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9.4 — Overloading the comparison operators

Overloading the comparison operators is simple once you’ve learned how to overload the arithmetic operators.

Because the comparison operators are all binary operators that do not modify their operands, we will make our overloaded comparison operators friend functions.

Here’s an example Point class from the previous lesson with an overloaded operator== and operator!=.

The code here should be straightforward. Because the result of operator!= is the opposite of operator==, we define operator!= in terms of operator==, which helps keep things simpler, more error free, and reduces the amount of code we have to write.

It doesn’t really make sense to overload operator> or operator< for the Point class. What does it mean for a 3d point to be greater or less than another point? Greater than or less than isn't a concept we normally apply to 3d points, so it's better not to include those operators in the Point class, because the results of the operators (whatever you define them to be) would not be intuitive. Here's a different example with an overloaded operator>, operator<, operator>=, and operator<=:

This is also pretty straightforward. Note that there is some redundancy here as well. operator> and operator<= are logical opposites, so one could be defined in terms of the other. operator< and operator>= are also logical opposites, and one could be defined in terms of the other. In this case, I chose not to do so because the function definitions are so simple, and the comparison operator in the function name line up nicely with the comparison operator in the return statement.

9.5 -- Overloading unary operators +, -, and !
Index
9.3 -- Overloading the I/O operators

27 comments to 9.4 — Overloading the comparison operators

  • David

    operator> is defined twice; operator< is missing.

    [ Fixed! -Alex ]

  • Andrian

    Could you make an example to show how you can use one of the overloaded operators?

    • Sure.

  • Ameerah

    thanks Alex very mutch i love you totorile

    simple &easy

  • mslade

    Shouldn’t all of the arguments to the operator overload functions be const?

    • xgiggs

      Yes, you’re right, I think. If they aren’t const, then those operators can’t be used on const variables even though there is no real reason they shouldn’t, since they don’t actually change the variable values.

  • j.howard

    Great tutorial as usual. Just as an idea the < and > operators could be used to compare the magnitude of the vector from the origin to the point in the Point class.

  • Onix_GCI

    Wouldn’t the functions in the first example always return wrong result due to small rounding errors in double comparison?

  • int main()
    {
    using namespace std;
    Cents a(80);
    Cents b(100);
    if (a > b)
    cout << "A is GREATER THAN B";
    else
    cout << "B is GREATER THAN A" << endl;
    cout << "===================================" << endl;

    Cents x(80);
    Cents y(78);
    if (x <= y)
    cout << "X is LESS THAN OR EQUAL TO Y";
    else
    cout << "Y is LESS THAN OR EQUAL TO X" << endl;

    cout << "===================================" << endl;

    Cents ict1(80);
    Cents ict2(200);
    if (ict1 < ict2)
    cout << "ICT1 is LESS THAN ICT2" <<endl;
    else
    cout << "ICT2 is LESS THAN ICT1" << endl;

    cout << "===================================" <= ict4)
    cout << "ICT3 is GREATER THAN OR EQUAL TO ICT4";
    else
    cout << "ICT4 is GREATER THAN OR EQUAL TO ICT3";
    cin.get();
    return 0;
    // working okay by ifycent2 de ict
    }

  • Rahul

    Why can’t I overload operator ‘~=’ same way I overload ‘==’? I want it to have same definition as ‘!=’.

    It gives error "declaration of ‘operator~’ as non-function". Is it an error with my syntax?

    I read somewhere that C++ overloads operators whose definitions are already known? Is that true? For example can’t overload definition of ‘#’ operator. Or define my own operator.

    my code:

    • Alex

      In C++, you can only overload operators that already exist in C++. ~= isn’t an operator that naturally exists in C++, so you can’t overload it.

  • "operator> and operator<= are logical opposites, so one could be defined in terms of the other. operator< and operator>= are also logical opposites"

    shouldn’t it be:

    "operator> and operator< are logical opposites, so one could be defined in terms of the other. operator<= and operator>= are also logical opposites"

    (last paragraph)

  • N.Abb

    Why we have to overload relational operators in pairs ?
    like if I overload == I must overload !=
    Thanks

    • Alex

      You don’t have to. However, since one can be defined in terms of the other, it’s almost trivial to write the second after you have the first.

  • j_nick

    This is also pretty straightforward. Note that there is some redundancy here as well. operator> and operator<= are logical opposites, so one could be defined in terms of the other

  • j_nick

    operator> and operator<= are logical opposites, so one could be defined in terms of the other. operator< and operator>= are also logical opposites
    const ?

    • N.ab

      Hi Alex : is correct ?…

      • Alex

        Yes. If two operators are logical opposites, one can be defined as the negation of the other.

        If the implementation of the operator is trivial, like the examples in the lesson, this probably isn’t worth doing (it’s easier to just implement it).

        If the implementation of the operator is more complicated, then this can be worth doing as a way to reduce redundant code.

  • nick

    Hi Alex;
    Should not the ! operator overloaded

    • Alex

      It could be implemented that way, and if operator > is complicated, I’d recommend it, to reduce the amount of redundant code.

      However, when the implementation of the comparison operators is trivial, it’s easier (and more understandable) to just implement them individually.

  • Valentino

    Why does (Cents > int) use my overloaded operator (Cents > Cents)?
    Thanks

    • Alex

      Because one of your operands is an object of type Cents, C++ will first look to see if there is an operator where both operands match. In this case, there isn’t. Next, the compiler will try to see if it can promote or convert types to make the operands match. In this case, because you have a Cents constructor that takes an int, the compiler will use this constructor to create a Cents object from the int.

      It’s essentially implicitly converting the if statement to this:

      • Valentino

        Thank you for your response. Is it possible to tell C++ that it should not have to do that so I will get an error on compile?

        • Alex

          Yes, put the explicit keyword before your constructor name. This will prevent C++ from using it for type conversions.

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