How can I initialize base class member variables in derived class constructor

Question

Why can t I do this   class A   public      int a  b      class B   public A       B     A    a 0   b 0

User · Answer

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

User · Answer

include lt stdio h gt    include lt iostream gt    include lt conio h gt   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 lt  lt   int i     lt  lt i lt  lt endl lt  lt  float f     lt  lt f lt  lt endl  lt  lt  double d     lt  lt d lt  lt endl                int main        Base   b   new Derived 10  1 2  3 89       b- gt 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

User · Answer

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

User · Answer

If you don t specify visibility for a class member  it defaults to  private   You should make your members private or protected if you want to access them in a subclass

User · Answer

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

User · Answer

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

User · Answer

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

User · Answer

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

[c++] How can I initialize base class member variables in derived class constructor?

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