Question

I am reading through implementing smart pointers and I found the following code,

template <class T>
class SmartPtr
{
public:
explicit SmartPtr(T* pointee) : pointee_(pointee);
SmartPtr& operator=(const SmartPtr& other);
~SmartPtr();
T& operator*() const
{
...
return *pointee_;
}
T* operator->() const
{
...
return pointee_;
}
private:
T* pointee_;
...
};

I am not able to understand the following,

1. “SmartPtr& operator=(const SmartPtr& other)”: why the parameter is constant? Doesn’t it lose its ownership when the assignment is done?
2. And why do we need “T& operator*() const” and “T* operator->() const” methods?

Thx@

Answer 1

Point 1. Not necessarily, depends on the design of the smart pointer. Some like boost:shared_ptr do not transfer ownership on assignment.

Point 2. Those methods simulate normal pointer operations on the smart pointer.