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

Only < and >

Answer

Only == and !=

Answer

Only the relational operators, not equality

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

It is operator[] and enables array indexing

Answer

It is operator-> enabling pointer-like member access

Answer

It is operator* and dereferences the object

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

Implement operator+ first, then delegate operator+= to it

Answer

Implement them independently to avoid circular dependency

Answer

operator+ should always be a virtual member function

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

A reference to the contained object

Answer

A void pointer that is later cast

Answer

A const pointer only

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

A raw pointer to the object

Answer

A const void pointer

Answer

A copy of the contained value

Question 6

What is an implicit conversion sequence in C++?

Accepted Answer

The sequence of standard and user-defined conversions the compiler applies automatically to match types

Answer

A series of explicit casts written by the programmer

Answer

A cast that always requires a conversion operator

Answer

A template specialization that handles type mismatches

Question 7

When the `explicit` keyword is applied to a conversion operator, what effect does it have?

Accepted Answer

The conversion requires an explicit cast rather than happening implicitly

Answer

The conversion is disabled entirely

Answer

The operator becomes a static member

Answer

The operator can only convert to bool