Inheritance - C++ - c++

I am using a class say baseClass, from which I derive another class derivedClass. I have a problem definition that says, apart from others:
i) A member - object initialiser should be used to initialise a data member, say var1, that is declared in the base class.
ii) i) is done inside a base class constructor. It says, this has to be invoked only via a derived class constructor.
iii) The base class is an abstract class, whose objects cannot be created. But, I have a third class, inside which, I use:
baseClass *baseObjects[5];
The compiler does not report an error.
I do not understand, what i) and ii) really mean. An explanation in simple words would be fine. Also, any assistance on iii) is welcome.

Question 1:
Read about constructors : http://www.cprogramming.com/tutorial/constructor_destructor_ordering.html
Question 2:
Read about initialization list:
http://www.cprogramming.com/tutorial/initialization-lists-c++.html
Question 3:
Read about pointers to derived class:
http://www.learncpp.com/cpp-tutorial/121-pointers-and-references-to-the-base-class-of-derived-objects/
I think this way instead of just answering your question you could understand what's going on,

I think an illustration will be best.
i)
class A
{
int i;
public:
A(int ii) : i(ii) {}
}
The part i(ii) is an example of a member - object initialiser. Since C++ guarantees all constructors of members will be called before the constructor body is entered, this is your only way of specifying which constructor to call for each member.
ii) In C++ there is no super keyword. You must specify the base class as such:
class B : public A
{
public:
B(int i) : A(i) {}
}
That's partially due to the fact C++ allows multiple inheritence.
iii) Note that you haven't created any objects, only pointers to objects. And it's by this method polymorphism via inheritence is acheived in C++.

#include <iostream>
class Base
{
public:
Base(int i)
{}
virtual ~Base() = 0
{}
protected:
int i_;
};
class Derived: public Base
{
public:
Derived(int i, int j) : Base(i), j_(j)
{}
private:
int j_;
};
int main(int argc, char* argv[])
{
//Base b(1); object of abstract class is not allowed
Derived d(1, 2); // this is fine
}
As you can see i_ is being initalized by the Derived class by calling the Base class constructor. The = 0 on the destructor assures that the Base class is pure virtual and therefore we cannot instantiate it (see comment in main).

i) The following is what is known as an initializer list, you can use initializer lists to make sure that the data members have values before the constructor is entered. So in the following example, a has value 10 before you enter the constructor.
Class baseClass
{
int a;
public:
baseClass(int x):a(x)
{
}
}
ii) This is how you would explicitly call a base class constructor from a derived class constructor.
Class derivedClass : public baseClass
{
int a;
public:
derivedClass(int x):baseClass(x)
{
}
}
iii) You can't directly create instances of an abstract class. However, you can create pointers to an abstract base class and have those pointers point to any of its concrete implementations. So if you have an abstract base class Bird and concrete implementations Parrot and Sparrow then Bird* bird could point to either a Parrot or Sparrow instance since they are both birds.

Related

Problem with derived class in c++ with no data member

I'm learning c++ inheritance and facing problems with the following exercise to create a base class A and a derived class B with certain requirements. I have my answer written down below, but there seems to be some problems with it. I also have a few question at the end of this post.
class A {
private:
int x;
protected:
A (): x(0) { }
A (int n): x(n) { }
int get() const {return x;}
public:
virtual void foo() = 0;
};
class B : public A {
public:
B (): { A(); }
B (int n): { A(n); }
virtual void foo() { std::cout << get();}
};
My questions are:
I'm pretty sure my code is not correctly written, but can anyone tell me what's incorrect?
Since x is private in A, B wouldn't be able to inherit that data member. So how is B able to invoke the constructor?
I'm pretty sure that A is an abstract class, but is B an abstract class too?
It's almost okay, there's two thing that is wrong:
First you have an empty constructor initializer list in the B constructors. That will lead to build errors.
Then in the B constructor the statement A() creates a temporary A object, which is promptly discarded and destructed. You need to "call" the parent class constructor from the B constructor initializer list:
B() : A() { /* Empty */ }
You meed to do the same for the parameterized B constructor as well.
You can't access private members in the base-class, but protected is okay. That's now protected works with public inheritance: The child class can access the base class protected members.
Since you override foo with an implementation B is not abstract, there's no abstract members of B.

Can't access protected member variables of the most base class through std::unique_ptr in diamond

I am not an advanced programmer. Suppose there is a classic diamond inheritance:
class Base
class A: virtual public Base
class B: virtual public Base
class Last: public A, public B
Suppose Base has a variable, m_x, that is common to both A and B, such that only one of A, or B, can be called at a time, not both (which is what is needed). To get around this, this is used:
class Last: public A, public B
{
private:
std::unique_ptr<Base> m_p;
public:
Last(int i)
{
if (i)
m_p = std::unique_ptr<Base>(new A());
else
m_p = std::unique_ptr<Base>(new B());
}
};
This is fine, but now m_p->m_x cannot be accessed anymore because it says it's protected, but both A and B call m_x in their constructors directly, with no problems.
Is this a known limitation or is this the wrong way to do it? If it's wrong, what solutions are there?
Here is some code based on the diagram found here (a bit lower on the page):
#include <iostream>
#include <memory>
class Power
{
protected:
double m_x;
public:
Power() {}
Power(double x): m_x {x} {}
virtual ~Power() = default;
};
class Scanner: virtual public Power
{
public:
Scanner() {}
Scanner(double x): Power(x) {} // scan document
};
class Printer: virtual public Power
{
public:
Printer() {}
Printer(double x): Power(x) {} // print document
};
class Copier: public Scanner, public Printer
{
private:
std::unique_ptr<Power> m_p;
public:
Copier() {}
Copier(double x, int i)
{
if (i)
m_p = std::unique_ptr<Power>(new Scanner(x));
else
m_p = std::unique_ptr<Power>(new Printer(x));
}
void print() { std::cout << this->Power::m_x << '\n'; }
};
int main(int argc, char *argv[])
{
Copier *copier {new Copier(1.618, 0)};
copier->print();
copier = new Copier(3.14, 1);
copier->print();
return 0;
}
Using both this->m_p and this->Power::m_x (according to answers and comments) compiles, but the output is 0.
To be sure I spell it out all: not only I am quite a beginner, but, given the example above, it oesn't really have to stay that way if there is another alternative to call Scanner or Printer only one at a time from inside Copier. I am not asking for opinions, I understand it's forbidden, but I won't reject them coming from more experienced users. After all, I am learning.
Both virtual inheritance and std::unique_ptr are red herrings. The problem comes down to this:
class Base
{
protected:
int m_x;
};
class Last : public Base
{
public:
Last()
{
Base base;
base.m_x = 0; // error
m_x = 1; // no error
}
};
The error is something like error C2248: 'Base::m_x': cannot access protected member declared in class 'Base' or error: 'int Base::m_x' is protected within this context.
The explanation is that protected is a somewhat special case. It does not only work on class level but also on the object level. And you have two relevant objects here:
The Last object which is being created by the constructor, i.e. the one pointed to by this. It's also a Base object because of the is-a inheritance relationship.
The local object named base within the constructor.
Now, the problem is that in the line base.m_x = 0;, you are in the context of the first object and not the second one. In other words, you are trying to access the m_x of base from outside base. C++ simply does not allow this.
A very technical explanation can be found in the C++ standard at ยง11.4 [class.protected], a more easily understandable one in an excellent answer here on Stack Overflow.
protected doesn't mean quite what you think it does.
Although Last is derived from Base, member functions of Last don't have access to the protected members of any Base object - just those Base objects that are sub-objects of some Last object.
So you can write: this->Base::x because *this is a Last object, but not m_p->x, because *m_p is of static type Base.
As others have noted, I think this is actually an XY problem. Having an object which derives from two classes, and then also has a pointer to another object of one of those classes is very strange indeed. I think you need to clarify what you are trying to do.

Construction of polymorphic classes

I have a question about the way classes using inheritance are constructed. In the example below, the call to the Base constructor is dependent on a function implemented in the Derived class. This function is again dependent on the Derived member generator, which will not be initialized until after the Base constructor call.
If the Base class is constructed first, won't the variable Base::in_ contain garbage data?
class Derived
: public Base
{
Derived()
: Base(get_data()),
generator(5) {}
Generator generator;
int get_data() { return generator.get_some_data(); }
};
class Base
{
Base(int in)
: in_(in) {}
int in_;
}
First, nothing in your code is polymorphic. Polymorphism is about virtual functions.
Next, your class Base depends on nothing. Look between class Base { and the matching }, there's nothing in there that depends on anything outside.
What happens is the construction of the Base subobject within Derived depends on another member of Derived, which is constructed after Base. Your analysis is basically correct, it's a real problem and it needs to be solved by refactoring your classes.
The easiest way is two-stage initialization of Base (the following is pseudocode and probably will not compile):
class Base {
Base(int in = 0) : in_(in) {}
void set(int in) { in_ = in; }
int in_;
};
class Derived : public Base {
Derived() : Base(), generator(5) {
Base::set(generator);
}
...
};
Another method is moving the dependency to another base class:
class Base2 { ... generator implemented here ... };
class Derived : public Base2, public Base {
Derived() : Base2(5), Base(Base2::generator) {}
};
See Order of calling constructors/destructors in inheritance
The constructor for Base will be called before any members of Derived are initialized. So generator will not have been initialized when get_data() is called.
Basically, Base will be initialized with garbage.
I have no idea what the standard says, but this is clearly a matter of what is the value of generator when get_data() is called, as it would be if you call it from somewhere else then the constructor.
Code called from within a constructor is no way different to other code.
My guess is that get_data() is called before generator is constructed and thus you will get a NULL exception at runtime.
Another point to mention is that the base class is of course being constructed first. It is the base class. Derived IS A Base, it has a field in_.
As a rule of thumb it is always a good pattern to:
Avoid method calls from the constructor. Use an additional init method for that.
In any case beware of circular constructor calls.
You also could make get_data abstract in Base and then call it within the Init() of Base which would IMHO be the right way to implement a dependency to child classes as you call it.
class Derived
: public Base
{
Derived()
: Base(),
generator(5) {}
Generator generator;
virtual int get_data() { return generator.get_some_data(); }
};
class Base
{
Base()
{
}
Init() {
in_ = get_data();}
int in_;
virtual int get_data() = 0;
}

Why private members get inherited?

So private members in the base class are also in the inherited class but not accessible in it, right?
Are they actually in the memory allocated to the the inherited object?
Are they actually in the memory allocated to the the inherited object?
Yes, they need to exist. The private members are part of the implementation detail of the base class. Without them, in general, the base class wouldn't be able to function (which is why they exist in the first place).
Making them private just allows the base class to create its implementation however it chooses, without exposing that to anybody, including the subclass.
Yes. Just for example, you can use a public function from the base class that manipulates private data, even in an instance of the derived class:
class Base {
int x;
public:
Base() : x(0) {}
void inc() { ++x; }
void show() { std::cout << x << "\n"; }
};
class Derived : public Base {
};
int main() {
Derived d;
d.show();
d.inc();
d.show();
}
With a properly functioning compiler, this must display:
0
1
...showing that the data in the Base object is present in the Derived object, even though it's not (directly) accessible.
Of course with almost anything in C++, there's the "as-if" rule -- if the compiler can determine that it can somehow produce the correct observable behavior for the program, even without including the private part(s) of the base class, then it's free to do so. The most obvious example of this would be if you included something (member function or data) in the base class that was simply never used in practice.
Yes they are,
When object of the derived class is being constructed all of its base classes are first being constructed as well.
Consider this example:
class Base
{
int x;
public:
Base(int px)
: x(px)
{
}
};
class Derived : public Base
{
int y;
public:
Derived(int px, int py)
: y(py), Base(px)
{
}
};
This example compiles and works and Base is initialized (constructor is called) before you reach the body of the Derived constructor.

How is this initialization list implemented using a virtual class?

#include<iostream.h>
class A{
public:
int i;
A(int j=3):i(j){}
};
class B:virtual public A{
public:
B(int j=2):A(j){}
};
class C:virtual public A{
public:
C(int j=1):A(j){}
};
class D:public B, public C {
public:
D(int j=0):A(j), B(j+1), C(j+2){}
};
int main()
{
D d;
cout<<d.i;
return 0;
}
I am not being able to understand how the final output is zero. Every time j is initialized in default way to some fixed value, how is the value initialized in the constructor of class D being passed to class A?
Since A is a virtual base class, it should be constructed only once, so it is not possible to create it with different constructor parameters, and the C++ compiler has to choose one way of creating a base class.
The obvious question is: which one is used?
And the rule is: the one specified in the most derived class that inherits A directly.
The initialization order is simple: first A (with the parameter value from D constructor initialization list), then B (it is D's first ancestor; and it uses the instance of A created before), then C (and it shares the same A instance), finally D (and it also shares the same A object as B and C).
The rule with virtual base inheritance is:
"The most derived class in a hierarchy must construct a virtual base"
In your case, from the most derived class D You explicitly called the constructor of A by passing an argument 0 So it sets the i to 0. As mentioned in rule virtual base class is constructed through most derived class only and the other constructor calls through intermediate hierarchy have no effect since it is only constructed once.
The order of calling is:
A(int)
B(int)
C(int)
Good Read:
Why virtual base class constructors called first?