Dynamic binding inside constructor for virtual Function in C++ - c++

As per the standard, we know that constructor always go for an early binding of a virtual function inside them because they don't have complete idea the derived class hierarchy downside.
In this case if early binding is used inside my base constructor, I have passed a derived object to a base class pointer which is completely acceptable (an upcasting is done here). If early binding is used the selection of the virtual function should be based on the type of the pointer (which is Base * here) but not the content of the pointer(the object pointed by the pointer because we don't know the exact object being pointed). In that case since the pointer type is Base * we should have invoked only Base class virtual function in both cases. Could some one please clarify this?
I think dynamic binding is used here rather than early binding. Please correct me if my understanding is wrong.
The first line of the output which invokes base is completely fine
class Base
{
public:
Base(){
fun();
}
Base(Base *p)
{
p->fun();
}
virtual void fun()
{
cout<<"In Base"<<endl;
}
};
class Derived : public Base
{
public:
void fun()
{
cout<<"In Derived"<<endl;
}
};
int main()
{
Derived d;
Base b(&d);
}
O/P :
In Base
In Derived

The rule for virtual calls within a constructor body applies to an object currently being constructed, because that object is not yet considered to be an object of any derived class. (And there's a similar rule for an object currently being destroyed.) It has little to do with "early binding" in the sense of using a compile-time type, such as the syntax ClassName::member_func() forces.
Your code has two different objects d and b, and the constructor of d has entirely finished by the time you get to the p->fun(); line.
In detail:
The program enters main.
The object d is created using the implicitly declared default constructor of Derived.
The first thing Derived::Derived() does is create the base class subobject, by calling the default constructor Base::Base().
The body of Base::Base() calls fun(). Since we have not yet entered the body of the Derived::Derived() constructor, this virtual lookup invokes Base::fun().
Base::Base() finishes.
The body of Derived::Derived(), which is empty, executes and finishes.
Back in main, the object b is created by passing a pointer to d to the constructor Base::Base(Base*).
The body of Base::Base(Base *p) calls p->fun(). Since p is a pointer to d, which is already a completely constructed object of type Derived, this virtual lookup invokes Derived::fun().
Contrast with this slightly different example, where we define a default constructor of Derived to pass this (implicitly converted to Base*) to the constructor for the Base subobject. (This is valid, though could be risky if the pointer were used in other ways, such as in an initializer for a base or member of Base.)
#include <iostream>
using std::cout;
using std::endl;
class Base
{
public:
Base(){
fun();
}
Base(Base *p)
{
p->fun();
}
virtual void fun()
{
cout<<"In Base"<<endl;
}
};
class Derived
{
public:
Derived() : Base(this) {}
virtual void fun() override
{
cout << "In Derived" << endl;
}
};
int main()
{
Derived d;
}
This program will print just "In Base", since now in Base::Base(Base *p), p does point at the same object which is currently being constructed.

The reason is that C++ classes are constructed from Base classes to derived classes and virtual call table of the complete object is created when the object creation process is completed. Therefore, the base class function is called in the code excerpt above. Unless otherwise mandatory, you should never make virtual function calls in the constructor. Below is an excerpt from Bjarne Stroustrup's C++ Style and Technique FAQ:
In a constructor, the virtual call mechanism is disabled because
overriding from derived classes hasn’t yet happened. Objects are constructed from the base up, “base before derived”.
Destruction is done “derived class before base class”, so virtual
functions behave as in constructors: Only the local definitions are used –
and no calls are made to overriding functions to avoid touching the (now
destroyed) derived class part of the object.

Related

Problem with abstract class inheritance and constructor at C++

class Base {
public:
Base(int a) : a_(a) {
//do something
someMethod();
//do something else
};
protected:
int a_;
virtual void someMethod() = 0 {};
};
class Derived : Base {
public:
Derived() {
Base::Base(42);
}
protected:
void someMethod() override {
//realisation
}
};
int main() {
Derived *obj = new Derived();
delete obj;
}
This code doesn't work by two mistakes: base class's default constructor is needed and base class's constructor with parameters can't be called because of using abstract methods
My problem is that someMethod() realised in class Derived is not called at all when I create object of class Derived. Also I don't want to use default constructor of class Base, but compiler is swearing.
How can I correct my code to see functionality that I want?
How can I correct my code to see functionality that I want?
Remove the call to a pure virtual function in the constructor of Base.
Call someMethod in the constructor of the derived class that overrides it instead.
Provide an initialiser to the Base subobject in the member initiliser list. If you don't provide an initialiser to the base, it will be default initialised.
Why this cannot work ?
This design will not work because of the way obects are constructed.
When you construct a Derived, the first thing happening is that a Base object is constructed with a Base constructor. There is no Derived object at this moment, so if you'd invoke the virtual function in the Base constructor, it would be the virtual that would be valid for the Base class until you leave the Base constructor's body.
This is allowed by the standard, but with restrictions:
[base.class.init]/16: Member functions (including virtual member functions) can be called for an object under construction. (...) However,
if these operations are performed in a ctor-initializer (or in a
function called directly or indirectly from a ctor-initializer) before
all the mem-initializers for base classes have completed, the program
has undefined behavior.
These restriction do not apply to virtual functions that are called from the constructor body, since the body is executed after all initializers.
But in your case the virtual function is pure virtual for the Base. So it's UB according to the following clause:
[class.abstract]/6: Member functions can be called from a constructor (or destructor) of an abstract class; the effect of making
a virtual call to a pure virtual function directly or
indirectly for the object being created (or destroyed) from such a
constructor (or destructor) is undefined.
What is the alternative
There is unfortunately no other real alternative than using a two step initialization in which you first construct the object and then call an initialization function before using the object.
The only issue with this approach is the risk of forgetting the call to the initialization function. You can protect you against this:
using a flag to indicate if the intialization has taken place and check this flag in all member functions (yes, it's a little overhead).
if you have an abstract class, you may perhaps use a factory pattern (factory method or abstract factory). You could then let the factory do the call.
you may use the builder pattern, and make sure that the constructor is only visible to the builder who won't forget the initialization either.
Other problems with your code
You must be careful about how you "call" the base constructor from the derived constructor:
class Derived : Base {
public:
Derived() : Base(42) // this is the correct place !
{
//Base::Base(42); //<==== OUCH !!! NO !! This creates a temporary base object !!
}
...
};
You'd also need to be careful about the pure virtuals (I don't know if it's a typo or if your compiler could compile your code):
virtual void someMethod() = 0; // if it's abstract, no pending {} !

C++ 11 Delegated Constructor Pure Virtual Method & Function Calls -- Dangers?

Not a Duplicate of Invoking virtual function and pure-virtual function from a constructor:
Former Question relates to C++ 03, not new Constructor Delegation behavior in C++ 11, and the question does not address the mitigation of undefined behavior by using delegation to ensure proper construction before pure virtual implementations are executed.
In C++ 11, what are the dangers of invoking Pure Virtual functions in a class' constructor, during construction, but after the class/object has been "fully constructed" via constructor delegation?
Apparently, somewhere in the C++ 11 spec such a constraint exists,
Member functions (including virtual member functions, 10.3) can be
called for an object under construction. Similarly, an object under
construction can be the operand of the typeid operator ..
- 12.6.2 #13 of the [C++ Working Draft] (http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2011/n3242.pdf)
Can't find "Fair Use" version of Published Spec.
C++11 considers an object constructed once any constructor finishes
execution. Since multiple constructors will be allowed to execute,
this will mean that each delegate constructor will be executing on a
fully constructed object of its own type. Derived class constructors
will execute after all delegation in their base classes is complete.
- Wikipedia saying that this is a C++ 11 thing.
Actual C++ 11 Reference unknown.
Following Example Compiles AND RUNS in Nov CTP of Visual Studio 2012 C++ Compiler:
#include <string>
/**************************************/
class Base
{
public:
int sum;
virtual int Do() = 0;
void Initialize()
{
Do();
}
Base()
{
}
};
/**************************************/
// Optionally declare class as "final" to avoid
// issues with further sub-derivations.
class Derived final : public Base
{
public:
virtual int Do() override final
{
sum = 0 ? 1 : sum;
return sum / 2 ; // .5 if not already set.
}
Derived(const std::string & test)
: Derived() // Ensure "this" object is constructed.
{
Initialize(); // Call Pure Virtual Method.
}
Derived()
: Base()
{
// Effectively Instantiating the Base Class.
// Then Instantiating This.
// The the target constructor completes.
}
};
/********************************************************************/
int main(int args, char* argv[])
{
Derived d;
return 0;
}
With the updates, the example code looks okay to me, with the caveat that if you ever make a subclass of Derived, the subclass's override of Do() won't get called by Derived(const std::string &), rather Derived::Do() will still get called; which might not be what you wanted. In particular, when Initialize() is called from the Derived(const std::string &) constructor, the object is still "only" a Derived object and not a SubDerived object yet (because the SubDerived layer of construction-code hasn't started yet) and that is why Derived::Do() would be called and not SubDerived::Do().
Q: What if the subclass uses the same delegation pattern to ensure everything is instantiate in the same way?
A: That would mostly work, but only if it's okay for Derived::Do() to be called before SubDerived::Do() is called.
In particular, say you had class SubDerived that did the same things as Derived does above. Then when the calling code did this:
SubDerived foo("Hello");
the following sequence of calls would occur:
Base()
Derived()
Derived(const std::string &)
Base::Initialize()
Derived::Do()
SubDerived()
SubDerived(const std::string &)
Base::Initialize()
SubDerived::Do()
... so yes, SubDerived::Do() would eventually get called, but Derived::Do() would have been called also. Whether or not that will be a problem depends on what the various Do() methods actually do.
Some advice: Calling virtual methods from within a constructor is usually not the best way to go. You might want to consider simply requiring the calling code to call Do() manually on the object after the object is constructed. It's a bit more work for the calling code, but the advantage is that it you can avoid the not-very-obvious-or-convenient semantics that come into play when doing virtual method calls on partially-constructed objects.
In a typical single-constructor inheritance scenario, it is UB to call a pure virtual function in the base constructor:
[C++11: 10.4/6]: Member functions can be called from a constructor (or destructor) of an abstract class; the effect of making a virtual call (10.3) to a pure virtual function directly or indirectly for the object being created (or destroyed) from such a constructor (or destructor) is undefined.
struct Base
{
Base()
{
foo(); // UB
}
virtual void foo() = 0;
};
struct Derived : Base
{
virtual void foo() {}
};
There is no exemption here for such a call being made in a delegated constructor call, because at this point the more-derived part of the object still hasn't been constructed.
struct Base
{
Base()
{
foo(); // still UB
}
Base(int) : Base() {};
virtual void foo() = 0;
};
struct Derived : Base
{
virtual void foo() {}
};
Here's the Wikipedia passage that you cited:
C++11 considers an object constructed once any constructor finishes execution. Since multiple constructors will be allowed to execute, this will mean that each delegate constructor will be executing on a fully constructed object of its own type. Derived class constructors will execute after all delegation in their base classes is complete.
The key is the second bolded sentence, rather than the first, as one may misconstrue from a quick glance.
However, your posted code snippet is fine, and that's because the derived constructor body is undergoing execution, not the constructor for the abstract class, which has already been completely constructed. That said, the fact that you've had to ask for proof that it's safe should be some indication that this is not the most expressive or intuitive approach, and I would try to avoid it in your design.

C++ Type changes during derived class construction - virtual function question

C++ type change during deletion
I've read that when you construct a derived type, the type changes depending on which constructor is being called. So, if you create a derived object and call a virtual function using a base pointer, normally it would map to the implementation in the derived class. If you called the virtual function in the base class constructor though, it would use the base class implementation as the type of the object is technically that of the base class while in that function. For example (makeshift code, sorry if it doesn't compile):
class Base {
Base()
{
std::cerr << "Base Constructor.";
func();
}
virtual void func() {
std::cerr << "Func base called." << std::endl;
}
};
class Derived : public Base {
Derived()
{
std::cerr << "Derived Constructor.";
func();
}
void func() {
std::cerr << "Func derived called." << std::endl;
}
};
int main() {
Derived* d = new Derived;
delete d;
}
Should output:
Base Constructor.
Func base called.
Derived Constructor.
Func derived called.
First of all, is this always true or is it implementation dependent?
If I used RTTI and typeinfo, would the type printed in the base actually be that of the base, or is this more of an unwritten rule sort of situation?
Is it dangerous to call virtual functions from constructors with this in mind, or is it safe as long as you know what you're doing?
To keep it short and simple, you can have a Rule:
The virtual mechanism is disabled in Constructors and Destructors
A virtual function call in Base class will always call the base class version of the function, the same in derived class results in call to the Derived class version of the function.
First of all, is this always true or is it implementation dependent?
Yes this is always true. This is not implementation-dependent.
If I used RTTI and typeinfo, would the type printed in the base actually be that of the base?
Yes it would be of Base; Derived object doesn't even exist while you are in Base class constructor.
Is it dangerous to call virtual functions from constructors with this in mind, or is it safe as long as you know what you're doing?
No it is not dangerous to call virtual functions from constructor as long as you understand the semantics behind it.
This C++ FAQ should be a good read for you.
It's well-defined.
[n3290: 12.7/4]: Member functions, including virtual functions
(10.3), can be called during construction or destruction (12.6.2).
When a virtual function is called directly or indirectly from a
constructor or from a destructor, including during the construction or
destruction of the class’s non-static data members, and the object to
which the call applies is the object (call it x) under construction or
destruction, the function called is the final overrider in the
constructor’s or destructor’s class and not one overriding it in a
more-derived class. If the virtual function call uses an explicit
class member access (5.2.5) and the object expression refers to the
complete object of x or one of that object’s base class subobjects but
not x or one of its base class subobjects, the behavior is undefined.
There is an excellent article from Scott Meyers. It is from his book Effective C++.
The article can be found at:
Never Call Virtual Functions during Construction or Destruction
It also discusses an alternative implementation.
Recently I had a similar problem which I solved this way:
class EthernetFrame
{
protected:
/** ctor to be called from derived classes */
EthernetFrame(unsigned inPayloadLength)
{
calculatePadBytes(inPayloadLength);
}
private:
/** calculates needed required PadBytes for Frames < 64B
* #param inPayloadLength we need to know the length of the actual L3 frame
*/
void calculatePadBytes(unsigned inPayloadLength);
};
class IPv4Frame : public EthernetFrame
{
public:
/** create empty IPv4 packet */
IPv4Frame() :
EthernetFrame(cIPv4_MINIMUM_LENGTH)
{};
// IPv4 header + trailer in bytes
unsigned cIPv4_MINIMUM_LENGTH;
protected:
/** ctor to be called from derived classes */
IPv4Frame(unsigned inPayloadLength) :
EthernetFrame(cIPv4_MINIMUM_LENGTH+inPayloadLength)
{};
};

How to call a derived class method from a base class method within the constructor of base

I am wondering if it is possible to call a derived class´ function from within a function called by the base constructor (shouldn´t it already be created when the code in the brackets are executed?)
#pragma once
class ClassA
{
public:
ClassA(void);
virtual ~ClassA(void);
void Init();
protected:
short m_a;
short m_b;
virtual void SetNumbers(short s);
};
include "ClassA.h"
#include <iostream>
ClassA::ClassA(void) : m_a(0), m_b(0)
{
Init();
}
ClassA::~ClassA(void)
{
}
void ClassA::SetNumbers(short s)
{
std::cout << "In ClassA::SetNumbers()\n";
m_a = s;
m_b = s;
}
void ClassA::Init()
{
this->SetNumbers(2);
}
#pragma once
#include "ClassA.h"
class ClassB : public ClassA
{
public:
ClassB(void);
virtual ~ClassB(void);
virtual void SetNumbers(short);
int x;
};
#include "ClassB.h"
#include <iostream>
ClassB::ClassB(void)
{
}
ClassB::~ClassB(void)
{
}
void ClassB::SetNumbers(short s)
{
std::cout << "In ClassB::SetNumbers()\n";
m_a = ++s;
m_b = s;
ClassA::SetNumbers(s);
}
Any suggestions how to do it?...
Thank You in advance :)...
No. All parts of B (starting with A, as it's base) are constructed before B's constructor is called. So, by the time SetNumbers is called, no part of B (except for the A part) has been constructed --- and that may include the v-table, so there's no way to know where that call is going to go.
Of course, there is a simple solution to this: Call B::SetNumber() from within B's constructor (That is, after all, the purpose of B's constructor)
You can't do this for the simple logical reason that while the base class is being constructed, the derived class hasn't even begun to be constructed. You can't call a member function on an object that doesn't exist (yet).
In practice, even if you managed to call SetNumbers and assign to the member variables of the derived class before they were initialized they would surely be overwritten when they finally get initialized. I admit it's a bit pointless to reason about this as we would be well outside defined behaivour.
No, sorry. :( It might compile in one or two C++ compilers, but it's not recommended. From the C++ FAQ Lite section 10.7:
[10.7] Should you use the this pointer
in the constructor?
[...snip...]
Here is something that never works:
the {body} of a constructor (or a
function called from the constructor)
cannot get down to a derived class by
calling a virtual member function that
is overridden in the derived class. If
your goal was to get to the overridden
function in the derived class, you
won't get what you want. Note that you
won't get to the override in the
derived class independent of how you
call the virtual member function:
explicitly using the this pointer
(e.g., this->method()), implicitly
using the this pointer (e.g.,
method()), or even calling some other
function that calls the virtual member
function on your this object. The
bottom line is this: even if the
caller is constructing an object of a
derived class, during the constructor
of the base class, your object is not
yet of that derived class. You have
been warned.
NOTE: Emphasis mine.
More details at the link
The only time you can do this is when something is derived from a template that is parameterised by itself:
template<typename T> class base
{
T* down_cast() throw()
{
return static_cast<Derived*>(this);
}
const T* down_cast() const throw()
{
return static_cast<const Derived*>(this);
}
public:
base()
{
down_cast()->doSomething();
}
/* … */
};
class derived : private base<derived>
{
public:
void doSomething()
{
}
};
Note that doSomething is public and not virtual.
We can static_cast to derived, because it's known that derived is the derived type.
Deriving something from a base parameterised by itself is a strange thing to be doing at the best of times. It's said that when the ATL team in microsoft used it they asked the C++ compiler team if it was valid and nobody was sure, though it is valid because template construction depends on names as follows:
First the template is available, but not used in a class. Then, the name derived available. Then it instantiates the layout of base<derived> — this requires knowledge of the member variables and virtual functions, as long as none of that depends upon knowledge of derived’s layout (pointers and references are fine) this will all go okay. Then it will create the layout of derived, and finally it will create derived’s member functions, which may include creating member functions for base<derived>. So as long as base<derived> doesn’t contain a derived member variable (base classes can never contain a member variable of a type derived from themselves) or a virtual function that requires knowledge of derived’s layout we can indeed do the dicey-looking piece of inheritance above.
This includes being able to call non-virtual public members of derived from base during construction, because it's already part of base. There are strong limitations on this. In particular, if doSomething() depends on anything constructed in derived's constructor it won't work as derived hasn't been constructed yet.
Now, is this actually a good idea? No.
A simple design solution is to use aggregation instead of inheritance.

Why base class destructor (virtual) is called when a derived class object is deleted?

A difference between a destructor (of course also the constructor) and other member functions is that, if a regular member function has a body at the derived class, only the version at Derived class gets executed. Whereas in case of destructors, both derived as well as base class versions get executed?
It will be great to know what exactly happens in case of destructor (maybe virtual) & constructor, that they are called for all its base classes even if the most derived class object is deleted.
Thanks in advance!
The Standard says
After executing the body of the destructor and destroying any automatic objects allocated within the body,
a destructor for class X calls the destructors for X’s direct non-variant members,the destructors for X’s direct
base classes and, if X is the type of the most derived class (12.6.2), its destructor calls the destructors for
X’s virtual base classes. All destructors are called as if they were referenced with a qualified name, that is,
ignoring any possible virtual overriding destructors in more derived classes. Bases and members are destroyed
in the reverse order of the completion of their constructor (see 12.6.2). A return statement (6.6.3) in a
destructor might not directly return to the caller; before transferring control to the caller, the destructors
for the members and bases are called. Destructors for elements of an array are called in reverse order of
their construction (see 12.6).
Also as per RAII resources need to be tied to the lifespan of suitable objects and the destructors of respective classes must be called upon to release the resources.
For example the following code leaks memory.
struct Base
{
int *p;
Base():p(new int){}
~Base(){ delete p; } //has to be virtual
};
struct Derived :Base
{
int *d;
Derived():Base(),d(new int){}
~Derived(){delete d;}
};
int main()
{
Base *base=new Derived();
//do something
delete base; //Oops!! ~Base() gets called(=>Memory Leak).
}
Constructor and destructor are different from the rest of regular methods.
Constructor
can't be virtual
in derived class you either call explicitly constructor of base class
or, in case where you don't call base class constructor compiler will insert the call. It will call the base constructor without parameters. If no such constructor exists then you get compiler error.
struct A {};
struct B : A { B() : A() {} };
// but this works as well because compiler inserts call to A():
struct B : A { B() {} };
// however this does not compile:
struct A { A(int x) {} };
struct B : A { B() {} };
// you need:
struct B : A { B() : A(4) {} };
Destructor:
when you call destructor on derived class over a pointer or a reference, where the base class has virtual destructor, the most derived destructor will be called first and then the rest of derived classes in reversed order of construction. This is to make sure that all memory has been properly cleaned. It would not work if the most derived class was called last because by that time the base class would not exists in memory and you would get segfault.
struct C
{
virtual ~C() { cout << __FUNCTION__ << endl; }
};
struct D : C
{
virtual ~D() { cout << __FUNCTION__ << endl; }
};
struct E : D
{
virtual ~E() { cout << __FUNCTION__ << endl; }
};
int main()
{
C * o = new E();
delete o;
}
output:
~E
~D
~C
If the method in base class is marked as virtual all the inherited methods are virtual as well so even if you don't mark the destructors in D and E as virtual they will still be virtual and they still get called in the same order.
This is by design. The destructor on the base class must be called in order for it to release its resources. Rule of thumb is that a derived class should only clean up its own resources and leave the base class to clean up itself.
From C++ spec:
After executing the body of the
destructor and destroying any
automatic objects allocated within the
body, a destructor for class X calls
the destructors for X’s direct
members, the destructors for X’s
direct base classes and, if X is the
type of the most derived class
(12.6.2), its destructor calls the
destructors for X’s virtual base
classes. All destructors are called as
if they were referenced with a
qualified name, that is, ignoring any
possible virtual overriding
destructors in more derived classes.
Bases and members are destroyed in the
reverse order of the completion of
their constructor (see
12.6.2).
Also, because there is only one destructor, there is no ambiguity as to which destructor a class must call. This is not the case for constructors, where a programmer must pick which base class constructor should be called if there isn't an accessible default constructor.
Because that's how dtor's work. When you create an object, ctors are invoked starting from the base, and going all the way to the most derived. When you destroy objects (correctly) the reverse happens. The time that making a dtor virtual makes a difference is if/when you destroy an object via a pointer (or reference, though that's fairly unusual) to the base type. In that case, the alternative isn't really that only the derived dtor gets invoked -- rather, the alternative is simply undefined behavior. That make happen to take the form of invoking only the derived dtor, but it might take an entirely different form as well.
A base class destructor may be responsible for cleaning up resources that were allocated by the base class constructor.
If your base class has a default constructor (one that doesn't take parameters or has defaults for all its parameters) that constructor is automatically called upon construction of a derived instance.
If your base class has a constructor that requires parameters, you must call it manually in the initializer list of the derived class constructor.
Your base class destructor will always be automatically called upon deletion of the derived instance since destructors don't take parameters.
If you're using polymorphism and your derived instance is pointed to by a base class pointer, then the derived class destructor is only called if the base destructor is virtual.
As Igor says constructors must be called for base classes. Consider what would happen if it wouldn't be called:
struct A {
std::string s;
virtual ~A() {}
};
struct B : A {};
If the destructor for A would not be called when deleting a B instance, A would never be cleaned up.
When any object is destroyed, destructors run for all sub-objects. This includes both reuse by containment and reuse by inheritance.