Question

I have a class called user which has a lname field. Is this the right way to overload the “<” operator?

bool User::operator<(const User& other)
{
    std::cout << "< operator was called" << std::endl;
    if (this != &other)
    {
        if (lname.compare(other.lname) == 0)
        {
            return true;
        }
    }
    return false;
}

I am trying to use this in a more complicated set of things and it is failing – just want to make sure this much is right.

Answer 1

As others have pointed out, your operator< doesn’t allow the left side to be const. Changing the function signature to

bool User::operator<(const User& other) const

is an improvement. But I would actually recommend making it a non-member function instead:

class User {
public:
    friend bool operator<(const User& u1, const User& u2);
    // ...
};

bool operator<(const User& u1, const User& u2)
{
    // ...
}

For one thing, it’s a little more legible in my opinion.

But also, it sometimes makes a technical difference. With a non-member function, the expression a < b attempts implicit conversions on both a and b to see if your operator< is a viable overload. But with a member function, implicit conversions can apply to b, but not to a: a must be of type User or a derived type. This can lead to surprising situations where a < b compiles but b < a doesn’t.