c++ - 从 BoostCon 谈话中重载取消引用运算符的奇怪方式

Question

我在 youtube 上观看了名为“现代 C++ 技术简介(第一部分)”的 boostcon 演讲。在第 22 分钟左右，演讲者展示了一个重载解引用运算符的类。

template<typename T,
         typename CheckingPolicy = NoChecking,
         typename BadPointerPolicy = BadPointerDoNothing>
class pointer_wrapper
{
public:
    pointer_wrapper() : value_(0) {}
    explicit pointer_wrapper(T* p) : value_(p) {}

    operator T*()
    {
        if ( ! CheckingPolicy::check_pointer(value_) )
        {
            return BadPointerPolicy::handle_bad_pointer(value_);
        }
        else
        {
            return value_;
        }
    }

private:
    T* value_;
};

我从未见过这种重载取消引用运算符的方法。为什么没有返回类型，为什么 T 出现在“operator”关键字之后？我一直认为重载这个运算符的方式是这样的:

T& operator *()
{
    // ...
    return *value_
}

如果有人感兴趣，here is the talk

Answer 1

它是类型T*的隐式转换运算符。n3337 12.3.2/1

A member function of a class X having no parameters with a name of the form

conversion-function-id:

operator conversion-type-id

conversion-type-id:

type-specifier-seqconversion-declaratoropt

conversion-declarator:

ptr-operator conversion-declaratoropt

specifies a conversion from X to the type specified by the conversion-type-id. Such functions are calledconversion functions. No return type can be specified.

If a conversion function is a member function, thetype of the conversion function (8.3.5) is “function taking no parameter returning conversion-type-id”.