Can a Specialized Template Class Inherit from Another Specialized Template Class? - c++

Question

I first defined
class Hash
{
};
Then a specialization of Hash.
template <class T>
class Hash<int, T>
{
public:
Hash(int slotN = 11);
bool insert(int key, T val);
bool remove(int key);
bool contains(int key);
bool query(int key, T& val) ;
protected:
// Basic Variables of the Hash Model.
list<int>* slot;
list<T>* slotVal;
int slotN;
};
I want to use this specialized version of Hash
to implement another specialization: Hash of
String-Valued Keys.
template <class T>
class Hash<string, T> : public Hash<int, T>
{
public:
Hash(int slotN);
bool insert(string key, T val);
bool remove(string key);
bool contains(string key);
bool query(string key, T& val) ;
private:
// Calculate the String's Hash Key.
int str2key( string key);
};
But it seemed I cannot access fields in the class Hash. Why?

Answer 1

When you say "I cannot access fields in the class Hash" I guess you mean, that you when you are using Hash<string, T> (for some type T) that you cannot call the overloaded functions from Hash<int, T>. The reason for this is name hiding: when you overload a member function in a derived class, all members with the same name in the base class are hidden unless you make them explicitly available. The way to do it is a using declaration:
template <class T>
class Hash<string, T> : public Hash<int, T>
{
public:
Hash(int slotN);
using Hash<int, T>::insert;
using Hash<int, T>::remove;
using Hash<int, T>::contains;
using Hash<int, T>::query;
bool insert(string key, T val);
bool remove(string key);
bool contains(string key);
bool query(string key, T& val) ;
private:
// Calculate the String's Hash Key.
int str2key( string key);
};
If you just need to access the base class members from your derived class's implementation, you can also access the names using qualification with the class name. For example:
template <typename T>
bool Hash<string, T>::insert(string key, T val) {
return this->Hash<int, T>::insert(this->str2key(key, val);
}
Thinking a bit more about the question, there is another potential issue: If you access the data members in the base class you need to make sure that the compiler considers the name a dependent name. Otherwise it is looked up in phase one and won't the names in the base because the can only be found in phase two:
template <typename T>
bool Hash<string, T>::insert(string key, T val) {
int n0 = slotN; // doesn't work: looked up in phase 1
int n1 = this->slotN; // OK: name is dependent
int n2 = Hash<int, T>::slotN; // OK, too
}
Personally, I wouldn't publicly derive from a class with a different key but I assume you have your reasons. BTW, I assume that your primary declaration of Hash looks something like this although it doesn't matter for the problem, really:
template <typename K, typename T>
class Hash;
(if it doesn't have any members, I would rather not define it, either).