CPP Operator Overloading & Type Conversions 3

Question 1

In C++20, defining operator<=> with a default implementation allows the compiler to automatically synthesize which operators?

Accepted Answer

All six comparison operators (<, >, <=, >=, ==, !=)

Answer

Providing a defaulted operator<=> (and operator==) lets the compiler synthesize all six comparison operators for the class.

Question 2

What is the function call operator and how is it overloaded?

Accepted Answer

It is operator() and makes class objects callable like functions

Answer

Overloading `operator()` makes a class object a functor — it can be called using function call syntax (e.g., `obj(args)`).

Question 3

What is the canonical C++ pattern for implementing operator+ in terms of operator+=?

Accepted Answer

Implement operator+= as a member, then define operator+ as a non-member returning a modified copy

Answer

The canonical pattern: implement `operator+=` as a member modifying *this, then implement `operator+` as a non-member creating a copy and applying +=.

Question 4

What must operator-> return when overloaded in a smart pointer class?

Accepted Answer

A raw pointer or an object that itself has operator-> defined

Answer

operator-> must return a raw pointer (which the compiler auto-dereferences to apply ->) or an object with its own operator-> so the chain continues.

Question 5

What does overloaded operator* typically return in a smart pointer or iterator class?

Accepted Answer

A reference to the pointed-to object

Answer

operator* typically returns a reference (T&) to the pointed-to object, allowing it to be used as an lvalue (e.g., `*ptr = value`).

(CPP) C++ Programming Professional Certificate Practice Test

(CPP) C++ Programming Professional Certificate Practice Test

CPP Operator Overloading & Type Conversions 3