Maybe I am wrong, but this seems to be a very basic question. Suddenly my inheritance chain stopped working. Writing a small basic test application proved that it was me that was wrong (so I can't blame the compiler).
I have a base class, with the default behavior in a virtual function. A child class derives from that and changes the behavior.
#include <iostream>
class Base
{
public:
Base() { print(); }
~Base() {}
protected:
virtual void print() { std::cout << "base\n"; }
};
class Child : public Base
{
public:
Child() {}
~Child() {}
protected:
virtual void print() { std::cout << "child\n"; }
};
int main()
{
Base b;
Child c;
}
This prints:
base
base
When a Child instance is created, why is Base::print() called? I thought that by using the virtual keyword, the function can be replaced for the derived class.
At what point did I get myself confused?
You are calling a virtual method in the constructor, which is not going to work, as the child class isn't fully initialized yet.
See also this StackOverflow question.
Although your current problem is the virtual method call from a constructor that others have mentioned, I noticed that you didn't make the destructors virtual. That's normally a bad thing. In your case, the destructors are nops and there are no members that are objects with destructors,... but if your code changes then it's easy for bad things to happen.
Related
This question already has answers here:
Private virtual method in C++
(5 answers)
Closed 2 years ago.
considering below example
#include <iostream>
#include <string>
class A
{
public:
virtual void foo() { std::cout<< "FOO A\n"; }
private:
void bar() { std::cout<< "BAR A\n"; }
virtual void vbar() { std::cout<< "VBAR A\n"; }
};
class B : public A
{
public:
void foo() { std::cout<< "FOO B\n"; bar(); vbar(); }
private:
void bar() { std::cout<< "BAR B\n"; }
virtual void vbar() { std::cout<< "VBAR B\n"; }
};
int main()
{
A* b = new B();
b->foo();
}
The output will give us
FOO B
BAR B
VBAR B
Since its simple example first that come to my mind I cant figure out any private virtual method use case. In case of public virtual method, the base pointer class interface will adapt to its defined vtable, but as in given example for private virtuals it doesnt matter
One possible use is for letting a base class define a structure, and having derived classes implement the behaivour of the components of said structure (the template method pattern). For example,
struct foo
{
void do_stuff() {
// defines order in which some operations are executed
do_op1();
do_op1();
do_op3();
}
private:
// These don't have to be pure virtual. A base,
// default implementation could also be provided.
virtual void do_op1() = 0;
virtual void do_op2() = 0;
virtual void do_op3() = 0;
};
// implements the operations
struct foo1 : foo
{
private:
void do_op1() override { ... }
void do_op2() override { ... }
void do_op3() override { ... }
};
The virtual methods are private because it does not make sense to call them in isolation. The base class knows when and how to call them.
There are probably simpler and better ways of implementing this in "modern C++", but this kind of thing might have been seen in the 90s and 00s.
There are situations where it might be useful, some argue, that it should be the prefered method, when possible, like Herb Sutter:
Guideline #2: Prefer to make virtual functions private.
...This lets the derived classes override the function to customize the behavior as needed, without further exposing the virtual functions directly by making them callable by derived classes (as would be possible if the functions were just protected). The point is that virtual functions exist to allow customization; unless they also need to be invoked directly from within derived classes' code, there's no need to ever make them anything but private. But sometimes we do need to invoke the base versions of virtual functions (see the article "Virtually Yours"[5] for an example), and in that case only it makes sense to make those virtual functions protected, thus:
Guideline #3: Only if derived classes need to invoke the base implementation of a virtual function, make the virtual function protected...
http://www.gotw.ca/publications/mill18.htm
So I recently accidentally called some virtual functions from the constructor of a base class, i.e. Calling virtual functions inside constructors.
I realise that I should not do this because overrides of the virtual function will not be called, but how can I achieve some similar functionality? My use-case is that I want a particular function to be run whenever an object is constructed, and I don't want people who write derived classes to have to worry about what this is doing (because of course they could call this thing in their derived class constructor). But, the function that needs to be called in-turn happens to call a virtual function, which I want to allow the derived class the ability to override if they want.
But because a virtual function gets called, I can't just stick this function in the constructor of the base class and have it get run automatically that way. So I seem to be stuck.
Is there some other way to achieve what I want?
edit: I happen to be using the CRTP to access other methods in the derived class from the base class, can I perhaps use that instead of virtual functions in the constructor? Or is much the same issue present then? I guess perhaps it can work if the function being called is static?
edit2: Also just found this similar question: Call virtual method immediately after construction
If really needed, and you have access to the factory.
You may do something like:
template <typename Derived, typename ... Args>
std::unique_ptr<Derived> Make(Args&&... args)
{
auto derived = std::make_unique<Derived>(std::forward<Args>(args));
derived->init(); // virtual call
return derived;
}
There is no simple way to do this. One option would be to use so-called virtual constructor idiom, hide all constructors of the base class, and instead expose static 'create' - which will dynamically create an object, call your virtual override on it and return (smart)pointer.
This is ugly, and what is more important, constrains you to dynamically created objects, which is not the best thing.
However, the best solution is to use as little of OOP as possible. C++ strength (contrary to popular belief) is in it's non-OOP specific traits. Think about it - the only family of polymorphic classess inside standard library are streams, which everybody hate (because they are polymorphic!)
I want a particular function to be run whenever an object is constructed, [... it] in-turn happens to call a virtual function, which I want to allow the derived class the ability to override if they want.
This can be easily done if you're willing to live with two restrictions:
the constructors in the entire class hierarchy must be non-public, and thus
a factory template class must be used to construct the derived class.
Here, the "particular function" is Base::check, and the virtual function is Base::method.
First, we establish the base class. It has to fulfill only two requirements:
It must befriend MakeBase, its checker class. I assume that you want the Base::check method to be private and only usable by the factory. If it's public, you won't need MakeBase, of course.
The constructor must be protected.
https://github.com/KubaO/stackoverflown/tree/master/questions/imbue-constructor-35658459
#include <iostream>
#include <utility>
#include <type_traits>
using namespace std;
class Base {
friend class MakeBase;
void check() {
cout << "check()" << endl;
method();
}
protected:
Base() { cout << "Base()" << endl; }
public:
virtual ~Base() {}
virtual void method() {}
};
The templated CRTP factory derives from a base class that's friends with Base and thus has access to the private checker method; it also has access to the protected constructors in order to construct any of the derived classes.
class MakeBase {
protected:
static void check(Base * b) { b->check(); }
};
The factory class can issue a readable compile-time error message if you inadvertently use it on a class not derived from Base:
template <class C> class Make : public C, MakeBase {
public:
template <typename... Args> Make(Args&&... args) : C(std::forward<Args>(args)...) {
static_assert(std::is_base_of<Base, C>::value,
"Make requires a class derived from Base");
check(this);
}
};
The derived classes must have a protected constructor:
class Derived : public Base {
int a;
protected:
Derived(int a) : a(a) { cout << "Derived() " << endl; }
void method() override { cout << ">" << a << "<" << endl; }
};
int main()
{
Make<Derived> d(3);
}
Output:
Base()
Derived()
check()
>3<
If you take a look at how others solved this problem, you will notice that they simply transferred the responsibility of calling the initialization function to client. Take MFC’s CWnd, for instance: you have the constructor and you have Create, a virtual function that you must call to have a proper CWnd instantiation: “these are my rules: construct, then initialize; obey, or you’ll get in trouble”.
Yes, it is error prone, but it is better than the alternative: “It has been suggested that this rule is an implementation artifact. It is not so. In fact, it would be noticeably easier to implement the unsafe rule of calling virtual functions from constructors exactly as from other functions. However, that would imply that no virtual function could be written to rely on invariants established by base classes. That would be a terrible mess.” - Stroustrup. What he meant, I reckon, is that it would be easier to set the virtual table pointer to point to the VT of derived class instead of keep changing it to the VT of current class as your constructor call goes from base down.
I realise that I should not do this because overrides of the virtual function will not be called,...
Assuming that the call to a virtual function would work the way you want, you shouldn't do this because of the invariants.
class B // written by you
{
public:
B() { f(); }
virtual void f() {}
};
class D : public B // written by client
{
int* p;
public:
D() : p( new int ) {}
void f() override { *p = 10; } // relies on correct initialization of p
};
int main()
{
D d;
return 0;
}
What if it would be possible to call D::f from B via VT of D? You will use an uninitialized pointer, which will most likely result in a crash.
...but how can I achieve some similar functionality?
If you are willing to break the rules, I guess that it might be possible to get the address of desired virtual table and call the virtual function from constructor.
Seems you want this, or need more details.
class B
{
void templateMethod()
{
foo();
bar();
}
virtual void foo() = 0;
virtual void bar() = 0;
};
class D : public B
{
public:
D()
{
templateMethod();
}
virtual void foo()
{
cout << "D::foo()";
}
virtual void bar()
{
cout << "D::bar()";
}
};
Let's say I have the following code:
class BaseMember
{
};
class DerivedMember : public BaseMember
{
};
class Base
{
private:
BaseMember* mpMember;
protected:
virtual BaseMember* initializeMember(void)
{
return new BaseMember[1];
}
virtual void cleanupMember(BaseMember* pMember)
{
delete[] pMember;
}
public:
Base(void)
: mpMember(NULL)
{
}
virtual ~Base(void)
{
cleanupMember(mpMember);
}
BaseMember* getMember(void)
{
if(!mpMember)
mpMember = initializeMember();
return mpMember;
}
};
class Derived : public Base
{
protected:
virtual BaseMember* initializeMember(void)
{
return new DerivedMember;
}
virtual void cleanupMember(BaseMember* pMember)
{
delete pMember;
}
};
Base and BaseMember are parts of an API and may be subclassed by the user of that API, like it is done via Derived and DerivedMember in the example code.
Base initializes mpBaseMember by a call to it's virtual factory function initializeMember(), so that the derived class can override the factory function to return a DerivedMember instance instead of a BaseMember instance.
However, when calling a virtual function from within a base class constructor, the base implementation and not the derived class override gets called.
Therefor I am waiting with the initialization of mpMember until it gets accessed for the first time (which of course implies, that the base class and any derived class, that may could get derived further itself, are not allowed to access that member from inside the constructor).
Now the problem is: Calling a virtual member function from within the base base destructor will result in a call of the base class implementation of that function, not of the derived class override.
That means that I can't simply call cleanupMember() from within the base class destructor, as that would call it's base class implementation, which may not be able to correctly cleanup the stuff, that the derived implementation of initializeMember() has initialized.
For example the base class and the derived class could use incompatible allocators that may result in undefined behavior when getting mixed (like in the example code - the derived class allocates the member via new, but the base class uses delete[] to deallocate it).
So my question is, how can I solve this problem?
What I came up with is:
a) the user of the API has to explicitly call some cleanup function before the Derived instance gets destructed. That can likely be forgotten.
b) the destructor of the (most) derived class has to call a cleanup function to cleanup stuff which initialization has been triggered by the base class. That feels ugly and not well designed as ownership responsibilities are mixed up: base class triggers allocation, but derived class has to trigger deallocation, which is very counter-intuitive and can't be known by the author of the derived class unless he reads the API documentation thoroughly enough to find that information.
I would really like to do this in a more fail-proof way than relying on the users memory or his reliability to thoroughly read the docs.
Are there any alternative approaches?
Note: As the derived classes may not exist at compile time of the base classes, static polymorphism isn't an option here.
What about a modification of the factory pattern that would include the cleanup method? Meaning, add a attribute like memberFactory, an instance of a class providing creation, cleanup, as well as access to the members. The virtual initialization method would provide and initialize the right factory, the destructor ~Base would call the cleanup method of the factory and destruct it.
(Well, this is quite far from the factory pattern... Perhaps it is known under another name?)
If you really want to do this sort of thing you can do it like this:
class Base {
BaseMember* mpMember;
protected:
Base(BaseMember *m) : mpMember(m) {}
virtual void doCleanupMember(BaseMember *m) { delete [] m; }
void cleanupMember() {
// This gets called by every destructor and we only want
// the first call to do anything. Hopefully this all gets inlined.
if (mpMember) {
doCleanupMember(pmMember);
mpMember = nullptr;
}
}
public:
Base() : mpMember(new BaseMember[1]) { }
virtual ~Base(void) { cleanupMember(); }
};
class Derived : public Base {
virtual void doCleanupMember(BaseMember *m) override { delete m; }
public:
Derived() : Base(new DerivedMember) {}
~Derived() { cleanupMember(); }
};
However there are reasons this is a bad idea.
First is that the member should be owned an exclusively managed by Base. Trying to divide up responsibility for Base's member into the derived classes is complicated and just asking for trouble.
Secondly the ways you're initializing mpMember mean that the member has a different interface depending on who initialized it. Part of the problem you've already run into is that the information on who initialized the member has been destroyed by the type you get to ~Base(). Again, trying to have different interfaces for the same variable is just asking for trouble.
We can at least fix the first problem by using something like shared_ptr which lets up specify a deleter:
class Base {
std::shared_ptr<BaseMember> mpMember;
public:
Base(std::shared_ptr<BaseMember> m) : mpMember(m) { }
Base() : mpMember(std::make_shared<BaseMember>()) { }
virtual ~Base() {}
};
class Derived : virtual public Base {
public:
Derived()
: Base(std::shared_ptr<BaseMember>(new DerivedMember[1],
[](BaseMember *m){delete [] m;} ) {}
};
This only hides the difference in the destruction part of the member's interface. If you had an array of more elements the different users of the member would still have to be able to figure out if mpMember[2] is legal or not.
First of all, you must use RAII idiom when developing in C++. You must free all your resources in destructor, of course if you don't wish to fight with memory leaks.
You can create some cleanupMember() function, but then you should check your resources in destructor and free them if they are not deleted (as cleanupMember can be never called, for example because of an exception). So add destructor to your derived class:
virtual ~Derived()
{
Derived::cleanupMember(mpMember);
}
and manage the member pointer in the class itself.
I also recommend you to use smart pointers here.
Never never never call virtual methods in constructor/destructor because it makes strange results (compiler makes dark and weird things you can't see).
Destructor calling order is child and then parent
You can do like this (but there is probalby a better way) :
private:
// private destructor for prevent of manual "delete"
~Base() {}
public:
// call this instead use a manual "delete"
virtual void deleteMe()
{
cleanupMember(mpMember);
delete this; // commit suicide
}
More infos about suicide :
https://stackoverflow.com/a/3150965/1529139 and http://www.parashift.com/c++-faq-lite/delete-this.html
PS : Why destructor is virtual ?
Let mpMember be protected and let it be initialized in derived class constructor and deallocated in derived destructor.
Inspired by the ideas from https://stackoverflow.com/a/19033431/404734 I have come up with a working solution :-)
class BaseMember
{
};
class DerivedMember : public BaseMember
{
};
class BaseMemberFactory
{
public:
virtual ~BaseMemberFactory(void);
virtual BaseMember* createMember(void)
{
return new BaseMember[1];
}
virtual void destroyMember(BaseMember* pMember)
{
delete[] pMember;
}
};
class DerivedMemberFactory : public BaseMemberFactory
{
virtual BaseMember* createMember(void)
{
return new DerivedMember;
}
virtual void destroyMember(BaseMember* pMember)
{
delete pMember;
}
};
class Base
{
private:
BaseMemberFactory* mpMemberFactory;
BaseMember* mpMember;
protected:
virtual BaseMemberFactory* getMemberFactory(void)
{
static BaseMemberFactory fac;
return &fac;
}
public:
Base(void)
: mpMember(NULL)
{
}
virtual ~Base(void)
{
mpMemberFactory->destroyMember(mpMember);
}
BaseMember* getMember(void)
{
if(!mpMember)
{
mpMemberFactory = getMemberFactory();
mpMember = mpMemberFactory->createMember();
}
return mpMember;
}
};
class Derived : public Base
{
protected:
virtual BaseMemberFactory* getMemberFactory(void)
{
static DerivedMemberFactory fac;
return &fac;
}
};
Wikipedia has the following example on the C++11 final modifier:
struct Base2 {
virtual void f() final;
};
struct Derived2 : Base2 {
void f(); // ill-formed because the virtual function Base2::f has been marked final
};
I don't understand the point of introducing a virtual function and immediately marking it as final. Is this simply a bad example, or is there more to it?
Typically final will not be used on the base class' definition of a virtual function. final will be used by a derived class that overrides the function in order to prevent further derived types from further overriding the function. Since the overriding function must be virtual normally it would mean that anyone could override that function in a further derived type. final allows one to specify a function which overrides another but which cannot be overridden itself.
For example if you're designing a class hierarchy and need to override a function, but you do not want to allow users of the class hierarchy to do the same, then you might mark the functions as final in your derived classes.
Since it's been brought up twice in the comments I want to add:
One reason some give for a base class to declare a non-overriding method to be final is simply so that anyone trying to define that method in a derived class gets an error instead of silently creating a method that 'hides' the base class's method.
struct Base {
void test() { std::cout << "Base::test()\n"; }
};
void run(Base *o) {
o->test();
}
// Some other developer derives a class
struct Derived : Base {
void test() { std::cout << "Derived::test()\n"; }
};
int main() {
Derived o;
o.test();
run(&o);
}
Base's developer doesn't want Derived's developer to do this, and would like it to produce an error. So they write:
struct Base {
virtual void test() final { ... }
};
Using this declaration of Base::foo() causes the definition of Derived to produce an error like:
<source>:14:13: error: declaration of 'test' overrides a 'final' function
void test() { std::cout << "Derived::test()\n"; }
^
<source>:4:22: note: overridden virtual function is here
virtual void test() final { std::cout << "Base::test()\n"; }
^
You can decide if this purpose is worthwhile for yourself, but I want to point out that declaring the function virtual final is not a full solution for preventing this kind of hiding. A derived class can still hide Base::test() without provoking the desired compiler error:
struct Derived : Base {
void test(int = 0) { std::cout << "Derived::test()\n"; }
};
Whether Base::test() is virtual final or not, this definition of Derived is valid and the code Derived o; o.test(); run(&o); behaves exactly the same.
As for clear statements to users, personally I think just not marking a method virtual makes a clearer statement to users that the method is not intended to be overridden than marking it virtual final. But I suppose which way is clearer depends on the developer reading the code and what conventions they are familiar with.
For a function to be labelled final it must be virtual, i.e., in C++11 §10.3 para. 2:
[...] For convenience we say that any virtual function overrides itself.
and para 4:
If a virtual function f in some class B is marked with the virt-specifier final and in a class D derived from
B a function D::f overrides B::f, the program is ill-formed. [...]
i.e., final is required to be used with virtual functions (or with classes to block inheritance) only. Thus, the example requires virtual to be used for it to be valid C++ code.
EDIT: To be totally clear: The "point" asked about concerns why virtual is even used. The bottom-line reason why it is used is (i) because the code would not otherwise compile, and, (ii) why make the example more complicated using more classes when one suffices? Thus exactly one class with a virtual final function is used as an example.
It doesn't seem useful at all to me. I think this was just an example to demonstrate the syntax.
One possible use is if you don't want f to really be overrideable, but you still want to generate a vtable, but that is still a horrible way to do things.
Adding to the nice answers above - Here is a well-known application of final (very much inspired from Java). Assume we define a function wait() in a Base class, and we want only one implementation of wait() in all its descendants. In this case, we can declare wait() as final.
For example:
class Base {
public:
virtual void wait() final { cout << "I m inside Base::wait()" << endl; }
void wait_non_final() { cout << "I m inside Base::wait_non_final()" << endl; }
};
and here is the definition of the derived class:
class Derived : public Base {
public:
// assume programmer had no idea there is a function Base::wait()
// error: wait is final
void wait() { cout << "I am inside Derived::wait() \n"; }
// that's ok
void wait_non_final() { cout << "I am inside Derived::wait_non_final(); }
}
It would be useless (and not correct) if wait() was a pure virtual function. In this case: the compiler will ask you to define wait() inside the derived class. If you do so, it will give you an error because wait() is final.
Why should a final function be virtual? (which is also confusing) Because (imo) 1) the concept of final is very close to the concept of virtual functions [virtual functions has many implementations - final functions has only one implementation], 2) it is easy to implement the final effect using vtables.
I don't understand the point of introducing a virtual function and immediately marking it as final.
The purpose of that example is to illustrate how final works, and it does just that.
A practical purpose might be to see how a vtable influences a class' size.
struct Base2 {
virtual void f() final;
};
struct Base1 {
};
assert(sizeof(Base2) != sizeof(Base1)); //probably
Base2 can simply be used to test platform specifics, and there's no point in overriding f() since it's there just for testing purposes, so it's marked final. Of course, if you're doing this, there's something wrong in the design. I personally wouldn't create a class with a virtual function just to check the size of the vfptr.
While refactoring legacy code (e.g. removing a method that is virtual from a mother class), this is useful to ensure none of the child classes are using this virtual function.
// Removing foo method is not impacting any child class => this compiles
struct NoImpact { virtual void foo() final {} };
struct OK : NoImpact {};
// Removing foo method is impacting a child class => NOK class does not compile
struct ImpactChildClass { virtual void foo() final {} };
struct NOK : ImpactChildClass { void foo() {} };
int main() {}
Here is why you might actually choose to declare a function both virtual and final in a base class:
class A {
void f();
};
class B : public A {
void f(); // Compiles fine!
};
class C {
virtual void f() final;
};
class D : public C {
void f(); // Generates error.
};
A function marked final has to be also be virtual. Marking a function final prevents you from declaring a function with the same name and signature in a derived class.
Instead of this:
public:
virtual void f();
I find it useful to write this:
public:
virtual void f() final
{
do_f(); // breakpoint here
}
protected:
virtual void do_f();
The main reason being that you now have a single place to breakpoint before dispatching into any of potentially many overridden implementations. Sadly (IMHO), saying "final" also requires that you say "virtual."
I found another case where for virtual function is useful to be declared as final. This case is part of SonarQube list of warnings. The warning description says:
Calling an overridable member function from a constructor or destructor could result in unexpected behavior when instantiating a subclass which overrides the member function.
For example:
- By contract, the subclass class constructor starts by calling the parent class constructor.
- The parent class constructor calls the parent member function and not the one overridden in the child class, which is confusing for child class' developer.
- It can produce an undefined behavior if the member function is pure virtual in the parent class.
Noncompliant Code Example
class Parent {
public:
Parent() {
method1();
method2(); // Noncompliant; confusing because Parent::method2() will always been called even if the method is overridden
}
virtual ~Parent() {
method3(); // Noncompliant; undefined behavior (ex: throws a "pure virtual method called" exception)
}
protected:
void method1() { /*...*/ }
virtual void method2() { /*...*/ }
virtual void method3() = 0; // pure virtual
};
class Child : public Parent {
public:
Child() { // leads to a call to Parent::method2(), not Child::method2()
}
virtual ~Child() {
method3(); // Noncompliant; Child::method3() will always be called even if a child class overrides method3
}
protected:
void method2() override { /*...*/ }
void method3() override { /*...*/ }
};
Compliant Solution
class Parent {
public:
Parent() {
method1();
Parent::method2(); // acceptable but poor design
}
virtual ~Parent() {
// call to pure virtual function removed
}
protected:
void method1() { /*...*/ }
virtual void method2() { /*...*/ }
virtual void method3() = 0;
};
class Child : public Parent {
public:
Child() {
}
virtual ~Child() {
method3(); // method3() is now final so this is okay
}
protected:
void method2() override { /*...*/ }
void method3() final { /*...*/ } // this virtual function is "final"
};
virtual + final are used in one function declaration for making the example short.
Regarding the syntax of virtual and final, the Wikipedia example would be more expressive by introducing struct Base2 : Base1 with Base1 containing virtual void f(); and Base2 containing void f() final; (see below).
Standard
Referring to N3690:
virtual as function-specifier can be part of decl-specifier-seq
final can be part of virt-specifier-seq
There is no rule having to use the keyword virtual and the Identifiers with special meaning final together. Sec 8.4, function definitions (heed opt = optional):
function-definition:
attribute-specifier-seq(opt) decl-specifier-seq(opt) declarator virt-specifier-seq(opt) function-body
Practice
With C++11, you can omit the virtual keyword when using final. This compiles on gcc >4.7.1, on clang >3.0 with C++11, on msvc, ... (see compiler explorer).
struct A
{
virtual void f() {}
};
struct B : A
{
void f() final {}
};
int main()
{
auto b = B();
b.f();
}
PS: The example on cppreference also does not use virtual together with final in the same declaration.
PPS: The same applies for override.
I think most of the answers miss an important point. final means no more override after it has been specified. Marking it on a base class is close to pointless indeed.
When a derived class might get derived further, it can use final to lock the implementation of a given method to the one it provided.
#include <iostream>
class A {
public:
virtual void foo() = 0;
virtual void bar() = 0;
};
class B : public A {
public:
void foo() final override { std::cout<<"B::foo()"<<std::endl; }
void bar() override { std::cout<<"B::bar()"<<std::endl; }
};
class C : public B {
public:
// can't do this as B marked ::foo final!
// void foo() override { std::cout<<"C::foo()"<<std::endl; }
void bar() override { std::cout<<"C::bar()"<<std::endl; }
};
Update: This issue is caused by bad memory usage, see solution at the bottom.
Here's some semi-pseudo code:
class ClassA
{
public:
virtual void VirtualFunction();
void SomeFunction();
}
class ClassB : public ClassA
{
public:
void VirtualFunction();
}
void ClassA::VirtualFunction()
{
// Intentionally empty (code smell?).
}
void ClassA::SomeFunction()
{
VirtualFunction();
}
void ClassB::VirtualFunction()
{
// I'd like this to be called from ClassA::SomeFunction()
std::cout << "Hello world!" << endl;
}
The C# equivalent is as follows: Removed C# example, as it's not relevant to the actual problem.
Why isn't the ClassB::VirtualFunction function being called when called from ClassA::SomeFunction? Instead ClassA::VirtualFunction is being called...
When I force implementation of the virtual function ClassA::VirtualFunction, like so:
class ClassA
{
public:
virtual void VirtualFunction() = 0;
void SomeFunction();
}
class ClassB : public ClassA
{
public:
void VirtualFunction();
}
void ClassA::SomeFunction()
{
VirtualFunction();
}
void ClassB::VirtualFunction()
{
// I'd like this to be called from ClassA::SomeFunction()
std::cout << "Hello world!" << endl;
}
The following error occurs at runtime, despite the derrived function deffinately being declared and defined.
pure virtual method called
terminate called without an active exception
Note: It seems like the error can be caused even by bad memory usage. See self-answer for details.
Update 1 - 4:
Comments removed (not releavnt).
Solution:
Posted as an answer.
class Base {
public:
virtual void f() { std::cout << "Base" << std::endl; }
void call() { f(); }
};
class Derived : public Base {
public:
virtual void f() { std::cout << "Derived" << std::endl; }
};
int main()
{
Derived d;
Base& b = d;
b.call(); // prints Derived
}
If in the Base class you do not want to implement the function you must declare so:
class Base {
public:
virtual void f() = 0; // pure virtual method
void call() { f(); }
};
And the compiler won't allow you to instantiate the class:
int main() {
//Base b; // error b has a pure virtual method
Derived d; // derive provides the implementation: ok
Base & b=d; // ok, the object is Derived, the reference is Base
b.call();
}
As a side note, be careful not to call virtual functions from constructors or destructors as you might get unexpected results.
If you're getting that 'pure virtual method called
terminate called without an active exception' error message, that means you're calling the virtual function from the constructor or destructor of classA (the base class), which you should not do.
on the pure virtual method called error:
You should create a different question as it is in fact different than the other. The answer to this question is on the very last paragraph of my previous answer to your initial question:
Do not call virtual functions from constructors or destructors
class Base
{
public:
Base() { f(); }
virtual void f() = 0;
};
class Derived : public Base
{
public:
virtual void f() {}
};
int main()
{
Derived d; // crashes with pure virtual method called
}
The problem in the code above is that the compiler will allow you to instantiate an object of type Derived (as it is not abstract: all virtual methods are implemented). The construction of a class starts with the construction of all the bases, in this case Base. The compiler will generate the virtual method table for type Base, where the entry for f() is 0 (not implemented in base). The compiler will execute the code in the constructor then. After the Base part has completely been constructed, construction of the Derived element part starts. The compiler will change the virtual table so that the entry for f() points to Derived::f().
If you try calling the method f() while still constructing Base, the entry in the virtual method table is still null and the application crashes.
When A calls VirtualFunction() it will automatically call the version on B. That is the point of virtual functions.
I am not as familiar with the C++ syntax tho. Do you have to declare the function to be virtual at the point of the body as well as in the header?
Alsop, in class B you probably need to mark it as override
in C# its easy. I just don't know the c++ syntax.
public class ClassA
{
public **virtual** void VirtualFunction(){}
public void FooBar()
{
// Will call ClassB.VirtualFunction()
VirtualFunction();
}
}
public class ClassB
{
public **overide** void VirtualFunction()
{
// hello world
}
}
If you want to force the derived classes to implement the VirtualFunction:
class ClassA
{
public:
virtual void VirtualFunction()=0;
void SomeFunction();
}
This is C++. Default the derived function will be called.
If you want to call the base-class function do:
void ClassA::SomeFunction()
{
// ... various lines of code ...
ClassA::VirtualFunction();
}
There's nothing wrong with your code but your sample is incomplete. You do not state where you are calling SomeFunction from.
As has already been pointed out by dribeas you must be careful calling virtual functions from your constructor as the virtual tables are only built up as each class in the hierarchy completes construction.
Edit: The following paragraph of my reply was incorrect. Apologies. It is fine to call SomeFunction from the constructor of ClassB as the vtable is in place (at least) by the end of the initialiser list i.e. once you are in the body of the constructor. It is not fine to call it from ClassA's constructor of course.
Original paragraph:
I suspect you must be calling SomeFunction from the constructor of ClassB at which point only the vtable up to type ClassA will be complete i.e. to the virtual dispatch mechanism your class is still of type ClassA. It only becomes an object of type ClassB when the constructor completes.
To call a virutal function function you need to call via a pointer or a reference.
void ClassA::SomeFunction()
{
VirtualFunction(); // Call ClassA::VirtualFunction
this->VirtualFunction(); // Call Via the virtual dispatch mechanism
// So in this case call ClassB::VirtualFunction
}
You need to be able to distinguish the two different types of call otherwise the classA::VirtualFunction() becomes inaccessible when it is overridden.
As pointed out by others if you want to make the base class version abstract then use the = 0 rather than {}
class A
{
virtual void VirtualFunction() =0;
....
But sometimes it is legitimate to have an empty definition. This will depend on your exact usage.
You aren't defining the function in ClassB correctly, it should be:
public class ClassB
{
public void override AbstractFunction()
{
// hello world
}
}
Then, any call from the base class to virtual/abstract methods will call the implementation on the derived instance.