out.
So, say I have some classes derived from stl list, i.e:
class Class1: public list<T>{
virtual func1();
}
class Class2 : public Class1<w/e>{
func1();
}
The problem I'm having is when passing these to a function. How can I properly pass an instance of one of these to a function so that it will use the proper virtual function if it is possible for an instance of either type to be passed? I haven't done inheritance and virtual functions in awhile so I am a bit rusty here.
(This assuming that the function is not a member function of the classes).
You will need to pass them to the function as a reference
int function(Class1& obj);
or as a pointer;
int function(Class1* obj);
and it should just plain call the correct version of the virtual method.
If you pass them without either of those
int function(Class1 obj);
The object will be copy constructed to always be a Class1 in the function, and will always call Class1::func1()
Related
Here are my code
#include <iostream>
using namespace std;
class BaseClass
{
public:
BaseClass() {}
void init(const int object) { cout<<"BaseClass::init"<<endl; }
void run(const int object) { cout<<"BaseClass::run calls =>"; init(object); }
};
class Derived : public BaseClass {
public:
Derived() {}
void init(const int object) { cout<<"Derived::init"<<endl; }
};
int main() {
BaseClass b;
b.init('c');
b.run('c');
Derived d;
d.init(5); // Calls Derived::init
d.run(5); // Calls Base::init. **I expected it to call Derived::init**
}
And here is generated output
BaseClass::init
BaseClass::run calls =>BaseClass::init
Derived::init
BaseClass::run calls =>BaseClass::init
With call d.run(5), Why "BaseClass::init" is being called instead of "Derived::init" ?
I though we need virtual functions only when calling through a pointer.
What is the rationale behind keeping such behavior ?
Why "BaseClass::init" is being called instead of "Derived::init" ?
Because init is a non-virtual member function. To have the desired effect, you need to make init a virtual member function as shown below:
class BaseClass
{
public:
BaseClass() {}
//NOTE THE VIRTUAL KEYWORD HERE
virtual void init(const int object) { cout<<"BaseClass::init"<<endl; }
//other member function here as before
};
Demo
I though we need virtual functions only when calling through a pointer.
Note that the statement init(object); is equivalent to writing
this->init(object); //here `this` is a pointer
When you wrote:
d.run(5);
In the above statement, first the address of object d is implicitly passed as the first argument to the implicit this parameter of member function run. The type of this implicit this parameter is BaseClass* and this happens due to derived to base conversion. Now, the call init(object); is equivalent to this->init(object);. But since init is a non-virtual member function, the call is resolved at compile time meaning the base class init will be called.
Basically when a member function is called with a derived class object, the compiler first looks to see if that member exists in the derived class. If not, it begins walking up the inheritance chain and checking whether the member has been defined in any of the parent classes. It uses the first one it finds. This means for the call d.run(5) the search for a member function named run starts inside the derived class. But since there is no function named run inside derived class, the compiler looks into the direct base class BaseClass and finds the member function named run. So it stops its search and uses this found run member function. And as i said, in your example, init is non-virtual and so the call is resolved at compile time to the base class run.
On the other hand, if we make init to be a virtual member function, then this call will be resolved at run-time meaning the derived class init will be called.
I though we need virtual functions only when calling through a pointer.
No. Thats not right. You need to make a function virtual when you want to enable calling it based on the dynamic type of the object. And once you do have a virtual function you can use pointers or references to make use of the virtual dispatch.
You need to make a method virtual when you want to override it. As BaseClass::run is not overridden by Derived::run, there is no virtual dispatch and calling init from BaseClass::run calls BaseClass::init. If you want to enable virtual dispatch for BaseClass::init you need to declare it virtual.
Further, consider that your code is equivalent to:
void run(const int object) {
cout<<"BaseClass::run calls =>";
this->init(object);
}
this is BaseClass*, hence BaseClass::init is called. You are calling init via a pointer, but that does not matter because init is not virtual. If you want to call init based on the dynamic type of the object, thats exactly what virtual is good for. The "overhead" you refer to in comments is not really overhead, but just the minimum needed to get the behavior you want. If you can change the design and do not need runtime polymorphism you can take a look at CRTP wich is a form of static polymorphism.
I have a template method in a parent class, that should call another method from the base class. It works, if the method is explicitly defined in the base class, but it doesn't work if the method is inherited. I can't figure out what's exactly wrong with this code (although I know it's a little weird :)
class A
{
protected:
virtual void someMethod()
{
}
template <class TBaseClass>
void templateMethod()
{
TBaseClass::someMethod();
}
};
class B : public A
{
};
class C : public B
{
protected:
virtual void someMethod()
{
templateMethod<A>(); // this works
templateMethod<B>(); // this doesn't
}
};
This ends up with compiler error:
error C2352: 'A::someMethod' : illegal call of non-static member function
What exactly is wrong? I'm not looking for workarounds, that's easy. I'd like to know why is this incorrect.
In template <TBaseClass> void A::templateMethod() the invocant, this, is of type A *. So when you try to call B::someMethod on it, the compiler won't recognize it a object method call, because B is not a base class, but it can still be a static method call, so the compiler will try that, find B::someMethod inherited via A and complain it is not static. The fact that A is a base class of this is not relevant; only that B is not.
The issue is that class A is not automatically granted access to class B. The methods within the class don't know that the current instance of A is actually of type B.
To do what you're trying, you need to cast A to the target type:
template <class TBaseClass>
void templateMethod()
{
static_cast<TBaseClass*>(this)->someMethod();
}
This will work if the particular instance of A you're calling really is a B (as it is in the example you give). There is no type checking done here to be sure it is. If you pass a class that is not in the inheritance hierarchy you will be in the work of undefined behavior.
That said, your example has "someMethod()" as virtual. If you're doing what I just did here then making that method virtual may not make sense, as you are explicitly saying which instance you want. If you do leave it virtual, then just calling someMethod() directly in A will get you the most derived instance.
I think it's not working because you are trying to call a static method.
Static methods can be inherited but they can't be virtual.
Thats why A::someMethod will work and B::someMethod won"t work.
I have a function like this:
void something(void *obj)
{
obj->Set();
}
The compiler says that left of the dereference operator has to be a pointer to a class/struct/union/generic (translated from german not sure about wording).
The idea is that I want to call the something function of obj no matter what is passed to something. It is ensured that it has this function. How can i achieve that?
--EDIT--
I started to work on an existing Software which has like > 100 Classes for datatypes. In one part of the code there is a big switch statement which depending on an id creates an instance of one of these classes and calls the Set function for that one. Now i want to do multiple of these calls parallel, and because of this i want to bring the ->Set() call to a seperate function which i then can call in a new thread. Sadly there is no baseclass and i cant change too much in the "big picture". What is the best way to do this?
C++ doesn’t allow this (for good reasons: even if you can ensure that the object always has a function, C++ cannot, and since you can make mistakes, C++ is justified in distrusting you).
The proper way to do this is to have a common base class which defined this method for all types that you want to use here, and then use this common base class as the argument of this function.
Alternatively, if it’s known at compile time which type is used here, then the appropriate implementation uses templates:
template <typename T>
void f(T const& obj) {
obj.something();
}
Whatever you do, void* is not appropriate. There are very rare legitimate use-cases for it in C++.
You need a base class or interface for whatever is passed into doSth:
class Base
{
public:
virtual void something() = 0; //override this in derived classes
}
doSth(Base* obj)
{
obj->something();
}
You can also cast the void* back to the original type:
doSth(void* obj)
{
((Base*)obj)->something();
}
but passing a void* as parameter suggests a faulty design. What exactly are you trying to achieve?
You need to implement pure virtual Base class with this function:
class Base
{
public:
virtual ~Base(){}
virtual void somefunction()=0;
}
class Derived1: public Base
{
public:
void somefunction()
{
//do something
}
}
class Derived2: public Base
{
public:
void somefunction()
{
//do something
}
}
And than use dynmic cast to get Base* from void*
doSth(void *obj)
{
Base *bobj=dynamic_cast<Base*>(obj);
if ( bobj )
bobj->somefunction();
}
Or mor simplier:
doSth(Base *obj)
{
obj->somefunction();
}
And usage is like:
Base *p1 = new Derived1();
Base *p2 = new Derived2();
doSth(p1); // cals somefunction in Derived1 class
doSth(p2); // cals somefunction in Derived2 class
The doSth method could take a function pointer as a parameter.
doSth( (*someFunc)() ) {
obj->*someFunc();
}
The call would look like:
doSth( &function );
When passing function pointers between different classes you should create a typedef for each function pointer and use qualifiers for each function identifier.
Just define an interface that lists all the functions of all the objects that you want to reference by the pointer, but the type of this pointer should not be void, but the name of this interface instead.
Then you will be able to call every function of every object that you want by this pointer, but make sure that all structures and classes of the objects implement all the functions of the interface!
This is also important to write the : public and then the name of the interface in the header of every structure and class!
I'm writing a little program in C++ and I what to play with polymorphism.
So the code goes like this:
//Base.h
class Base{
public:
void method(const objectType1& stuff1);
//objectType1 is whatever type you what (int, string, ...)
protected:
Base(void);
virtual ~Base(void);
virtual void method(void) = 0;
};
//Derived.h
/*FINAL - DO NOT INHERIT THIS CLASS OR YOU WILL die :)*/
class Derived : public Base{
public:
Derived(void);
~Derived(void);
private:
void method(void);
};
//Main.cpp
int main(){
objectType1 ot1;
//this code works
Base *bd = new Derived;
bd->method(ot1);
//this dosen't
Derived *dd = new Derived;
dd->method(ot1);
// he doesn't call Base::method(const objectType1& stuff1),
// he calls Derived::method(void)
return 0;
}
I've solved the problem by renaming the virtual method void method(void) in something else and all is well.
My questions are:
Why doesn't the compiler know to call the method from the base class?
And why do I see void method(void) in main since it is declared as protected in Base and private in Derived ? (is this some sort of side effect)
Thanks :) .
The member method in the base class is public, which means that any piece of code can call it. Besides being public, it is virtual, which means that the execution will be dynamically dispatched at runtime to the final overrider.
Access to the method is checked statically in the static type of the reference/pointer through which the call is dispatched, even if it will be dynamically dispatched.
On the other hand in the second case, the call is through the derived type, and the access specifier is checked at that level, where the compiler finds the member function to be private and thus complains.
In plain english, when the object is used as a base, it behaves as a base, but when used directly it behaves as derived.
The method definition in Derived hides the definition in Base. To avoid that, add using Base::method; to Derived.
This happens because when calling functions in Derived, the compiler looks into Base only if it didn't fine the name in Derived. Note that it doesn't look for the signature but only for the name in that stage. Overload resolution based on the signature is done only afterwards. This is actually not much different to the following situation:
void function(int);
void function();
int main()
{
void function(); // this declaration hides *both* global declarations
function(3); // error
}
In your case Base plays the role of the global scope and Derived the role of the function scope.
Also note that this is not related to virtaul fgunctions; indeed the same would happen without virtual (except that you then couldn't call Derived's method via a pointer to Base, of course, but would call the Base version instead).
could anyone explain function overriding in c++ please! also am confused about virtual function concept. some tutorials say without the keyword virtual, both the derived class and base class objects invoke the base class function. then how does overriding take place?
There's no need to redocument what is already good and out there. I think this and this are a wonderful explanations about virtual functions. Then you probably also need to hear about abstract classes. Let us know if you have further questions.
Read this please. Concentrate on C++ sections. Then ask specific questions you have after reading.
Let me try to post an example (This is out of my head, so there may be slight syntax errors :) )
Baseclass:
class BaseClass
{
public:
void normalFunction();
virtual void virtualFunction();
}
Derived class:
class DerivedClass : public BaseClass
{
public:
void normalFunction();
virtual void virtualFunction();
}
Okay, so we got our classes defined. Now, some examples:
void main()
{
BaseClass base;
DerivedClass derived;
base.normalFunction(); //Invokes BaseClass::normalFunction();
base.virtualFunction(); //Invoked BaseClass::virtualFunction();
derived.normalFunction();//Invokes DerivedClass::normalFunction();
derived.virtualFunction();//Invokes DerivedClass::virtualFunction();
// Okay, nothing special yet, here comes the fun:
BaseClass *basePtr = &base;
BaseClass *derivedPtr = &derived;
basePtr->normalFunction(); //Invokes BaseClass::normalFunction();
basePtr->virtualFunction();//Invokes BaseClass::virtualFunction();
derivedPtr->normalFunction(); //Invokes BaseClass::normalFunction(); !! this is because it's a BaseClass pointer.
derivedPtr->virtualFunction();//Invokes DerivedClass::virtualFunction();
}
.. So, in conclusion, without virtual, the type of the pointer dictates which method will be invoked, with virtual, any type overriding the virtual method will have it's method called regardless of the pointer type :)
This is at a cose of a very minor overhead in the form of a vtable (virtual table), a compiler-detail which will map each method to the different derived types.
Each C++ class that has at least one virtual function contains a "virtual table" or VTABLE which is used to dynamically look up the address of a function at runtime.
Suppose you have two classes: Base and Derived. Further suppose that Derived derives from Base, and that "b" and "d" are instances of Derived, but b's compile-time type is Base and d's is Derived.
Now suppose that both Base and Derived declare a function "foo". Now, if "foo" is declared to be virtual in Base, then "foo" will have an entry in Base's VTABLE, and when you call "b.foo()" or "d.foo()", the compiler will know that it is a virtual function, and it will inject code that will look up the address of "foo" in the VTABLE at runtime.... which will find the address of the definition of "foo" as given in class Derived (i.e. Derived::foo).
Now suppose that both Base and Derived declare a function "foo", but "foo" has not been declared virtual in Base. When you call "b.foo()" or "d."foo()" the compiler will attempt to call Base::foo and Derived::foo directly, bypassing the virtual table lookup. Even though b's runtime type may be Derived, b's compile-time type is Base, and so calling "b.foo()" will result in "Base::foo" rather than "Derived::foo" being called.
In C++, the term "overriding" is used to refer to the former case; that is, when the function is declared as virtual, and another function in a derived class replaces the original function that was defined in the base class. By contrast, the term "overshadowing" is used to refer to the latter case; that is, a function in the derived class is called instead of the one in the base class, simply because it is closer in scope, but it has not truly replaced the function that was defined in the base class.
The override can only be done if you declare the function in the super virtual. Because it is virtual so it's not true ... meaning that is someone invoke it with that signature it may not invoke that function. We have to wait and see at runtime if someone overridden it. That is the meaning of virtual.
class A {
void F_A() { cout << "A' A"; }
virtual void F_B() { cout << "A' B"; }
}
class B : public A {
void F_B() { cout << "B' B"; }
}
A o = new B();
o.F_A(); // Non-virtual so the compiler knows that it can only be the one in class A
o.F_B(); // Virtual so the compiler does not know if at runtime ... o is instance of A or B. So it have to wait and see.
// In this case, it's B at runtime ('`new B()`'), so it run the one in B.
To sum up, if a function (method to be more precise) is declared 'virtual', it can be overridden. Otherwise, it can't be; hence, any invocation always goes to the one in the super class.
Hope this help clarifying.
For Virtual function always remember this thumb rule :
When function is normal then function
of type of object will be invoked,
When function is Virtual then function
of type of actual instance will be
called.
I like to illustrate this example with a Chess board;
class ChessPiece
{
public:
void moveTo(Pos dst);
virtual checkValidAndMoveTo(Pos dst) = 0;
};
class King: public ChessPieve
{ virtual checkValidAndMoveTo(Pos dst);}
class Queen: public ChessPieve
{ virtual checkValidAndMoveTo(Pos dst);}
// etc
ChessPiece* board[8][8];
So when a move is made to square the game will call checkValidAndMoveTo() on the piece.
board[pickedUpPiece.x][pickedUpPiece.y]->checkValidAndMoveTo()
This will now call the appropriate checkValidAndMoveTo() for the particular piece. Once this move is finished we expect it to call MoveTo(). Since at the point it knows what type is is it will drill to down and get the most overidden version of MoveTo() below it current type.