What does < (less-than sign) between a function identifier and a function parameter mean?

Question

struct point {
    int x, y;
    bool operator<(const point &p) {
    if (x == p.x) return y < p.y;
    else return x < p.x; 
    } 
}; 

What does < between operator and (const point &p) signify?

Answer 1

What does < between operator and (const point &p) signify?

It's a part of the member function name. It is a definition of a member function which is an overload of the < operator. It is what makes this work:

point a{};
point b{};

if(a < b) { // point::operator<(const point&) is used here.
    // 
}

Note: The member function should really be const qualified because calling the function does not change *this (the lefthand side of <).

bool operator<(const point &p) const {
//                             ^^^^^

---

... and it's also preferable if it's not even a member function, but a free function (taking two const point&):

bool operator<(const point& lhs, const point &rhs) {
    return std::tie(lhs.x, lhs.y) < std::tie(rhs.x, rhs.y);
} 

This is because if point has an implicit conversion constructor, it will allow an instance of a type that is implicitly convertible to point to appear on both the lefthand and righthand side of <. Example:

struct point {
    point(int X, int Y) :x{X}, y{Y} {}
    point(const std::pair<int, int>& pair) : x{pair.first}, y{pair.second} {}

    int x, y;
};

bool operator<(const point& lhs, const point &rhs) {
    return std::tie(lhs.x, lhs.y) < std::tie(rhs.x, rhs.y);
} 

int main() {
    std::pair<int, int> ip{10,20};
    point pnt{10,19};

    // This works with operator< as a member function or a free function:
    if(pnt < ip) std::cout << "pnt < ip\n";

    // This will not work if `operator<` is a member function:
    if(ip < pnt) std::cout << "ip < pnt\n";
//     ^^
// implicitly converted to `point`
}