Initializer List

u_peerless · #1 11-Jul-2008, 06:51

Hi C++ Coders,
I have a query. I was going through practising some C++ programs. We know that base class constructor is invoked first before derived class constructor.

Now suppose I have a program:

CPP / C++ / C Code:

class foo
{
      public:
             foo(int i)
             {
                     cout<<"Foo's Constructoe Called With:"<< i<<":\n";
                     }
      };
class bar:public foo
{
      public:
             bar():foo(10)
             {
                          cout<<"Bar's Constructr:\n";
                          }
      };

int main()
{
//    foo f(12);
    bar b;
    getch();
}

I have created an initializer list to initialize i as 10. But the foo class constructor should be invoked first before bar constructor. When I am running the program, then foo constructor is displaying as 10. My qs is how 10 value is getting when derived class constructor has not been invoked yet?

L7Sqr · #2 11-Jul-2008, 09:07

CPP / C++ / C Code:

#include <iostream>
using namespace std;

class foo1 {
        public:
                foo1 (int i) { cout << "FOO1 [" << i << "]" << endl; }
}; 
class foo2 {
        public:
                foo2 (int i) { cout << "FOO2 [" << i << "]" << endl; }
}; 

class bar1 : public foo1, foo2 {
        public:
                bar1 () : foo1 (42), foo2 (13) { cout << "BAR1" << endl ; }
};

class bar2 : public foo2, foo1 {
        public:
                bar2 () : foo1 (42), foo2 (13) { cout << "BAR2" << endl ; }
};

int main () {
        bar1 b1;
        bar2 b2;
        return 0;
}

Quote:

Originally Posted by output

FOO1 [42]
FOO2 [13]
BAR1
FOO2 [13]
FOO1 [42]
BAR2

Realize that when you declare the class, you have to declare any base classes at the same time. This allows the compiler to know what constructors are needed before the constructor of the current class is invoked. Also notice that the order of the initializer list is not necessarily important, while the order in the declaration is.

Quote:

Originally Posted by Stroustroup

... constructors are called in the order they are declared in the class rather than the order they appear in the initializer list.

ocicat · #3 11-Jul-2008, 09:15

Quote:

Originally Posted by u_peerless

My qs is how 10 value is getting when derived class constructor has not been invoked yet?

The initialization list specifies which base class constructor is to be executed.

Consider the following changes to your posted code:

CPP / C++ / C Code:

#include <iostream>

using namespace std;

class foo {
protected:
    int v;
public:
    foo()  { 
        cout << "foo::foo() default constructor" << endl;
        v = 0;  
        cout << "v = " << v << endl; 
    }
    foo(int i) { 
        cout << "foo::foo(" << i << ") constructor" << endl;
        v = i;  
        cout << "v = " << v << endl; 
    }
};

class bar : public foo {
public:
    bar() : foo() { 
        cout << "bar::bar() default constructor" << endl; 
        cout << "v = " << v << endl;
    }
    bar(int i) : foo(i)  { 
        cout << "bar::bar(" << i << ") constructor" << endl;
        cout << "v = " << v << endl;
    }
};

int main()
{
    bar a(1);
    bar b;

    return 0;
}

Execution yields:

Code:

foo::foo(1) constructor
v = 1
bar::bar(1) constructor
v = 1
foo::foo() default constructor
v = 0
bar::bar() default constructor
v = 0

Peter_APIIT · #4 12-Jul-2008, 23:28

Initializer List called before enter into {}.

I hope this help.

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Initializer List

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