Compile-time constant id
Asked Answered
T

17

49

Given the following:

template<typename T>
class A
{
public:
    static const unsigned int ID = ?;
};

I want ID to generate a unique compile time ID for every T. I've considered __COUNTER__ and the boost PP library but have been unsuccessful so far. How can I achieve this?

Edit: ID has to be usable as the case in a switch statement

Edit2: All the answers based on the address of a static method or member are incorrect. Although they do create a unique ID they are not resolved in compile time and therefore can not be used as the cases of a switch statement.

Troopship answered 26/9, 2011 at 22:26 Comment(18)
Oh, I'd love to see an answer to this one. AFAIK, there is no solution with the current state of affairs in C++.Sidwohl
Macros aren't going to be reevaluated when the template is instantiated, so that's never going to work.Dias
If you can manage to find a compile-time hash function, make a hash of __FILE__ and __LINE__.Keyhole
Do the ids need to linearly increase from zero? Or just unique ids?Ousel
@tenfour: That would get evaluated the same for all T.Twist
True; I'll leave the comment though in case anyone else makes the same mistake I did. Well, I doubt it's possible to hash FILE at compile-time anyway.Keyhole
Does this have to be int, or can it be a different type (such as intptr_t)?Gantt
Also, why do you need this compile-time counter?Gantt
This is not possible.Proust
Is c++11 acceptable? Or must it be c++03, or c++98?Basicity
@AaronMcDaid c++11, 14, 17, all good, so long as it works.Troopship
What are you looking to do? Are you just looking for "uniqueness" of class A based on different typename? Or unique including subsequent compilations? For example A<int> would have a diff id each compilation? And curious why is it so important for it to be in a case statement?Cathey
You could do thisManfred
@MobileBen uniqueness based on T, doesn't have to change each compilation. At the time this was written people weren't understanding what a 'compile time constant' was, so it was easiest to use the switch example (cases in a switch need to be compile time constant). The result needs to be a compile time constant.Troopship
Instead of automating the traits class, just require that it be defined for every T.Dorthydortmund
@Manfred I think that's right... I think that's the only compile-time answer so far which doesn't require defining IDs per type or listing all the types which can be used... can you make it an answer with some demo code?Troopship
@Troopship I posted a solution using this method....tested it: works with switch statementsManfred
Check this post: #46872769Petrapetracca
A
11

This is sufficient assuming a standards conforming compiler (with respect to the one definition rule):

template<typename T>
class A
{
public:
    static char ID_storage;
    static const void * const ID;
};

template<typename T> char A<T>::ID_storage;
template<typename T> const void * const A<T>::ID= &A<T>::ID_storage;

From the C++ standard 3.2.5 One definition rule [basic.def.odr] (bold emphasis mine):

... If D is a template and is defined in more than one translation unit, then the last four requirements from the list above shall apply to names from the template’s enclosing scope used in the template definition (14.6.3), and also to dependent names at the point of instantiation (14.6.2). If the definitions of D satisfy all these requirements, then the program shall behave as if there were a single definition of D. If the definitions of D do not satisfy these requirements, then the behavior is undefined.

Ayo answered 26/9, 2011 at 23:37 Comment(16)
I would make that a void * - you're only interested in equality comparison, and so there's no need to expose the implementation detail that you're pointing to charsGantt
Is not sufficient, each compilation unit will get its own copy of the values. To work they should be defined in a CPP, per the one definition rule, but it can't be done since they are templates.Sidwohl
@K-ballo, do you have a standard reference for that?Gantt
@bdonlan: For what? The one definition rule? Or the fact that static objects defined within a header file will be defined in each translation unit that includes them? Remember that #include is no different than copy & paste the file contents...Sidwohl
The latter. AIUI templates get special handling here.Gantt
I've just tested this in a multiple compilation unit test - and it worked fine. My compiler was g++ 4.0.1 on OS X, with default compilation flags. (Yes its an old compiler, and yes it doesn't mean its correct by the standard - but at least its one measurement.)Showmanship
I found the relevant text in the C++11 draft standard. So yay! I'm totally right!Ayo
@MSN: Nice... but not usable at compile-time. AFAIK the compile-time generation of unique IDs is impossible. Note that using a pointer to a private static function (somewhat like the Safe Bool idiom) would work equally well, and still not answer the question.Darkness
@MatthieuM., Ah, you can't use it in a switch statement. But, it's certainly usable as a template parameter, assuming you use ID_storage instead of ID. And it even works with partial template specialization!Ayo
@MSN: I am not sure where you're going. I don't see why you would use ID_storage as a template parameter since its value would not identify a particular template instance anyway. And I don't get your comment on partial specialization, do you mean to specialize on A<T>::ID_storage ? It would be a non-deducible context.Darkness
@MatthieuM., you can do: template <char *a, char *b> struct foo { int x; }; template <char *a> struct foo<a, a> { int y }; foo<&A<int>::ID_storage, &A<char>::ID_storage>().x; foo<&A<int>::ID_storage, &A<int>::ID_storage>().y; Not necessarily useful, but possible.Ayo
@MSN: why not directly use A<int>::ID if all you want is an address :x ?Darkness
@MatthieuM., That's not a compile time constant expression. Like I said, not really useful.Ayo
No address is fully evaluated in compile time right? This is not compile time constant and cannot be used in a switch.Troopship
Just to clarify: does everyone agree that, given multiple definitions of a given function/method across many translations units, the linker must select exactly one to be used at runtime? (Perhaps there are other limitations, but this is interesting in isolation)Basicity
You can not take the memory address of something at compile time, thus can not be used with constexpr ids. Plus it creates static objects. Also, why couldn't the value of ID be its own memory address?Denote
S
10

What I usually use is this:

template<typename>
void type_id(){}

using type_id_t = void(*)();

Since every instantiation of the function has it's own address, you can use that address to identify types:

// Work at compile time
constexpr type_id_t int_id = type_id<int>;

// Work at runtime too
std::map<type_id_t, std::any> types;

types[type_id<int>] = 4;
types[type_id<std::string>] = "values"s

// Find values
auto it = types.find(type_id<int>);

if (it != types.end()) {
    // Found it!
}
Satrap answered 22/9, 2016 at 13:51 Comment(8)
this is not compile timeTroopship
The address of a function is known at compile time. So yeah, it work at runtime with std::map but you could use it in constexpr functions too.Satrap
@Troopship The constexpr means that it definitely at compile timeManfred
I tried this one, but was unable to get past "cannot use const void * in a switch statement".Petrapetracca
@NarftheMouse indeed, you won't be able to switch on pointers. My solution is not meant to be used in switch. You might be able to do it by creating a compile time string of the type name and hash the value in a std::size_t. Then switch will work. Using my solution, multiple if statement or a compile time table might work too.Satrap
It may not work on Windows. Address of function can be the same after some Visual Studio optimizations. See: github.com/gelldur/EventBus/issues/19Footed
@DawidDrozd clang on windows also can let duplicates of inline functions. This is why I tried this instead: codereview.stackexchange.com/questions/209950/…Satrap
@GuillaumeRacicot thanks for the link. It would be nice if you could update your answer :)Footed
G
5

It is possible to generate a compile time HASH from a string using the code from this answer.

If you can modify the template to include one extra integer and use a macro to declare the variable:

template<typename T, int ID> struct A
{
    static const int id = ID;
};

#define DECLARE_A(x) A<x, COMPILE_TIME_CRC32_STR(#x)>

Using this macro for the type declaration, the id member contains a hash of the type name. For example:

int main() 
{
    DECLARE_A(int) a;
    DECLARE_A(double) b;
    DECLARE_A(float) c;
    switch(a.id)
    {
    case DECLARE_A(int)::id:
        cout << "int" << endl;
        break;
    case DECLARE_A(double)::id:
        cout << "double" << endl;
        break;
    case DECLARE_A(float)::id:
        cout << "float" << endl;
        break;
    };
    return 0;
}

As the type name is converted to a string, any modification to the type name text results on a different id. For example:

static_assert(DECLARE_A(size_t)::id != DECLARE_A(std::size_t)::id, "");

Another drawback is due to the possibility for a hash collision to occur.

Gamez answered 16/9, 2016 at 23:57 Comment(2)
To reduce collision probability to comfortable level you can use constexpr MD5 function with __int128 type for ID.Etzel
I can submit famous DJB2 hash function that works at compile time too, here it is: wandbox.org/permlink/zcghdF5jXQiGJ6CM, I mean as an more sober alternative to COMPILE_TIME_CRC32_STRPiceous
S
4

This seems to work OK for me:

template<typename T>
class Counted
{
  public:
  static int id()
  {
    static int v;
    return (int)&v;
  }
};

#include <iostream>

int main()
{
  std::cout<<"Counted<int>::id()="<<Counted<int>::id()<<std::endl;
  std::cout<<"Counted<char>::id()="<<Counted<char>::id()<<std::endl;

}
Showmanship answered 27/9, 2011 at 0:10 Comment(6)
That works ok for your use case, now move the definition of Counted to a header of its own, and include it from different translation units.Sidwohl
@Sidwohl Seems to work OK for my simple test case with multiple translation units. Do you have a case where it fails?Showmanship
Mmmh, actually this approach seems to work. The standard guarantees that all translation units will point to the same v object.Sidwohl
I think the result of id() is evaluated in run time.Troopship
@Dave - Ah yeah, it mat not be a compile time constant. I missed that part in the question.Showmanship
This works well but it is not compile time. Try to initialize a static member in a struct.Valadez
N
3

Use the memory address of a static function.

template<typename T>
class A  {
public:
    static void ID() {}
}; 

(&(A<int>::ID)) will be different from (&(A<char>::ID)) and so on.

Nim answered 26/9, 2011 at 22:37 Comment(5)
Compiler complains: "‘ID’ was not declared in this scope" + "‘&’ cannot appear in a constant-expression"Izettaizhevsk
In order to take the address of a static constant integral (note your missing the const modifier), such integral has to be defined in an appropiate CPP file which cannot be done with templates. Is undefined behavior otherwise; and if the static is declared within the header file you'll get different addresses at different compilation units.Sidwohl
Does the C++ standard really guarantee this for ints? This can certainly fail on 64-bit platforms at leastGantt
Also, does the C++ standard even guarantee that template function pointers will compare equal like this?Gantt
@bdonlan: Actually that's not guaranteed since it will get different addresses in different compilation units. Besides, even within the same compilation unit, compilers like MSVC refuse that as they optimize all functions to point to the same one.Sidwohl
R
3

Using this constant expression counter:

template <class T>
class A
{
public:
    static constexpr int ID() { return next(); }
};
class DUMMY { };
int main() {
    std::cout << A<char>::ID() << std::endl;
    std::cout << A<int>::ID() << std::endl;
    std::cout << A<BETA>::ID() << std::endl;
    std::cout << A<BETA>::ID() << std::endl;
    return 0;
}

output: (GCC, C++14)

1
2
3
3

The downside is you will need to guess an upper bound on the number of derived classes for the constant expression counter to work.

Redpoll answered 30/11, 2016 at 5:8 Comment(2)
This looks like it might be a helpful solution for me but I don't see where next is definedQuickstep
I can't get the "solution" from the linked webpage to work... I suppose it was too much of a hack.Shilling
F
2

I encountered this exact problem recently. My solution:

counter.hpp

class counter
{
    static int i;
    static nexti()
    {
        return i++;
    }
};

Counter.cpp:

int counter::i = 0;

templateclass.hpp

#include "counter.hpp"

    template <class T>
    tclass
    {
        static const int id;
    };

    template <class T>
    int tclass<T>::id = counter::nexti();

It appers to work properly in MSVC and GCC, with the one exception that you can't use it in a switch statement.

For various reasons I actually went further, and defined a preprocessor macro that creates a new class from a given name parameter with a static ID (as above) that derives from a common base.

Favien answered 27/9, 2011 at 12:45 Comment(0)
H
2

Here is a possible solution mostly based on templates:

#include<cstddef>
#include<functional>
#include<iostream>

template<typename T>
struct wrapper {
    using type = T;
    constexpr wrapper(std::size_t N): N{N} {}
    const std::size_t N;
};

template<typename... T>
struct identifier: wrapper<T>... {
    template<std::size_t... I>
    constexpr identifier(std::index_sequence<I...>): wrapper<T>{I}... {}

    template<typename U>
    constexpr std::size_t get() const { return wrapper<U>::N; }
};

template<typename... T>
constexpr identifier<T...> ID = identifier<T...>{std::make_index_sequence<sizeof...(T)>{}};

// ---

struct A {};
struct B {};

constexpr auto id = ID<A, B>;

int main() {
    switch(id.get<B>()) {
    case id.get<A>():
        std::cout << "A" << std::endl;
        break;
    case id.get<B>():
        std::cout << "B" << std::endl;
        break;
    }
}

Note that this requires C++14.

All you have to do to associate sequential ids to a list of types is to provide that list to a template variable as in the example above:

constexpr auto id = ID<A, B>;

From that point on, you can get the given id for the given type by means of the get method:

id.get<A>()

And that's all. You can use it in a switch statement as requested and as shown in the example code.

Note that, as long as types are appended to the list of classes to which associate a numeric id, identifiers are the same after each compilation and during each execution.
If you want to remove a type from the list, you can still use fake types as placeholders, as an example:

template<typename> struct noLonger { };
constexpr auto id = ID<noLonger<A>, B>;

This will ensure that A has no longer an associated id and the one given to B won't change.
If you won't to definitely delete A, you can use something like:

constexpr auto id = ID<noLonger<void>, B>;

Or whatever.

Howund answered 22/9, 2016 at 20:39 Comment(3)
constexpr auto id = ID<A, B>; so you have to know every possible type which you want a compile time ID of up front and list them?Troopship
@Troopship Well, yes, you have. This is a kind of an implicit requirement if you want to assign them an id at compile time. Either if you list them somehow or if you decide to base it on the CRTP idiom, you are explicitly marking your types.Howund
@Troopship Added more details. Hoping it can help you.Howund
M
2

Ok.....so this is a hack that I found from this website. It should work. The only thing you need to do is add another template parameter to your struct that takes a counter "meta-object". Note that A with int, bool and char all have unique IDs, but it is not guaranteed that int's will be 1 and bool will be 2, etc., because the order in which templates are initiated is not necessarily known.

Another note:

This will not work with Microsoft Visual C++

#include <iostream>
#include "meta_counter.hpp"

template<typename T, typename counter>
struct A
{
    static const size_t ID = counter::next();
};

int main () {
    typedef atch::meta_counter<void> counter;
    typedef A<int,counter> AInt;
    typedef A<char,counter> AChar;
    typedef A<bool,counter> ABool;
    switch (ABool::ID)
    {
        case AInt::ID:
            std::cout << "Int\n";
            break;
        case ABool::ID:
            std::cout << "Bool\n";
            break;
        case AChar::ID:
            std::cout << "Char\n";
            break;
    }

    std::cout << AInt::ID << std::endl;
    std::cout << AChar::ID << std::endl;
    std::cout << ABool::ID << std::endl;
    std::cout << AInt::ID << std::endl;
    while (1) {}
}

Here is meta_counter.hpp:

// author: Filip Roséen <[email protected]>
// source: http://b.atch.se/posts/constexpr-meta-container

#ifndef ATCH_META_COUNTER_HPP
#define ATCH_META_COUNTER_HPP

#include <cstddef>

namespace atch { namespace {

  template<class Tag>
  struct meta_counter {
    using size_type = std::size_t;

    template<size_type N>
    struct ident {
      friend constexpr size_type adl_lookup (ident<N>);
      static constexpr size_type value = N;
    };

    // - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 

    template<class Ident>
    struct writer {
      friend constexpr size_type adl_lookup (Ident) {
        return Ident::value;
      }

      static constexpr size_type value = Ident::value;
    };

    // - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 

    template<size_type N, int = adl_lookup (ident<N> {})>
    static constexpr size_type value_reader (int, ident<N>) {
      return N;
    }

    template<size_type N>
    static constexpr size_type value_reader (float, ident<N>, size_type R = value_reader (0, ident<N-1> ())) {
      return R;
    }

    static constexpr size_type value_reader (float, ident<0>) {
      return 0;
    }

    // - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 

    template<size_type Max = 64>
    static constexpr size_type value (size_type R = value_reader (0, ident<Max> {})) {
      return R;
    }

    template<size_type N = 1, class H = meta_counter>
    static constexpr size_type next (size_type R = writer<ident<N + H::value ()>>::value) {
      return R;
    }
  };
}}

#endif /* include guard */
Manfred answered 23/9, 2016 at 22:42 Comment(6)
question: Why does A have a template argument counter? Could it not just use meta_counter<void> directly? And is the only purpose to the meta_counter Tag to support having multiple counts?Troopship
@Troopship The only purpose of the Tag argument is to support multiple counts (like I said I did not write a lot of this code). No if you just have meta_counter<void> all the classes have the same ID (I tried that initially though). That said, you could probably do some template trickery so that is transparent if the need was there to do that.Manfred
innterrestttinnnng. I'll have to read about why having meta_counter not as a template argument wouldn't work. That doesn't make a lot of sense to me. Something about template instantiation?Troopship
@Troopship I don't know the answer to that question. All I did was experiment a little.Manfred
Copied and pasted your code, it doesn't seem to work. All the IDs are the same. Using clang / xcode 8 / c++14Troopship
@Troopship using c++ online shell it works......I don't have access to clang so I don't know what to sayManfred
C
2

using template and if constexpr, need c++17

#include <iostream>

template <typename Type, typename... Types>
struct TypeRegister{
    template<typename Queried_type>
    static constexpr int id(){
        if constexpr (std::is_same_v<Type, Queried_type>) return 0;
        else{
            static_assert((sizeof...(Types) > 0), "You shan't query a type you didn't register first");
            return 1 + TypeRegister<Types...>::template id<Queried_type>();
        }
    }
};

int main(){
    using reg_map = TypeRegister<int, float, char, const int&>;
    std::cout << reg_map::id<const int&>() << std::endl;// 3
    // std::cout << reg_map::id<const int>() << std::endl;// error
}
Circumscribe answered 6/4, 2021 at 7:54 Comment(1)
This is exactly what std::tuple is doing from decades agoRazzledazzle
S
1

This can't be done. An address to a static object is the closest you can get to a unique id, however in order to take addresses of such objects (even static const integrals) they must be defined somewhere. Per the one definition rule, they should be defined within a CPP file, which cannot be done since they are templates. If you define the statics within a header file, then each compilation unit will get its own version of it implemented of course at different addresses.

Sidwohl answered 26/9, 2011 at 23:58 Comment(1)
I just posted an answer that does do just thisManfred
H
1

I had a similar problem a few months ago. I was looking for a technique to define identifiers that are the same over each execution.
If this is a requirement, here is another question that explores more or less the same issue (of course, it comes along with its nice answer).
Anyway I didn't use the proposed solution. It follows a description of what I did that time.


You can define a constexpr function like the following one:

static constexpr uint32_t offset = 2166136261u;
static constexpr uint32_t prime = 16777619u;

constexpr uint32_t fnv(uint32_t partial, const char *str) {
    return str[0] == 0 ? partial : fnv((partial^str[0])*prime, str+1);
}

inline uint32_t fnv(const char *str) {
    return fnv(offset, str);
}

Then a class like this from which to inherit:

template<typename T>
struct B {
    static const uint32_t id() {
        static uint32_t val = fnv(T::identifier);
        return val;
    }
};

CRTP idiom does the rest.
As an example, you can define a derived class as it follows:

struct C: B<C> {
    static const char * identifier;
};

const char * C::identifier = "ID(C)";

As long as you provide different identifiers for different classes, you will have unique numeric values that can be used to distinguish between the types.

Identifiers are not required to be part of the derived classes. As an example, you can provide them by means of a trait:

template<typename> struct trait;
template<> struct trait { static const char * identifier; };

// so on with all the identifiers

template<typename T>
struct B {
    static const uint32_t id() {
        static uint32_t val = fnv(trait<T>::identifier);
        return val;
    }
};

Advantages:

  • Easy to implement.
  • No dependencies.
  • Numeric values are the same during each execution.
  • Classes can share the same numeric identifier if needed.

Disadvantages:

  • Error-prone: copy-and-paste can quickly become your worst enemy.

It follows a minimal, working example of what has been described above.
I adapted the code so as to be able to use the ID member method in a switch statement:

#include<type_traits>
#include<cstdint>
#include<cstddef>

static constexpr uint32_t offset = 2166136261u;
static constexpr uint32_t prime = 16777619u;

template<std::size_t I, std::size_t N>
constexpr
std::enable_if_t<(I == N), uint32_t>
fnv(uint32_t partial, const char (&)[N]) {
    return partial;
}

template<std::size_t I, std::size_t N>
constexpr
std::enable_if_t<(I < N), uint32_t>
fnv(uint32_t partial, const char (&str)[N]) {
    return fnv<I+1>((partial^str[I])*prime, str);
}

template<std::size_t N>
constexpr inline uint32_t fnv(const char (&str)[N]) {
    return fnv<0>(offset, str);
}

template<typename T>
struct A {
    static constexpr uint32_t ID() {
        return fnv(T::identifier);
    }
};

struct C: A<C> {
    static constexpr char identifier[] = "foo";
};

struct D: A<D> {
    static constexpr char identifier[] = "bar";
};

int main() {
    constexpr auto val = C::ID();

    switch(val) {
    case C::ID():
        break;
    case D::ID():
        break;
    default:
        break;
    }
}

Please, note that if you want to use ID in a non-constant expression, you must define somewhere the identifiers as it follows:

constexpr char C::identifier[];
constexpr char D::identifier[];

Once you did it, you can do something like this:

int main() {
    constexpr auto val = C::ID();
    // Now, it is well-formed
    auto ident = C::ID();

    // ...
}
Howund answered 17/9, 2016 at 23:3 Comment(2)
If you're requiring every T to have a unique identifier static member string, that defeats the point - they each already have a unique compile time identifierTroopship
@Troopship If your requirement is to have a numeric identifier (differences between compare integers vs compare strings, likely I have not to provide more details about this), but still you want to use a handy and easy to use identifier that is thrown away at compile time... Et voilà!!Howund
C
1

Here is a C++ code that uses __DATE__ and __TIME__ macro to get unique identifiers for types <T>

Format:

// __DATE__ "??? ?? ????"
// __TIME__ "??:??:??"

This is a poor quality hash function:

#define HASH_A 8416451
#define HASH_B 11368711
#define HASH_SEED 9796691    \
+ __DATE__[0x0] * 389        \
+ __DATE__[0x1] * 82421      \
+ __DATE__[0x2] * 1003141    \
+ __DATE__[0x4] * 1463339    \
+ __DATE__[0x5] * 2883371    \
+ __DATE__[0x7] * 4708387    \
+ __DATE__[0x8] * 4709213    \
+ __DATE__[0x9] * 6500209    \
+ __DATE__[0xA] * 6500231    \
+ __TIME__[0x0] * 7071997    \
+ __TIME__[0x1] * 10221293   \
+ __TIME__[0x3] * 10716197   \
+ __TIME__[0x4] * 10913537   \
+ __TIME__[0x6] * 14346811   \
+ __TIME__[0x7] * 15485863

unsigned HASH_STATE = HASH_SEED;
unsigned HASH() {
    return HASH_STATE = HASH_STATE * HASH_A % HASH_B;
}

Using the hash function:

template <typename T>
class A
{
public:
    static const unsigned int ID;
};

template <>
const unsigned int A<float>::ID = HASH();

template <>
const unsigned int A<double>::ID = HASH();

template <>
const unsigned int A<int>::ID = HASH();

template <>
const unsigned int A<short>::ID = HASH();

#include <iostream>

int main() {
    std::cout << A<float>::ID << std::endl;
    std::cout << A<double>::ID << std::endl;
    std::cout << A<int>::ID << std::endl;
    std::cout << A<short>::ID << std::endl;
}
Callum answered 22/9, 2016 at 12:6 Comment(0)
G
0

If non-monotonous values and an intptr_t are acceptable:

template<typename T>
struct TypeID
{
private:
    static char id_ref;
public:
    static const intptr_t ID;
};

template<typename T>
  char TypeID<T>::id_ref;
template<typename T>
  const intptr_t TypeID<T>::ID = (intptr_t)&TypeID<T>::id_ref;

If you must have ints, or must have monotonically incrementing values, I think using static constructors is the only way to go:

// put this in a namespace
extern int counter;

template<typename T>
class Counter {
private:
  Counter() {
    ID_val = counter++;
  }
  static Counter init;
  static int ID_val;
public:
  static const int &ID;
};

template<typename T>
  Counter<T> Counter<T>::init;
template<typename T>
  int Counter<T>::ID_val;
template<typename T>
  const int &Counter<T>::ID = Counter<T>::ID_val;

// in a non-header file somewhere
int counter;

Note that neither of these techniques is safe if you are sharing them between shared libraries and your application!

Gantt answered 26/9, 2011 at 23:53 Comment(1)
Could you explain what you mean by shared libraries (dll, lib or else) and why it's not safe to share between them.Lithesome
S
0

Another alternative is to consider the following class Data with the unique, static member field type:

template <class T>
class Data
{
public:
    static const std::type_index type;
};
// do [static data member initialization](https://mcmap.net/q/277166/-static-data-member-initialization/3041008)
// by [generating unique type id](https://mcmap.net/q/277167/-how-can-i-find-out-if-std-type_index-is-unique-for-my-compiler/3041008)
template <class T>
std::type_index const Data<T>::type = std::type_index(typeid(T));

produces the output (MinGWx64-gcc4.8.4 -std=c++11 -O2)

printf("%s %s\n", Data<int>::type.name(), Data<float>::type.name())
//prints "i f"

It's not exactly an integer id or pretty-printable string, nor a constexpr, but can be used as an index in (un)ordered associative containers.
It also appears to work if the Data.h header is included in multiple files (same hashCode() values).

Skit answered 1/11, 2015 at 1:15 Comment(1)
Right, to quote yourself: "All the answers based on [static initialization] are incorrect. Although they do create a unique ID they are not resolved in compile time and therefore can not be used as the cases of a switch statement"Skit
P
0

Here is a pragmatic solution, if you are ok with writing a single additional line DECLARE_ID(type) for each type you want to use:

 #include <iostream>

 template<class> struct my_id_helper;
 #define DECLARE_ID(C) template<> struct my_id_helper<C> { enum {value = __COUNTER__ }; }

 // actually declare ids:
 DECLARE_ID(int);
 DECLARE_ID(double);
 // this would result in a compile error: redefinition of struct my_id_helper<int>’
 // DECLARE_ID(int);

 template<class T>
 class A
 {
 public:
     static const unsigned int ID = my_id_helper<T>::value;
 };

 int main()
 {
     switch(A<int>::ID)
     {
     case A<int>::ID:    std::cout << "it's an int!\n"; break;
     case A<double>::ID: std::cout << "it's a double!\n"; break;
     // case A<float>::ID: // error: incomplete type ‘my_id_helper<float>’
     default: std::cout << "it's something else\n"; break;
     }
 }
Pellegrini answered 23/9, 2016 at 17:38 Comment(0)
P
0
template<typename T>
static void get_type_id() { void* x; new (x) T(); }
using type_id_t = void(*)();

works fine with optimizations

Peseta answered 27/10, 2018 at 15:31 Comment(0)

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