operator overload

Why << need to be friend? You need to consider: Because you need to write cout<< obj. If you declare << as a member operator, you have to write obj<<cout; it looks weird.
If you want to overload + operator, and support time t; 3+t; Previous method can’t be work. You need to define a friend function. In order to improve a t+3 efficiency, you also can define a member function time operator+(int i) member function.
time operator+(const time &t) const;
//member function. why there are three const?
//why return a obj not a reference.
// t = t1+t2

friend time operator+(int, const time &t)
nonmember friend function
// 3+t

time operator+(int i)
member function
// t+3,
//to avoid implicit conversion from 3 to obj.
ob = ob1+ob2 will changed to: ob1.operator+(ob2); so, it should be defined as the first const to avoid ob1+obj2= ob3, the last const make the ob to invoke this operator doesn’t change the value in this class.
assignment operator overload. 1) return a non-const reference, and 2) to avoid assign self.
class & operator=(const class& rhs)
{
   if(this ==&t) return *this;
   delete [] mem_array;
   //Do some things other
   return *this;
}