Upcasting and the copy-constructor
If you allow the compiler to synthesize a
copy-constructor for a derived class, it will automatically call the base-class
copy-constructor, and then the copy-constructors for all the member objects (or
perform a bitcopy on built-in types) so you’ll get the right
behavior:
//: C14:CopyConstructor.cpp
// Correctly creating the copy-constructor
#include <iostream>
using namespace std;
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;
}
};
class Member {
int i;
public:
Member(int ii) : i(ii) {
cout << "Member(int ii)\n";
}
Member(const Member& m) : i(m.i) {
cout << "Member(const Member&)\n";
}
friend ostream&
operator<<(ostream& os, const Member& m) {
return os << "Member: " << m.i << endl;
}
};
class Child : public Parent {
int i;
Member m;
public:
Child(int ii) : Parent(ii), i(ii), m(ii) {
cout << "Child(int ii)\n";
}
friend ostream&
operator<<(ostream& os, const Child& c){
return os << (Parent&)c << c.m
<< "Child: " << c.i << endl;
}
};
int main() {
Child c(2);
cout << "calling copy-constructor: " << endl;
Child c2 = c; // Calls copy-constructor
cout << "values in c2:\n" << c2;
} ///:~
The operator<< for
Child is interesting because of the way that it calls the
operator<< for the Parent part within it: by casting the
Child object to a Parent& (if you cast to a base-class
object instead of a reference you will usually get undesirable
results):
return os << (Parent&)c << c.m
Since the compiler then sees it as a
Parent, it calls the Parent version of
operator<<.
You can see that Child has no
explicitly-defined copy-constructor. The compiler then synthesizes the
copy-constructor (since that is one of the four functions it will
synthesize, along with the
default constructor – if you don’t create any constructors –
the operator= and the destructor) by calling the Parent
copy-constructor and the Member copy-constructor. This is shown in the
output
Parent(int ii)
Member(int ii)
Child(int ii)
calling copy-constructor:
Parent(const Parent&)
Member(const Member&)
values in c2:
Parent: 2
Member: 2
Child: 2
However, if you try to write your own
copy-constructor for Child and you make an innocent mistake and do it
badly:
Child(const Child& c) : i(c.i), m(c.m) {}
then the default constructor will
automatically be called for the base-class part of Child, since
that’s what the compiler falls back on when it has no other choice of
constructor to call (remember that some constructor must always be called
for every object, regardless of whether it’s a subobject of another
class). The output will then be:
Parent(int ii)
Member(int ii)
Child(int ii)
calling copy-constructor:
Parent()
Member(const Member&)
values in c2:
Parent: 0
Member: 2
Child: 2
This is probably not what you expect,
since generally you’ll want the base-class portion to be copied from the
existing object to the new object as part of copy-construction.
To repair the problem you must remember
to properly call the base-class copy-constructor (as the compiler does) whenever
you write your own copy-constructor. This can seem a little strange-looking at
first but it’s another example of upcasting:
Child(const Child& c)
: Parent(c), i(c.i), m(c.m) {
cout << "Child(Child&)\n";
}
The strange part is where the
Parent copy-constructor is called: Parent(c). What does it mean to
pass a Child object to a Parent constructor? But Child is
inherited from Parent, so a Child reference is a
Parent reference. The base-class copy-constructor call upcasts a
reference to Child to a reference to Parent and uses it to perform
the copy-construction. When you write your own copy constructors you’ll
almost always want to do the same
thing.
