need some help with copy constuctor

Code:

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class Parent {
int i;
public:
Parent(int ii) : i(ii) {
cout << "Parent(int ii)\n";
}
Parent(const Parent& b) : i(b.i) {
cout << "Parent(const Parent&)\n";
}
Parent() : i(0) { cout << "Parent()\n"; }
friend ostream&
operator<<(ostream& os, const Parent& b) {
return os << "Parent: " << b.i << endl;
}
};

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Hi,

Im having difficulty in understanding copy constructors so the operator overloading.can any one explain me in detail the following program and how copy constructor works.I also dont understand why b.i is used in the constructor initializer list.


Thanks in advance,

It is used to copy one object to another.
This means copying the data members of the class.

So if you call it like so -
Parent p1(100);
Parent p2(p1);

The second statement tells that p2 must be copied from p1.
The only data member of the class is i.

That is why b.i is used. b represents the object p1.

thank you for the reply.For example in operator overloading "+"they use i+b.i" does it mean current val +prev value??

Quote:

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Originally Posted by brett01

For example in operator overloading "+"they use i+b.i" does it mean current val +prev value??

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It means "perform some operation on i and b.i". Hopefully that operation has something akin with addition, but it could also say... test for equality, print a novel, etc. In the context of your question, the result of that operation is probably used to initialise a member variable, bit that is unusual for a copy constructor since the copy is usually identical in the values of its members (or the value of what they point to, in the case of pointer members) with the original.

Quote:

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Originally Posted by laserlight

It means "perform some operation on i and b.i". Hopefully that operation has something akin with addition, but it could also say... test for equality, print a novel, etc. In the context of your question, the result of that operation is probably used to initialise a member variable, bit that is unusual for a copy constructor since the copy is usually identical in the values of its members (or the value of what they point to, in the case of pointer members) with the original.

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Hi thank you for the reply mate.What i dont understand is that since constructor is always associated with the object so generally we declare only "i" and not "b.i" but when it comes to copy constructor why do we declare b.i.For example in the following code

Code:

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class Integer {
int i;
public:
Integer(int ii) : i(ii) {}
const Integer
operator+(const Integer& rv) const {
cout << "operator+" << endl;
return Integer(i + rv.i);
}
Integer&
operator+=(const Integer& rv) {
cout << "operator+=" << endl;
i += rv.i;
return *this;
}
};
int main() {
cout << "built-in types:" << endl;
int i = 1, j = 2, k = 3;
k += i + j;
cout << "user-defined types:" << endl;
Integer ii(1), jj(2), kk(3);
kk += ii + jj;
}

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Quote:

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Originally Posted by brett01

What i dont understand is that since constructor is always associated with the object so generally we declare only "i" and not "b.i" but when it comes to copy constructor why do we declare b.i.

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You declare a Parent object named b. Since Parent has an i member, you can then use b.i.

Quote:

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Originally Posted by laserlight

You declare a Parent object named b. Since Parent has an i member, you can then use b.i.

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Is there any advantage for declaring "b.i" ??

Quote:

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Originally Posted by brett01

Is there any advantage for declaring "b.i" ?

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Advantage over what? Since the compiler generated copy constructor will suffice, there is no advantage in implementing it yourself.