Template metaprogram converting type to unique number

Question

I just started playing with metaprogramming and I am working on different tasks just to explore the domain. One of these was to generate a unique integer and map it to type,

Answer 1

I'm not aware of a way to map a compile-time constant integer to a type, but I can give you the next best thing. This example demonstrates a way to generate a unique identifier for a type which - while it is not an integral constant expression - will generally be evaluated at compile time. It's also potentially useful if you need a mapping between a type and a unique non-type template argument.

struct Dummy
{
};

template<typename>
struct TypeDummy
{
    static const Dummy value;
};

template<typename T>
const Dummy TypeDummy<T>::value = Dummy();

typedef const Dummy* TypeId;

template<typename T, TypeId p = &TypeDummy<T>::value>
struct TypePtr
{
    static const TypeId value;
};

template<typename T, TypeId p>
const TypeId TypePtr<T, p>::value = p;

struct A{};

struct B{};

const TypeId typeA = TypePtr<A>::value;
const TypeId typeB = TypePtr<B>::value;

I developed this as a workaround for performance issues with ordering types using typeid(A) == typeid(B), which a certain compiler fails to evaluate at compile time. It's also useful to be able to store TypeId values for comparison at runtime: e.g. someType == TypePtr<A>::value

Answer 2

I think it is possible to do it for a fixed set of types, but quite a bit of work. You'll need to define a specialisation for each type, but it should be possible to use compile-time asserts to check for uniqueness. I'll assume a STATIC_ASSERT(const_expr), like the one in Boost.StaticAssert, that causes a compilation failure if the expression is false.

Suppose we have a set of types that we want unique IDs for - just 3 for this example:

class TypeA;
class TypeB;
typedef int TypeC;

We'll want a way to compare types:

template <typename T, typename U> struct SameType
{
    const bool value = false;
};

template <typename T> struct SameType<T,T>
{
    const bool value = true;
};

Now, we define an ordering of all the types we want to enumerate:

template <typename T> struct Ordering {};

template <> struct Ordering<void>
{
    typedef TypeC prev;
    typedef TypeA next;
};

template <> struct Ordering<TypeA>
{
    typedef void  prev;
    typedef TypeB next;
};

template <> struct Ordering<TypeB>
{
    typedef TypeA prev;
    typedef TypeC next;
};

template <> struct Ordering<TypeC>
{
    typedef TypeB prev;
    typedef void  next;
};

Now we can define the unique ID:

template <typename T> struct TypeInt
{
    STATIC_ASSERT(SameType<Ordering<T>::prev::next, T>::value);
    static int value = TypeInt<T>::prev::value + 1;
};

template <> struct TypeInt<void>
{
    static int value = 0;
};

NOTE: I haven't tried compiling any of this. It may need typename adding in a few places, and it may not work at all.

You can't hope to map all possible types to an integer field, because there are an unbounded number of them: pointer types with arbitrary levels of indirection, array types of arbitrary size and rank, function types with arbitrary numbers of arguments, and so on.

Answer 3

In principle, this is possible, although the solution probably isn't what you're looking for.

In short, you need to provide an explicit mapping from the types to the integer values, with one entry for each possible type:

template< typename T >
struct type2int
{
   // enum { result = 0 }; // do this if you want a fallback value
};

template<> struct type2int<AClass> { enum { result = 1 }; };
template<> struct type2int<BClass> { enum { result = 2 }; };
template<> struct type2int<CClass> { enum { result = 3 }; };

const int i = type2int<T>::result;

If you don't supply the fallback implementation in the base template, this will fail for unknown types if T, otherwise it would return the fallback value.

Depending on your context, there might be other possibilities, too. For example, you could define those numbers within within the types themselves:

class AClass {
  public:
    enum { inta_val = 1 };
  // ...
};

class BClass {
  public:
    enum { inta_val = 2 };
  // ...
};

// ... 

template< typename T >
struct type2int
{
   enum { result = T::int_val }; // will fail for types without int_val
};

If you give more context, there might be other solutions, too.

Edit:

Actually there isn't any more context to it. I was looking into if it actually was possible, but without assigning the numbers itself.

I think Mike's idea of ordering is a good way to do this (again, for a fixed set of types) without having to explicitly assign numbers: they're implicitly given by the ordering. However, I think that this would be easier by using a type list. The index of any type in the list would be its number. I think something like the following might do:

// basic type list manipulation stuff
template< typename T1, typename T2, typename T3...>
struct type_list;

// meta function, List is assumed to be some instance of type_list
template< typename T, class List >
struct index_of {
  enum { result = /* find index of T in List */ };
};

// the list of types you support
typedef type_list<AClass, BClass, CClass> the_type_list;

// your meta function
template< typename T >
struct type2int
{
   enum { result = index_of<T, the_type_list>::result };
};