I'd like to create a type with a compile-time guarantee that instances are one of N possible values. The values are int types. No arithmetic needed.
One possible solution is to use an enum class, but this scales poorly with N.
I think the solution would look most like a fixed range int. I imagine the usage would look something like this, but i'm not too sure how to implement this.Is something like this possible or am I limited to enum classes?
constexpr auto N = 60;
MyType<N> foo(30); // okay
MyType<N> foo(0); // okay
MyType<N> foo(59); // okay
MyType<N> bar(60); // compile error
MyType<N> boo(-1); // compile error
I don't think it's possible to get compile error, however you can limit it using custom class something like this:
#include <stdexcept>
class MyIntger {
public:
MyIntger(int num){
if(num < Min || num > Max){
throw std::runtime_error("Number out of range");
}
value = num;
}
private:
int value;
const static int Min = 0;
const static int Max = 60;
};
int main(){
MyIntger a(5); // OK
MyIntger b(61); // Throws std::runtime_error
}
You can have compile-time ints in C++ (and thus compile-time limit checking), but syntax is not the same as for runtime ints:
#include <type_traits>
template<size_t N>
struct MyInteger
{
// you might have it as T or limit it to size_t
template<typename T, T I>
MyInteger(std::integral_constant<T, I>)
: value(I) { static_assert(I < N); }
size_t value;
};
int main()
{
MyInteger<60> a(std::integral_constant<size_t, 59>()); // ok
MyInteger<60> b(std::integral_constant<size_t, 60>()); // error
}
If you don't like the long name, you can always have a template type alias, e.g.:
template<size_t K>
std::integral_constant<size_t, K> I = {};
int main()
{
MyInteger<60> a(I<59>);
}
I am looking for a convenient to create a C++ class where some member variables are only present if a template flag is set. As a simple example, let's assume I want to toggle an averageSum in an performance sensitive calculation, i.e.
struct Foo {
// Some data and functions..
void operator+=(const Foo& _other) {}
};
template<bool sumAverages>
class Calculator {
public:
// Some member variables...
// Those should only be present if sumAverages is true
int count = 0;
Foo resultSum;
void calculate(/* some arguments */) {
// Calculation of result...
Foo result;
// This should only be calculated if sumAverages is true
++count;
resultSum += result;
// Possibly some post processing...
}
};
One way would be using preprocessor defines, but those are rather inconvenient especially if I need both versions in the same binary. So I am looking for an alternative using templates and if constexpr and something like the following Conditional class:
template<bool active, class T>
struct Conditional;
template<class T>
struct Conditional<true, T> : public T {};
template<class T>
struct Conditional<false, T> {};
My first shot was this:
template<bool sumAverages>
class Calculator {
public:
int count = 0;
Conditional<sumAverages, Foo> resultSum;
void calculate(/* some arguments */) {
Foo result;
if constexpr(sumAverages) {
++count;
resultSum += result;
}
}
};
The if constexpr should incur no run time cost and as it is dependent on a template variable should allow non-compiling code in the false case (e.g. in this example Conditional<false, Foo> does not define a += operator, still it compiles). So this part is more or less perfect. However the variables count and resultSum are still somewhat present. In particular, as one can not derive from a fundamental type, the Conditional class does not allow to toggle the int dependent on the template. Furthermore every Conditional<false, T> variable still occupies one byte possibly bloating small classes. This could be solvable by the new [[no_unique_address]] attribute, however my current compiler chooses to ignore it in all my tests, still using at leas one byte per variable.
To improve things I tried inheriting the variables like this
struct OptionalMembers {
int count;
Foo resultSum;
};
template<bool sumAverages>
class Calculator : public Conditional<sumAverages, OptionalMembers> {
public:
void calculate(/* some arguments */) {
Foo result;
if constexpr(sumAverages) {
++OptionalMembers::count;
OptionalMembers::resultSum += result;
}
}
};
This should come at no space cost as inheriting from am empty class should do literally nothing, right? A possible disadvantage is that one cannot freely set the order of the variables (the inherited variables always come first).
My questions are:
Do you see any problems using the approaches described above?
Are there better options to de(activate) variables like this?
There are a different ways to solve this, one straightforward one would be using template specialization:
#include <iostream>
template <bool b> struct Calculator {
int calculate(int i, int j) { return i + j; }
};
template <> struct Calculator<true> {
int sum;
int calculate(int i, int j) { return sum = i + j; }
};
int main(int argc, char **argv) {
Calculator<false> cx;
cx.calculate(3, 4);
/* std::cout << cx.sum << '\n'; <- will not compile */
Calculator<true> cy;
cy.calculate(3, 4);
std::cout << cy.sum << '\n';
return 0;
}
Another solution would be to use mixin-like types to add features to your calculator type:
#include <iostream>
#include <type_traits>
struct SumMixin {
int sum;
};
template <typename... Mixins> struct Calculator : public Mixins... {
int calculate(int i, int j) {
if constexpr (is_deriving_from<SumMixin>()) {
return SumMixin::sum = i + j;
} else {
return i + j;
}
}
private:
template <typename Mixin> static constexpr bool is_deriving_from() {
return std::disjunction_v<std::is_same<Mixin, Mixins>...>;
}
};
int main(int argc, char **argv) {
Calculator<> cx;
cx.calculate(3, 4);
/* std::cout << cx.sum << '\n'; <- will not compile */
Calculator<SumMixin> cy;
cy.calculate(3, 4);
std::cout << cy.sum << '\n';
return 0;
}
I just want to ask if there are any uses for passing variables in cpp as template arguments
template<int a> struct foo {
int x = a;
};
int main() {
foo<2> bar;
std::cout << bar.x;
}
Something like this compiles, works and cout's 2 but the same thing can be done by doing
struct foo {
int x;
foo(int a) : x(a) {}
};
int main() {
foo bar(2);
std::cout << bar.x;
}
So what is the point of using variables in template arguments? I can also see a big flaw in using the first method: the variable a uses memory and isn't destructed after x is changed, as it would be after the constructor is called in the second example. It might be helpful if you showed some reasonable uses for that.
When you pass a variable through a template argument, it can be used in compile time.
For example, if you need to create a statically sized array in your class, you could use the template argument to pass the size of your array:
template <int TSize>
class Foo {
[...] // Do whatever you need to do with mData.
private:
std::array<int, TSize> mData;
};
There are many uses for constants in template parameters.
Static Sizes
This is how you would start implementing something like a std::array.
template <typename T, size_t SIZE>
struct Array {
T data[SIZE];
}
Template parameters are always usable in a constexpr context, so they can be used as sizes for statically sized arrays.
Providing Compile-Time Parameters to Algorithms
Another use is parametrizing algorithms like in the following code sample.
We have a uint32_t in ARGB order but to store it in a file, we might need to reorder it to BGRA or RGBA. We know the order at compile time, so we could use an ArgbOrder template variable.
enum class ArgbOrder { ARGB, RGBA, BGRA };
struct ChannelOffsets {
unsigned a;
unsigned r;
unsigned g;
unsigned b;
};
// and we can get a constexpr lookup table from this enum
constexpr ChannelOffsets byteShiftAmountsOf(ArgbOrder format)
{
...
}
template <ArgbOrder order>
void encodeArgb(uint32_t argb, uint8_t out[4])
{
// We can generate the shift amounts at compile time.
constexpr detail::ChannelOffsets shifts = shiftAmountsOf(order);
out[0] = static_cast<u8>(argb >> shifts.a);
out[1] = static_cast<u8>(argb >> shifts.r);
out[2] = static_cast<u8>(argb >> shifts.g);
out[3] = static_cast<u8>(argb >> shifts.b);
}
void example() {
encodeArgb<ArgbOrder::BGRA>(12345);
}
In this example, we can select the appropriate lookup table at compile time and have zero runtime cost. All that needs to happen at runtime is 4 shifts.
Feature Toggles
We can use bool template variables to toggle features in our code, like for example:
template <bool handleZeroSpecially>
int div(int x, int y) {
if constexpr (handleZeroSpecially) {
return y == 0 ? 0 : x / y;
}
else {
return x / y;
}
}
My question is about change/set dynamically a definition of a type checking a value of a variable which can change in run time like this simple example:
void changeMode(int mode)
{
if(mode == 1)
{
typedef MyType float;
}
else if(mode == 2)
{
typedef MyType double;
}
}
int main(int argc, char *argv[])
{
MyType (2);
MyType A = 3;
MyType B = 2;
MyType C = A + B;
return 0;
}
it's possible? It's a classic case of templates? it is possible to avoid using templates? My goal is centralize the type definition in order to be possible to switch at runtime, without the need to extend it to each class or use "templates" for each class/function that will use the given type.
typedef is evaluated at compile time, so no.
but you can do something like this using prepocessor:
#ifdef TYPE_FLOAT
typedef float Type;
#else
typedef int Type;
#endif
but keep in mind it's at compile time
You can extract your type-dependent code to function template (or class template in more complex cases):
template <typename MyType>
void ExecuteTypeDependentCode()
{
MyType A = 3;
MyType B = 2;
MyType C = A + B;
}
Then you can write the wrapper, its task is to switch between modes:
enum Modes
{
FloatMode = 1,
DoubleMode = 2
};
void ExecuteTypeDependentCode(Modes mode)
{
switch (mode)
{
case FloatMode:
ExecuteTypeDependentCode<float>();
break;
case DoubleMode:
ExecuteTypeDependentCode<double>();
break;
}
}
Example of using:
int main(int argc, char *argv[])
{
ExecuteTypeDependentCode(DoubleMode);
return 0;
}
The process of switching can be written only once (I described this method here).
Typedef is evaluated statically, so you need to do something like to have a "classic case of templates" (whatever that is :))
template <int Mode>
struct Dispatch { };
template <>
struct Dispatch <0> {
using U = int; // fancy way of saying typedef
};
template <>
struct Dispatch <1> {
using U = char; // feel free to use typedef
};
int main() {
Dispatch<0>::U x; // compile time constant in <> (zero)
x = 5;
Dispatch<1>::U y; // compile time constant in <> (one)
y = 'c';
}
As other answers have pointed out, this is not possible to achieve via typedef. However, you can emulate this behavior using polymorpic wrapper classes:
#include <cstdio>
class Gen
{
public:
static int mode;
static Gen* get()
{
switch(mode)
{
case 1:
return new GenFloat();
case 2:
return new GenDouble();
}
return NULL;
}
};
int Gen::mode = 0;
class DoubleGen : public Gen
{
public:
DoubleGen() : d(0) {}
DoubleGen(double val) : d(val) {}
double d;
};
class FloatGen : public Gen
{
public:
FloatGen() : f(0) {}
FloatGen(float val) : f(val) {}
float f;
};
int main(void)
{
Gen::mode = 1;
Gen* _float = gen::get();
gen::mode = 2;
Gen* _double = gen::get();
delete _float;
delete _double;
return 0;
}
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.
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.
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!
}
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.
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;
}
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.
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.
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.
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.
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 <filip.roseen#gmail.com>
// 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 */
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
}
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.
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();
// ...
}
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;
}
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!
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](http://stackoverflow.com/q/11300652/3041008)
// by [generating unique type id](http://stackoverflow.com/q/26794944/3041008)
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).
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;
}
}
template<typename T>
static void get_type_id() { void* x; new (x) T(); }
using type_id_t = void(*)();
works fine with optimizations