Rehashing requires recomputing the hash function for all items in the hash table. Since computing the hash function is expensive, suppose objects provide a hash member function of their own, and each object stores the result in an additional data member the first time the hash function is computed for it. Show how such a scheme would apply for the Employee class in Figure 5.8, and explain under what circumstances the remembered hash value remains valid in each Employee.

By "showing how such scheme would apply" it means update the program shown in Figure 5.8 with the new information/conditions given.

Figure 5.8:

#include #include using namespace std; // Example of an Employee class 4 Eclass Employee 10 public: const std::string &getName) const return name; 14 15 16 17 18 19 20 21 bool operator-(const Employee &rhs) const return getName() rns, getName(); bool operator!-(const Employee &rhs) const return (*thisrhs); 24 25 26 private: std::string name; double salary; int seniority; 28 29 30 31 32; // additional private members not shown template 35 Eclass hash 36 public: size t operatorO (const Employee &item) 38 39 static hash<:string> hf; return hf (item.getName()) 41 42 43