C ++：不寻常的操作符重载(C++: unusual operator overloading)

C ++：不寻常的操作符重载(C++: unusual operator overloading)

什么运营商在这里超载？

operator T * ()

我知道操作符方法具有以下结构：

type operator operator-symbol ( parameter-list )

假设我们有以下代码

template<typename T> class SmartPtr { public: SmartPtr(T* data): member(data) {} T* member; T& operator * () { return *member; } //usage: *TObj T*& operator () () { return member; } //usage: TObj() operator T * () { return member; } //usage: ??? };

没有编译错误，如果你尝试在ideone上 。 那么这里发生了什么？

ADD：我是否正确static_cast<T*>(TObj)调用operator T * ？ 我在这里试过。

What an operator is overloaded here?

operator T * ()

I know that the operator method has the following structure:

type operator operator-symbol ( parameter-list )

Assume we have the following code

template<typename T> class SmartPtr { public: SmartPtr(T* data): member(data) {} T* member; T& operator * () { return *member; } //usage: *TObj T*& operator () () { return member; } //usage: TObj() operator T * () { return member; } //usage: ??? };

No compilation errors if you try it on the ideone. So what is going on here?

ADD: Am I right that static_cast<T*>(TObj) makes a call of the operator T *? I've tried it here.

最满意答案

这是一个转换运算符 ，它允许将类转换为T* 。 用法：

T * p = TObj;

对于一个智能指针来说，这可能是一个坏主意，因为它容易意外地获得一个非智能指针。 标准智能指针通过get()函数提供显式转换，以防止意外转换。

That's a conversion operator, which allows the class to be converted to T*. Usage:

T * p = TObj;

It's probably a bad idea for a smart pointer to provide this, as it makes it easy to accidentally get a non-smart pointer. Standard smart pointers provide explicit conversion via a get() function instead, to prevent accidental conversions.