I am new to c++, and I was recently trying to understand overloading functions. I have questions about references in operator overloading. I get that CBoys& is reference to class CBoys, but i dont see difference in following 5 overload of operator =.
class CBoys{
//operator=(const CBoys&);
//operator=(const CBoys);
}
// CBoys& operator=(const CBoys&);
// CBoys operator=(const CBoys&);
// CBoys& operator=(const CBoys);
Thx for help
Well, for starters,
operator = (const CBoys&);
and
operator = (const CBoys);
are invalid. The only time when you omit the return type (operators are just like functions) is when you are overloading conversion operators. e.g.
operator bool();
when defined within your type will create a method to implicitly convert your type to bools.
As for the rest, they all differ based on their return types and the parameters they take in. The first one, CBoys& operator=(const CBoys&); takes in a reference to a constant CBoys and returns a reference to a CBoys. The second one returns a CBoys by value. The third one, while fine, makes a copy of the CBoys you pass into it (the object on the right hand side.)
Long story short, the constness and reference-ness of the parameters to an operator overload mean the same thing that they do in regular function declarations.
Implementing a copy assignment operator overload is typically pretty straight-forward:
class CBoys {
// stuff
public:
CBoys& operator = (const CBoys& rhs) {
CBoys temp(rhs); // make a copy.
using std::swap;
swap(*this, temp); // and swap the old *this with the new object.
return *this; // return the object by-reference.
}
};
This allows us to use the assignment operator in a straight-forward way with no surprises, as it behaves like the assignment operator for built-in types. For more information on copy assignment operators, see here: http://en.cppreference.com/w/cpp/language/as_operator
I hope this helps your understanding.
operator=must always be a member function, so the last three are invalid.