Typedef a template class without specifying the template parameters

问题

I'm trying to typedef either an unordered_map or std::map depending whether there are TR1 libraries available. But I don't want to specify the template parameters. From what i've read so far, typedef'ing templates without arguments is not possible until official c++0x standard is available. So does anyone know an elegant workaround for this?

#ifdef _TR1
#include <unordered_map> 
typedef std::tr1::unordered_map MyMap; //error C2976: too few template arguments
#else
#include <map> 
typedef std::map MyMap; //error C2976: too few template arguments
#endif

回答1:

The way I've seen this done is to wrap the typedef in a template-struct:

template<typename KeyType, typename MappedType>
struct myMap
{
#ifdef _TR1
    typedef std::tr1::unordered_map<KeyType, MappedType> type;
#else
    typedef std::map<KeyType, MappedType> type;
#endif
};

Then in your code you invoke it like so:

myMap<key, value>::type myMapInstance;

It may be a little more verbose than what you want, but I believe it meets the need given the current state of C++.

回答2:

You have to use full types for typedefs.

Use a #define macro instead.

来源：https://stackoverflow.com/questions/1474417/typedef-a-template-class-without-specifying-the-template-parameters