Lecture 7 - Operator Overloading

Does adding const to a method in the derived class (same return type, function name, and parameters) override the method?

DOESN'T override, but hides methods of same name

Nonmember operator function:

LHS: MAY be object of operators class, different class, or built-in type

Member operator function:

Left operand: must be class object or reference to class object If need private or protected data from RHS, function must be either - friend - use accessor or mutator methods

Which two operators does the compiler generate a default implementation if none is supplied?

- assignment operator (=) - member-wise assignment of class' data members - address operator (&) - returns address of object in memory (DO NOT OVERRIDE THIS)

Which operators cannot be overloaded? (give 5 examples)

. .* :: ?: sizeof

An overloaded operator must have at least one operand that is (1). This prevents what?

1) a user defined type 2) overloading operators for standard data types

To call the const operator[] from the non-const operator[] what steps do you need to take?

1) cast away const of op[] return type 2) add const to *this type

Non-member functions are made (1) if they need access to the internal state.

1) friends

Operator overloading is invoked using (1) syntax.

1) infix

Friendship is (1), but not (2).

1) inherited 2) commutative

Overloading an operator doesn't change what? (4)

1) operator precedence 2) associativity of operator (left vs right) 3) arity of operator (unary/binary) 4) meaning of operator for built-in types

The a++ operator injects a (1) that isn't in the ++a operator.

1) phantom parameter (int)

Is prefix or postfix notation preferred for incrementing? Why?

1) prefix 2) postfix calls prefix signature --> 2 copies created

In overriding, the (1) alone can't vary.

1) return type

operator[] for reading a value is (1) usage. operator[] for writing a value is (2) usage.

1) rvalue (value) 2) lvalue (reference)

Overloaded operator must adhere to the (1) of the original operator.

1) syntax

Operator overloading:

allows programmer to supply behavior that occurs when built-in operators are applied to user-defined types

Which of the Big 3 is an example of operator overloading?

assignment operator

Operator overloading can't be used for...

creating new operator symbols

Hiding:

derived class hides methods in the base if it has the same NAME as a method in the base class

Infix vs Prefix syntax:

infix: 1 + 2 + 3 prefix: (+ (+ 1 2) 3)

Preference for overloading unary operators?

make them class members

Operator functions are defined as what?

member functions or non-member functions

Overloading:

multiple functions with the same name occur in the same scope, but with different signatures

Unary operator:

operator doesn't take a parameter (involves one object, this)

Binary operator:

operator takes a parameter (involves two objects, this and param)

Each individual operator must be overloaded for use with user defined types (example?)

overloading operator= and operator- DOESN'T overload operator-=

Operator invocation syntax: (programmer preferred and computer preferred)

programmer: a = b computer: a.operator=(b)

How do you explicitly disable hiding?

under public: "using baseClass:methodName;"

Overriding:

when a derived class function has the same name AND signature as a base class virtual function (involves inheritance)

How do you write an operator overload?

write function definition same as function name of operator, followed by symbol for the operator to be overloaded: operator-