Not able to call base class constructor from dericed class with parameter in c++

Question

I have below scenario in c++ within DEV c++

class base{
    int i=0;
    public:
         base(){
            cout<<"base default constructor"<<endl;
         }
         base(int i){
             cout<<"base with int"<<endl;
         }
};

class derive : public base{
    int j;
    public:
        derive(){
            cout<<"derive default";
        }
        derive(int i){
            int k=5;
            base(k);//////error 
            cout<<"derived with int "<<i<<endl;
        }
        void fun(int i){
            cout<<"finction "<<i<<endl;
        }
};
int main()
{
    derive d(9);

}

After running following error is coming
[Error] conflicting declaration 'base k'
[Error] 'k' has a previous declaration as 'int k'
Why this error is coming.
Its working fine when i call base(5) with any parameter except declared in derived constructor.
Thanks in advance.

Hi, yes its temporary and i am not initializing constructor. But i want to know why base(5) or base(j) works inside derived() but not base(i). — Ryuk, Jun 08 '16 at 05:41

score 1 · Answer 1 · edited May 23 '17 at 10:34

1

If you call the base constructor within the derived class on the implicit parameter, it needs to be done in the field initialization list like so:

    derive(int i) : base(i) {
        cout<<"derived with int "<< i <<endl;
    }

live demo

edited May 23 '17 at 10:34

Community

1
1

answered Jun 08 '16 at 05:48

scohe001

15,110
2
31
51

score 0 · Answer 2 · answered Jun 08 '16 at 05:58

#include <iostream>

using namespace std;

class base{
public:
    base(){
        cout<<"base default constructor"<<endl;
    }
    base(int i){
        cout<<"base with int "<<i<<endl;
    }
};

class derive : public base{

public:
    int j;
    int k;
    derive(){
        cout<<"derive default\n";
    }
    derive(int i){
        k=5;
        base(this->k);//////error
        cout<<"derived with int "<<i<<endl;
    }
    void fun(int i){
        cout<<"finction "<<i<<endl;
    }
};
int main()
{
    derive p(10);


}

please explain this answer a bit. code only answers are mostly not of much value. thank u! keep up the good work! — Willi Mentzel, Jun 08 '16 at 15:26

score 0 · Answer 3 · answered Jun 08 '16 at 06:00

There are two interpretations of what base(k) could be:

it could be a the creation of temporary
it could be a variable declaration: the variable name can be in parenthesis

Out of these two choices, the compiler uses the second one: if something can be either a declaration or an expression, the declaration is chosen. This is related to the most vexing parse although it isn't that. The effect is that in your use base(k); tries to define a variable named k of type base.

You can disambiguate the meaning using one of

(base(k));
base{k};

If base also had a constructor taking an std::initializer_list<int> these two can have different meanings.

score -1 · Answer 4 · answered Jun 08 '16 at 05:49

In main()

    base b(3);
    derive d(9);
    base b1(2);

//create explicit object and pass integer parameter.then try also pass particular integer value in derive->base. in derive

base(8);

And always print the value of variable acting weirdly in a program

Not able to call base class constructor from dericed class with parameter in c++

4 Answers4