C++ less operator uses conversion operator

Question

I cannot understand why for the following dummy class:

class Foo {
public:
    operator double() const {
        return 3.14;
    }
};

when I try to compare the instances:

Foo f1;
Foo f2;

auto res = f1 < f2;

the less operator uses the existing

operator double() const

when comparing values. Where can I find the rules for this behaviour?

Answer 1

As mentioned in the comments, the compiler is allowed to make 1 user-defined conversion as it's doing here. It sees that it can convert both objects to double and double does have operator <, so it does that. If you only want to use your conversion operator when you (the programmer) ask for it you can add explicit:

explicit operator double() const {
    return 3.14;
}

Now, the compiler may only call it when you explicitly cast it to such a type:

Foo f1;
double d = (double)f1; // conversion operator