What overloaded version of the operator to call

Question

Suppose I declared indexes in a class

• char& operator[] (int index);
• const char operator[](int index) const;

In what state is the second overload called. Is this called only through a const object .

In the following scenarios, the statement version will be called.

 const char res1 = nonConstObject[10]; nonConstObject[10]; 

Answer 1

The first one is called. Do not confuse the return value; only arguments for choosing a method are considered. In this case, the implicit this not constant, so the non-constant version is called.

Answer 2

Persistent methods can only be called from persistent instances. Since nonConstObject is not defined as const, both calls will be made using a non-constant overloaded operator.