如何从模板化类方法返回依赖类型？

Question

Let's say I have a class based on a template ThingType . In the header, I use this to typedef a dependent type VectorThingType . I'd like to return this from a method GetVectorOfThings() . If I set VectorThingType as the return type, I get a Does not name a type error, as the type wasn't defined in this scope. Is there any way to do this without duplicating the code in the typedef ?

#include <vector>
#include <iostream>

template< typename ThingType >
class Thing
{
public:

 ThingType aThing;
 typedef std::vector< ThingType > VectorThingType;
 VectorThingType GetVectorOfThings();

Thing(){};
~Thing(){};

};

template< typename ThingType >
//VectorThingType // Does not name a type 
std::vector< ThingType > // Duplication of code from typedef
Thing< ThingType >
::GetVectorOfThings() {
  VectorThingType v;
  v.push_back(this->aThing);
  v.push_back(this->aThing);
  return v;
}

Answer 1

template< typename ThingType >
auto // <-- defer description of type until...
Thing< ThingType >
::GetVectorOfThings()
-> VectorThingType // <-- we are now in the context of Thing< ThingType >
{
  VectorThingType v;
  v.push_back(this->aThing);
  v.push_back(this->aThing);
  return v;
}

Answer 2

Just came across another answer to this question, which doesn't involve c++11.

template< typename ThingType >
typename Thing< ThingType >::VectorThingType
Thing< ThingType >
::GetVectorOfThings()
{
  VectorThingType v;
  v.push_back(this->aThing);
  v.push_back(this->aThing);
  return v;
}

Basically involves assuring the compiler that you are, in fact, dealing with a type via typename and then properly scoping the type using Thing< ThingType >:: . Could be useful if you are stuck with c++03 for some reason.