Mapping integers to types using C++ template fails in a specific case

Question

I am attempting to compile the following template based code in VC++ 2005.

    #include <iostream>
using namespace std;


/*
 * T is a template which maps an integer to a specific type.
 * The mapping happens through partial template specialization.
 * In the following T<1> is mapped to char, T<2> is mapped to long
 * and T<3> is mapped to float using partial template specializations
 */
template <int x>
struct T
{
public:
};

template<>
struct T<1>
{
public:
    typedef char xType;
};


template<>
struct T<2>
{
public:
    typedef long xType;
};

template<>
struct T<3>
{
public:
    typedef float xType;
};

// We can easily access the specific xType for a specific T<N>
typedef T<3>::xType x3Type;

/*!
 * In the following we are attempting to use T<N> inside another
 * template class T2<R>
 */

template<int r>
struct T2
{
    //We can map T<r> to some other type T3
    typedef T<r> T3;
    // The following line fails
    typedef T3::xType xType;
};

int main()
{
    T<1>::xType a1;
    cout << typeid(a1).name() << endl;
    T<2>::xType a2;
    cout << typeid(a2).name() << endl;
    T<3>::xType a3;
    cout << typeid(a3).name() << endl;
    return 0;
}

There is a particular line in the code which doesn't compile:

typedef T3::xType xType;

If I remove this line, compilation goes fine and the result is:

char
long
float

If I retain this line, compilation errors are observed.

main.cpp(53) : warning C4346: 'T<x>::xType' : dependent name is not a type
    prefix with 'typename' to indicate a type
    main.cpp(54) : see reference to class template instantiation 'T2<r>' being compiled
main.cpp(53) : error C2146: syntax error : missing ';' before identifier 'xType'
main.cpp(53) : error C4430: missing type specifier - int assumed. Note: C++ does not    support default-int

I am not able to figure out how to make sure that T::xType can be treated as a type inside the T2 template. Any help is highly appreciated.

Answer 1

Try

typedef typename  T3::xType xType;

Answer 2

Since T3 in your template class depends on the template parameter, the compiler can't known for sure what T3::xType will refer to (that might depend on the actual type r in each instantiation T2<r>).

To tell the compiler that T3::xType will be a type, you need to add the typename keyword:

typedef typename T3::xType xType;

Answer 3

The error message tells you exactly what you need to do: Add typename before the type.

typedef typename T3::xType xType;

The reason you need this is that if there's an identifier that can be treated as a variable or a type, the compiler will treat it as a variable, which is what's happening in this case. In order to let the compiler know that it's actually a type you use the typename keyword.