Is this a constructor operator or conversion operator?

Question

class B
{
    public:
        operator B() const{ }    // What is this and what is the purpose?

    private:
        int m_i;
};

So the question is, is that a conversion operator or constructor operator and what is the use of it? Where to use it?

Answer 1

It is a conversion function which will never be called implicitly. The Standard actually goes into some depth about this. 12.3.2/1:

A conversion function is never used to convert a (possibly cv-qualified) object to the (possibly cv-qualified) same object type (or a reference to it), to a (possibly cv-qualified) base class of that type (or a reference to it), or to (possibly cv-qualified) void.

And in a footnote,

These conversions are considered as standard conversions for the purposes of overload resolution (13.3.3.1, 13.3.3.1.4) and therefore initialization (8.5) and explicit casts (5.2.9). A conversion to void does not invoke any conversion function (5.2.9). Even though never directly called to perform a conversion, such conversion functions can be declared and can potentially be reached through a call to a virtual conversion function in a base class.

Also, conversion functions are still normal functions and can be called explicitly by name.

The note about virtual functions applies to code like this:

class B;

struct A {
    virtual operator B() const = 0;
};

struct B : A
{
    public:
        operator B() const{ return B(); } // virtual override

    private:
        int m_i;
};

A const & q = B(); // q has dynamic type B, static type A
B r = q; // Convert A to B using B::operator B()

Pedantic note: "conversion operator" is poor terminology. These are known as conversion functions and they are not considered to be a case of operator overloading, despite the operator keyword.