c++ - 这是构造函数运算符还是转换运算符？

Question

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

    private:
        int m_i;
};

那么问题来了，那是转换运算符还是构造运算符，它有什么用呢？在哪里使用？

Answer 1

这是一个永远不会被隐式调用的转换函数。该标准实际上对此进行了深入探讨。 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.

在脚注中，

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.

此外，转换函数仍然是普通函数，可以通过名称显式调用。

关于虚函数的注释适用于这样的代码:

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()

迂腐提示:“转换运算符”是一个糟糕的术语。这些被称为 转换函数，尽管有 operator 关键字，但它们不被视为运算符重载的情况。