How does rhs work?

Question

Here I understand rhs means right hand side but I don't understand how does compiler understand "rhs" refers to right hand side. And can someone explain in which case will this overloading be necessary?

MyArray<T>& operator=(const MyArray<T>& rhs); 

Answer 1

The compiler doesn't know that rhs stands for "right hand side", and in fact the name of that variable can be anything you like.

The compiler "knows" how to format this because the syntax of operator= requires it to be this way.

class A
{
public:
   A& operator=(const A& other);
};

The language defines the usage of this operator to take the form:

A a, b;
a = b;

The code above calls A::operator=(const &other) against the instance of A named a, and uses the instance of A named b as other.

Answer 2

The standard assignment operator to an instance of the same type has the prototype

MyArray<T>& operator=(const MyArray<T>&);

The name rhs is normally given to the function parameter since it appears on the right hand side of an assignment when the operator is invoked. It improves the legibility of your source code, that's all.

Answer 3

rhs is just a name people usually use for that operator, it has no special meaning. The way that operator is defined always makes the argument the right-hand element.

Answer 4

If you do something like:

int a = 5;
int b = 3;
a = b;

The assignment part is actually just a function call:

a.operator=(b);

Nothing special going on. The parameter name doesn't matter, just the signature which consists of the return type and parameters types, not names.