48
votes

I have a base class MyBase that contains a pure virtual function:

void PrintStartMessage() = 0

I want each derived class to call it in their constructor

then I put it in base class(MyBase) constructor

 class MyBase
 {
 public:

      virtual void PrintStartMessage() =0;
      MyBase()
      {
           PrintStartMessage();
      }

 };

 class Derived:public MyBase
 {     

 public:
      void  PrintStartMessage(){

      }
 };

void main()
 {
      Derived derived;
 }

but I get a linker error.

 this is error message : 

 1>------ Build started: Project: s1, Configuration: Debug Win32 ------
 1>Compiling...
 1>s1.cpp
 1>Linking...
 1>s1.obj : error LNK2019: unresolved external symbol "public: virtual void __thiscall MyBase::PrintStartMessage(void)" (?PrintStartMessage@MyBase@@UAEXXZ) referenced in function "public: __thiscall MyBase::MyBase(void)" (??0MyBase@@QAE@XZ)
 1>C:\Users\Shmuelian\Documents\Visual Studio 2008\Projects\s1\Debug\s1.exe : fatal error LNK1120: 1 unresolved externals
 1>s1 - 2 error(s), 0 warning(s)

I want force to all derived classes to...

A- implement it

B- call it in their constructor 

How I must do it?

8
@peachykeen What could you do with a class without a constructor? You couldn't construct it!curiousguy
@peachykeen "You can derive from it." Yes. But since you can't construct it, can't construct any instance of a derived class either.curiousguy
@peachykeen Of course not. Who told you this nonsense? You can't create a derived instance without a base class constructor.curiousguy
@peachykeen "You most definitely can construct the derived classes," You cannot construct a derived class as your base class is lacking any constructor (by hypothesis). In order to construct a derived instance you need to construct a base instance first.curiousguy

8 Answers

41
votes

There are many articles that explain why you should never call virtual functions in constructor and destructor in C++. Take a look here and here for details what happens behind the scene during such calls.

In short, objects are constructed from the base up to the derived. So when you try to call a virtual function from the base class constructor, overriding from derived classes hasn't yet happened because the derived constructors haven't been called yet.

20
votes

Trying to call a pure abstract method from a derived while that object is still being constructed is unsafe. It's like trying to fill gas into a car but that car is still on the assembly line and the gas tank hasn't been put in yet.

The closest you can get to doing something like that is to fully construct your object first and then calling the method after:

template <typename T>
T construct_and_print()
{
  T obj;
  obj.PrintStartMessage();

  return obj;
}

int main()
{
    Derived derived = construct_and_print<Derived>();
}
12
votes

You can't do it the way you imagine because you cannot call derived virtual functions from within the base class constructor—the object is not yet of the derived type. But you don't need to do this.

Calling PrintStartMessage after MyBase construction

Let's assume that you want to do something like this:

class MyBase {
public:
    virtual void PrintStartMessage() = 0;
    MyBase() {
        printf("Doing MyBase initialization...\n");
        PrintStartMessage(); // ⚠ UB: pure virtual function call ⚠
    }
};

class Derived : public MyBase {
public:
    virtual void PrintStartMessage() { printf("Starting Derived!\n"); }
};

That is, the desired output is:

Doing MyBase initialization...
Starting Derived!

But this is exactly what constructors are for! Just scrap the virtual function and make the constructor of Derived do the job:

class MyBase {
public:
    MyBase() { printf("Doing MyBase initialization...\n"); }
};

class Derived : public MyBase {
public:
    Derived() { printf("Starting Derived!\n"); }
};

The output is, well, what we would expect:

Doing MyBase initialization...
Starting Derived!

This doesn't enforce the derived classes to explicitly implement the PrintStartMessage functionality though. But on the other hand, think twice whether it is at all necessary, as they otherwise can always provide an empty implementation anyway.

Calling PrintStartMessage before MyBase construction

As said above, if you want to call PrintStartMessage before the Derived has been constructed, you cannot accomplish this because there is no yet a Derived object for PrintStartMessage to be called upon. It would make no sense to require PrintStartMessage to be a non-static member because it would have no access to any of the Derived data members.

A static function with factory function

Alternatively we can make it a static member like so:

class MyBase {
public:
    MyBase() {
        printf("Doing MyBase initialization...\n");
    }
};

class Derived : public MyBase {
public:
    static void PrintStartMessage() { printf("Derived specific message.\n"); }
};

A natural question arises of how it will be called?

There are two solution I can see: one is similar to that of @greatwolf, where you have to call it manually. But now, since it is a static member, you can call it before an instance of MyBase has been constructed:

template<class T>
T print_and_construct() {
    T::PrintStartMessage();
    return T();
}

int main() {
    Derived derived = print_and_construct<Derived>();
}

The output will be

Derived specific message.
Doing MyBase initialization...

This approach does force all derived classes to implement PrintStartMessage. Unfortunately it's only true when we construct them with our factory function... which is a huge downside of this solution.

The second solution is to resort to the Curiously Recurring Template Pattern (CRTP). By telling MyBase the complete object type at compile time it can do the call from within the constructor:

template<class T>
class MyBase {
public:
    MyBase() {
        T::PrintStartMessage();
        printf("Doing MyBase initialization...\n");
    }
};

class Derived : public MyBase<Derived> {
public:
    static void PrintStartMessage() { printf("Derived specific message.\n"); }
};

The output is as expected, without the need of using a dedicated factory function.

Accessing MyBase from within PrintStartMessage with CRTP

While MyBase is being executed, its already OK to access its members. We can make PrintStartMessage be able to access the MyBase that has called it:

template<class T>
class MyBase {
public:
    MyBase() {
        T::PrintStartMessage(this);
        printf("Doing MyBase initialization...\n");
    }
};

class Derived : public MyBase<Derived> {
public:
    static void PrintStartMessage(MyBase<Derived> *p) {
        // We can access p here
        printf("Derived specific message.\n");
    }
};

The following is also valid and very frequently used, albeit a bit dangerous:

template<class T>
class MyBase {
public:
    MyBase() {
        static_cast<T*>(this)->PrintStartMessage();
        printf("Doing MyBase initialization...\n");
    }
};

class Derived : public MyBase<Derived> {
public:
    void PrintStartMessage() {
        // We can access *this member functions here, but only those from MyBase
        // or those of Derived who follow this same restriction. I.e. no
        // Derived data members access as they have not yet been constructed.
        printf("Derived specific message.\n");
    }
};

No templates solution—redesign

Yet another option is to redesign your code a little. IMO this one is actually the preferred solution if you absolutely have to call an overridden PrintStartMessage from within MyBase construction.

This proposal is to separate Derived from MyBase, as follows:

class ICanPrintStartMessage {
public:
    virtual ~ICanPrintStartMessage() {}
    virtual void PrintStartMessage() = 0;
};

class MyBase {
public:
    MyBase(ICanPrintStartMessage *p) : _p(p) {
        _p->PrintStartMessage();
        printf("Doing MyBase initialization...\n");
    }

    ICanPrintStartMessage *_p;
};

class Derived : public ICanPrintStartMessage {
public:
    virtual void PrintStartMessage() { printf("Starting Derived!!!\n"); }
};

You initialize MyBase as follows:

int main() {
    Derived d;
    MyBase b(&d);
}
6
votes

You shouldn't call a virtual function in a constructor. Period. You'll have to find some workaround, like making PrintStartMessage non-virtual and putting the call explicitly in every constructor.

1
votes

If PrintStartMessage() was not a pure virtual function but a normal virtual function, the compiler would not complain about it. However you would still have to figure out why the derived version of PrintStartMessage() is not being called.

Since the derived class calls the base class's constructor before its own constructor, the derived class behaves like the base class and therefore calls the base class's function.

0
votes

I know this is an old question, but I came across the same question while working on my program.

If your goal is to reduce code duplication by having the Base class handle the shared initialization code while requiring the Derived classes to specify the code unique to them in a pure virtual method, this is what I decided on.

#include <iostream>

class MyBase
{
public:
    virtual void UniqueCode() = 0;
    MyBase() {};
    void init(MyBase & other)
    {
      std::cout << "Shared Code before the unique code" << std::endl;
      other.UniqueCode();
      std::cout << "Shared Code after the unique code" << std::endl << std::endl;
    }
};

class FirstDerived : public MyBase
{
public:
    FirstDerived() : MyBase() { init(*this); };
    void  UniqueCode()
    {
      std::cout << "Code Unique to First Derived Class" << std::endl;
    }
private:
    using MyBase::init;
};

class SecondDerived : public MyBase
{
public:
    SecondDerived() : MyBase() { init(*this); };
    void  UniqueCode()
    {
      std::cout << "Code Unique to Second Derived Class" << std::endl;
    }
private:
    using MyBase::init;
};

int main()
{
    FirstDerived first;
    SecondDerived second;
}

The output is:

 Shared Code before the unique code
 Code Unique to First Derived Class
 Shared Code after the unique code

 Shared Code before the unique code
 Code Unique to Second Derived Class
 Shared Code after the unique code
0
votes

Facing the same problem, I imaginated a (not perfect) solution. The idea is to provide a certificate to the base class that the pure virtual init function will be called after the construction.

class A
{
  private:
    static const int checkValue;
  public:
    A(int certificate);
    A(const A& a);
    virtual ~A();
    virtual void init() = 0;
  public:
    template <typename T> static T create();
    template <typeneme T> static T* create_p();
    template <typename T, typename U1> static T create(const U1& u1);
    template <typename T, typename U1> static T* create_p(const U1& u1);
    //... all the required possibilities can be generated by prepro loops
};

const int A::checkValue = 159736482; // or any random value

A::A(int certificate)
{
  assert(certificate == A::checkValue);
}

A::A(const A& a)
{}

A::~A()
{}

template <typename T>
T A::create()
{
  T t(A::checkValue);
  t.init();
  return t;
}

template <typename T>
T* A::create_p()
{
  T* t = new T(A::checkValue);
  t->init();
  return t;
}

template <typename T, typename U1>
T A::create(const U1& u1)
{
  T t(A::checkValue, u1);
  t.init();
  return t;
}

template <typename T, typename U1>
T* A::create_p(const U1& u1)
{
  T* t = new T(A::checkValue, u1);
  t->init();
  return t;
}

class B : public A
{
  public:
    B(int certificate);
    B(const B& b);
    virtual ~B();
    virtual void init();
};

B::B(int certificate) :
  A(certificate)
{}

B::B(const B& b) :
  A(b)
{}

B::~B()
{}

void B::init()
{
  std::cout << "call B::init()" << std::endl;
}

class C : public A
{
  public:
    C(int certificate, double x);
    C(const C& c);
    virtual ~C();
    virtual void init();
  private:
    double x_;
};

C::C(int certificate, double x) :
  A(certificate)
  x_(x)
{}

C::C(const C& c) :
  A(c)
  x_(c.x_)
{}

C::~C()
{}

void C::init()
{
  std::cout << "call C::init()" << std::endl;
}

Then, the user of the class can't construct an instance without giving the certificate, but the certificate can only be produced by the creation functions:

B b = create<B>(); // B::init is called
C c = create<C,double>(3.1415926535); // C::init is called

Moreover, the user can't create new classes inheriting from A B or C without implementing the certificate transmission in the constructor. Then, the base class A has the warranty that init will be called after construction.

-1
votes

I can offer a work around / "companion" to your abstract base class using MACROS rather than templates, or staying purely within the "natural" constraints of the language.

Create a base class with an init function e.g.:

class BaseClass
{
public:
    BaseClass(){}
    virtual ~BaseClass(){}
    virtual void virtualInit( const int i=0 )=0;
};

Then, add a macro for a constructor. Note there is no reason to not add multiple constructor definitions here, or have multiple macros to choose from.

#define BASECLASS_INT_CONSTRUCTOR( clazz ) \
    clazz( const int i ) \
    { \
        virtualInit( i ); \
    } 

Finally, add the macro to your derivation:

class DervivedClass : public BaseClass
{
public:
    DervivedClass();
    BASECLASS_INT_CONSTRUCTOR( DervivedClass )
    virtual ~DervivedClass();

    void virtualInit( const int i=0 )
    {
        x_=i;
    }

    int x_;
};