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CS253 Operator Functions

The List

• These operators can be overloaded:
  • unary: + - *
  • binary: + - * / % ˆ & | << >>
  • += -= *= /= %= ˆ= &= |= <<= >>=
  • ~ !
  • =
  • == != < > <= >=
  • && ||
  • ++ --
  • , ->* -> () []
  • new new[] delete delete[]
  • ""suffix

Yes, = is an operator, just like + is. Sure, = usually alters its left operand, whereas + doesn’t, but C++ doesn’t care about that.

What can’t you do?

You cannot:

• create new operators (sorry, no **, >>>, or ≠)
  • can’t change the lexical analyzer or parser
• redefine existing bindings (can’t make "zulu"+3 do concatenation)
  • because C++ is extensible, not mutable
• change precedence or grouping of existing operators
  • because C++ is extensible, not mutable

Spell it out

• The compiler doesn’t deduce methods from other mathematically-equivalent methods. That’s your job.
• You have to define every operator that you need.
• For example, if you define operator+(const Fraction &, double), that doesn’t let you do this:

    Fraction a, b;
    a = 1.2 + b;

• Defining operator+ does not create operator+=, or operator++.
• Defining preincrement doesn’t even define postincrement!

Use previous work

Consider these addition methods:

• Fraction operator+(const Fraction &)
• Fraction &operator+=(const Fraction &)
• Fraction &operator++() // pre
• Fraction operator++(int) // post

The smart programmer will have operator+= do the real work, have operator+ and preincrement invoke +=, and have postincrement call preincrement.

More re-use

For that matter, define operator-= in terms of operator+= (if negation is cheap for your class). Then, as before, you can have operator- and predecrement call operator-=, and have postdecrement call predecrement.

Methods

When possible, implement operator overloading as methods (as part of the class). When the method is called, the left operand is *this, and the right operand is the argument.

    Fraction a, b; a += b;

That will call a.operator+=(b).

    Fraction a, b, c
    a = b - c;

That will call a.operator=(b.operator-(c)).

Non-member functions

Sometimes, you can’t use a method (method ≠ function, right?):

    Fraction a, b;
    a = 1.2 + b;

If operator+ were a method, it would have to be a method of double. That wouldn’t work, so it must be a non-member function, a free function:

    Fraction operator+(double, const Fraction &);

Of course, you still need the method that handles Fraction+double. Function overloading is your friend.

Abuse

Java doesn’t allow operator overloading, supposedly because it’s too confusing. There is something to be said for that.

Resist the urge to redefine every 💣 ☠†※! operator possible. Only define those that:

• have clear mathematical analogues
• or mimic existing C++ operators (e.g., +=)
• or make sense by analogy (e.g., file+file could concatenate the contents of two files).

Example of abuse

A misguided programmer might define a-b, where a and b are strings, to mean “return a, without the chars that are also in b”.

string operator-(string a, const string &b) {
    for (size_t p=0; (p = a.find_first_of(b, p)) != a.npos;)
        a.erase(p, 1);
    return a;
}

int main() {
    const string name = "Bjarne Stroustrup";
    cout << name-"aeiou" << '\n';
}

Bjrn Strstrp

Maybe, maybe not. 🤷 Call it remove_chars(), instead.