How can I define a macro/typedef/etc for tr1::unordered_map that doesn't bind the template arguments?

Question

This might be a little silly question, but I just have to ask it. I am trying to use the unordered_map class in C++, but instead of referencing it everytime as tr1::unordered_map, I would like to just use the keyword hashMap.I know that

typedef tr1::unordered_map<string, int> hashMap 

works but that kind of fixes the datatype of the key and the value corresponding to the hashMap whereas I would like to have more like the following:

#define hashMap tr1::unordered_map

where I can just define the datatype of the key and the value depending on the the requirement, but this does not work. Did anyone face this problem before?

Thanks

Answer 1

This is something that was missing from C++ before C++11. In C++11, you can use template using:

template<typename Key, typename Value>
using hashMap = tr1::unordered_map<Key, Value>;

A usual workaround for C++03 is to make a template structure with a type member:

template<typename Key, typename Value>
struct hashMap {
  typedef tr1::unordered_map<Key, Value> type;
};
// then:
hashMap<string, int>::type myMap;

Inheriting from the class is possible in theory, but usually users refrain from doing so since the STL classes weren't meant to be inherited from.

Answer 2

One possibility would be to use inheritance to forward the key/value pair to unordered_map through a templated hashMap derrived class. IE:

template<typename key, typename value>
class hashMap : public tr1::unordered_map<key, value>
{
public:
     // Add constructors to forward to tr1::unordered_map's constructors as
     // needed
     hashMap() : tr1::unordered_map<key, value>() {} 
     //...
};

Then you can use hashMap as a template but actually be using unordered_map's public interface.

hashMap<string, int> foo;
foo["bar"] = 5;

Just don't do anything fancy other than forward, as STL types don't have virtual destructors.