Question

In the code below I would like to know what does the operator* overloadind do here.

// struct pointer to Shape
struct PtrToShape
{
Shape *ptr;

bool operator < (const PTRToShape & rhs) const
{ return *ptr < *rhs.ptr; }

const Shape & operator*() const
{ return *ptr; }

};

What will it overload? Will it overload the * operator for the struct or for the shape objects? Is this overloading used in:

return *ptr < *rhs.ptr

And in the line I previously mentioned the * (overloaded or not) refers to what? to rhs or to rhs.ptr ?

Thank you.

Answer 1

const Shape & operator*() const
{ return *ptr; }

This returns a reference dereferenced Shape object, the one called ptr, which is a member of your struct.

That overload will overload this *operator for your PtrToShape struct.

Also, no, it’s not used in return *ptr < *rhs.ptr