Why can't I do this?
class A
{
public:
int a, b;
};
class B : public A
{
B() : A(), a(0), b(0)
{
}
};
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11Are you asking why you can't do that, which is a language-design question, or are you asking how to work around that language limitation?Rob Kennedy– Rob Kennedy2011-09-13 17:14:08 +00:00Commented Sep 13, 2011 at 17:14
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I thought that there was some sort of a special way to do it that i'm not aware of, without having to use the base constructor.amrhassan– amrhassan2011-09-13 20:48:11 +00:00Commented Sep 13, 2011 at 20:48
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3The base class members are already initialized by the time your derived-class constructor gets to run. You can assign them, if you have access, or call setters for them, or you can supply values for them to the base class constructor, if there is one suitable. The one thing you cannot do in the devised class is initialize them.user207421– user2074212017-08-23 00:25:41 +00:00Commented Aug 23, 2017 at 0:25
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7 Answers
You can't initialize a and b in B because they are not members of B. They are members of A, therefore only A can initialize them. You can make them public, then do assignment in B, but that is not a recommended option since it would destroy encapsulation. Instead, create a constructor in A to allow B (or any subclass of A) to initialize them:
class A
{
protected:
A(int a, int b) : a(a), b(b) {} // Accessible to derived classes
// Change "protected" to "public" to allow others to instantiate A.
private:
int a, b; // Keep these variables private in A
};
class B : public A
{
public:
B() : A(0, 0) // Calls A's constructor, initializing a and b in A to 0.
{
}
};
9 Comments
R Samuel Klatchko
while your example is correct, your explanation is misleading. It's not that you can't initialize
a and b in B::B() because they are private. You can't initialize them because they are not members of class B. If you made them public or protected you could assign them in the body of B::B().
Gene Bushuyev
additionally, you solution makes class A non-aggregate, which might be important, so it must be mentioned.
In silico
@R Samuel Klatchko: Good point. When I was writing the answer I initially typed "You can't access
a and b..." and changed it to "You can't initialize..." without making sure the rest of the sentence made sense. Post edited.
David Rodríguez - dribeas
@Gene Bushuyev: The class in the original code in the question is not an aggregate (there are non-static private members)
Gene Bushuyev
@David -- correct, which is a user's error, and I'm trying to get to the user's intentions, skipping superficial.
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Leaving aside the fact that they are private, since a and b are members of A, they are meant to be initialized by A's constructors, not by some other class's constructors (derived or not).
Try:
class A
{
int a, b;
protected: // or public:
A(int a, int b): a(a), b(b) {}
};
class B : public A
{
B() : A(0, 0) {}
};
Comments
Somehow, no one listed the simplest way:
class A
{
public:
int a, b;
};
class B : public A
{
B()
{
a = 0;
b = 0;
}
};
You can't access base members in the initializer list, but the constructor itself, just as any other member method, may access public and protected members of the base class.
8 Comments
Wander3r
Nice. Is there any drawback in doing this way?
Violet Giraffe
@SaileshD: there may be, if you're initializing an object with a costly constructor. It will first be default-initialized when the instance of
B is allocated, then it will be assigned inside the B's constructor. But I also think the compiler can still optimize this.
Sparkofska
Inside
class A we can not rely on a and b being initialized. Any implementation of class C : public A, for example, might forget to call a=0; and leave a uninitialized.
Martin Pecka
@Wander3r Another drawback is that not all classes have assignment operators. Some can only be constructed, but not assigned to. Then you're done...
Ben Voigt
Of course this is not initialization. "The subclasses can still re-initialize them" is just nonsense, C++ has no such operation as re-initialization.
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# include<stdio.h>
# include<iostream>
# include<conio.h>
using namespace std;
class Base{
public:
Base(int i, float f, double d): i(i), f(f), d(d)
{
}
virtual void Show()=0;
protected:
int i;
float f;
double d;
};
class Derived: public Base{
public:
Derived(int i, float f, double d): Base( i, f, d)
{
}
void Show()
{
cout<< "int i = "<<i<<endl<<"float f = "<<f<<endl <<"double d = "<<d<<endl;
}
};
int main(){
Base * b = new Derived(10, 1.2, 3.89);
b->Show();
return 0;
}
It's a working example in case you want to initialize the Base class data members present in the Derived class object, whereas you want to push these values interfacing via Derived class constructor call.
Comments
Why can't you do it? Because the language doesn't allow you to initializa a base class' members in the derived class' initializer list.
How can you get this done? Like this:
class A
{
public:
A(int a, int b) : a_(a), b_(b) {};
int a_, b_;
};
class B : public A
{
public:
B() : A(0,0)
{
}
};
Comments
While this is usefull in rare cases (if that was not the case, the language would've allowed it directly), take a look at the Base from Member idiom. It's not a code free solution, you'd have to add an extra layer of inheritance, but it gets the job done. To avoid boilerplate code you could use boost's implementation
Comments
Aggregate classes, like A in your example(*), must have their members public, and have no user-defined constructors. They are intialized with initializer list, e.g. A a {0,0}; or in your case B() : A({0,0}){}. The members of base aggregate class cannot be individually initialized in the constructor of the derived class.
(*) To be precise, as it was correctly mentioned, original class A is not an aggregate due to private non-static members