Meta programming: Declare a new struct on the fly












14















Is it possible to declare a new type (an empty struct , or a struct without an implementation) on the fly?



E.g.



constexpr auto make_new_type() -> ???;

using A = decltype(make_new_type());
using B = decltype(make_new_type());
using C = decltype(make_new_type());

static_assert(!std::is_same<A, B>::value, "");
static_assert(!std::is_same<B, C>::value, "");
static_assert(!std::is_same<A, C>::value, "");


A "manual" solution is



template <class> struct Tag;

using A = Tag<struct TagA>;
using B = Tag<struct TagB>;
using C = Tag<struct TagC>;


or even



struct A;
struct B;
struct C;


but for templating / meta some magic make_new_type() function would be nice.



Can something like that be possible now that stateful metaprogramming is ill-formed?










share|improve this question


















  • 7





    Why would someone want to do this ? what is a typical use case?

    – Samer Tufail
    yesterday






  • 3





    Every lambda has a unique type :) As far as I know, they are the “just give me a unique type” idiom — the only one in C++11.

    – Kuba Ober
    yesterday













  • Related: unconstexpr. (no longer works as of GCC 8, and the code there is probably ill-formed NDR)

    – HolyBlackCat
    21 hours ago


















14















Is it possible to declare a new type (an empty struct , or a struct without an implementation) on the fly?



E.g.



constexpr auto make_new_type() -> ???;

using A = decltype(make_new_type());
using B = decltype(make_new_type());
using C = decltype(make_new_type());

static_assert(!std::is_same<A, B>::value, "");
static_assert(!std::is_same<B, C>::value, "");
static_assert(!std::is_same<A, C>::value, "");


A "manual" solution is



template <class> struct Tag;

using A = Tag<struct TagA>;
using B = Tag<struct TagB>;
using C = Tag<struct TagC>;


or even



struct A;
struct B;
struct C;


but for templating / meta some magic make_new_type() function would be nice.



Can something like that be possible now that stateful metaprogramming is ill-formed?










share|improve this question


















  • 7





    Why would someone want to do this ? what is a typical use case?

    – Samer Tufail
    yesterday






  • 3





    Every lambda has a unique type :) As far as I know, they are the “just give me a unique type” idiom — the only one in C++11.

    – Kuba Ober
    yesterday













  • Related: unconstexpr. (no longer works as of GCC 8, and the code there is probably ill-formed NDR)

    – HolyBlackCat
    21 hours ago
















14












14








14


1






Is it possible to declare a new type (an empty struct , or a struct without an implementation) on the fly?



E.g.



constexpr auto make_new_type() -> ???;

using A = decltype(make_new_type());
using B = decltype(make_new_type());
using C = decltype(make_new_type());

static_assert(!std::is_same<A, B>::value, "");
static_assert(!std::is_same<B, C>::value, "");
static_assert(!std::is_same<A, C>::value, "");


A "manual" solution is



template <class> struct Tag;

using A = Tag<struct TagA>;
using B = Tag<struct TagB>;
using C = Tag<struct TagC>;


or even



struct A;
struct B;
struct C;


but for templating / meta some magic make_new_type() function would be nice.



Can something like that be possible now that stateful metaprogramming is ill-formed?










share|improve this question














Is it possible to declare a new type (an empty struct , or a struct without an implementation) on the fly?



E.g.



constexpr auto make_new_type() -> ???;

using A = decltype(make_new_type());
using B = decltype(make_new_type());
using C = decltype(make_new_type());

static_assert(!std::is_same<A, B>::value, "");
static_assert(!std::is_same<B, C>::value, "");
static_assert(!std::is_same<A, C>::value, "");


A "manual" solution is



template <class> struct Tag;

using A = Tag<struct TagA>;
using B = Tag<struct TagB>;
using C = Tag<struct TagC>;


or even



struct A;
struct B;
struct C;


but for templating / meta some magic make_new_type() function would be nice.



Can something like that be possible now that stateful metaprogramming is ill-formed?







c++ templates metaprogramming stateful compile-time-constant






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked yesterday









kaykay

18.6k970118




18.6k970118








  • 7





    Why would someone want to do this ? what is a typical use case?

    – Samer Tufail
    yesterday






  • 3





    Every lambda has a unique type :) As far as I know, they are the “just give me a unique type” idiom — the only one in C++11.

    – Kuba Ober
    yesterday













  • Related: unconstexpr. (no longer works as of GCC 8, and the code there is probably ill-formed NDR)

    – HolyBlackCat
    21 hours ago
















  • 7





    Why would someone want to do this ? what is a typical use case?

    – Samer Tufail
    yesterday






  • 3





    Every lambda has a unique type :) As far as I know, they are the “just give me a unique type” idiom — the only one in C++11.

    – Kuba Ober
    yesterday













  • Related: unconstexpr. (no longer works as of GCC 8, and the code there is probably ill-formed NDR)

    – HolyBlackCat
    21 hours ago










7




7





Why would someone want to do this ? what is a typical use case?

– Samer Tufail
yesterday





Why would someone want to do this ? what is a typical use case?

– Samer Tufail
yesterday




3




3





Every lambda has a unique type :) As far as I know, they are the “just give me a unique type” idiom — the only one in C++11.

– Kuba Ober
yesterday







Every lambda has a unique type :) As far as I know, they are the “just give me a unique type” idiom — the only one in C++11.

– Kuba Ober
yesterday















Related: unconstexpr. (no longer works as of GCC 8, and the code there is probably ill-formed NDR)

– HolyBlackCat
21 hours ago







Related: unconstexpr. (no longer works as of GCC 8, and the code there is probably ill-formed NDR)

– HolyBlackCat
21 hours ago














3 Answers
3






active

oldest

votes


















19














You can almost get the syntax you want using



template <size_t>
constexpr auto make_new_type() { return (){}; }

using A = decltype(make_new_type<__LINE__>());
using B = decltype(make_new_type<__LINE__>());
using C = decltype(make_new_type<__LINE__>());


This works since every lambda expression results in a unique type. So for each unique value in <> you get a different function which returns a different closure.



If you introduce a macro you can get rid of having to type __LINE__ like



template <size_t>
constexpr auto new_type() { return (){}; }

#define make_new_type new_type<__LINE__>()

using A = decltype(make_new_type);
using B = decltype(make_new_type);
using C = decltype(make_new_type);





share|improve this answer





















  • 5





    You rely only on unicity of __LINE__ (so care with multiple TU, or several types on same line), so template <size_t> struct unique_tag {}; would be enough -> #define make_new_type unique_tag<__LINE__>. and using A = make_new_type;

    – Jarod42
    yesterday













  • @Jarod42 Good point. Mind if I add that to the answer as an alternative?

    – NathanOliver
    yesterday











  • Add it to the answer if you want..

    – Jarod42
    23 hours ago











  • Why not simply using A = (){};?

    – Mooing Duck
    19 hours ago






  • 1





    using A = decltype((){}); then.

    – Mooing Duck
    18 hours ago



















18














In C++20:



using A = decltype({}); // an idiom
using B = decltype({});
...


This is idiomatic code: that’s how one writes “give me a unique type” in C++20.



In C++11, the clearest and simplest approach uses __LINE__:



namespace {
template <int> class new_type {};
}

using A = new_type<__LINE__>; // an idiom - pretty much
using B = new_type<__LINE__>;


The anonymous namespace is the most important bit. It is a serious mistake not to put the new_type class in the anonymous namespace: the types then won't be unique anymore across translation units. All sorts of hilarity will ensue 15 minutes before you plan to ship :)



This extends to C++98:



namespace {
template <int> class new_type {};
}

typedef new_type<__LINE__> A; // an idiom - pretty much
typedef new_type<__LINE__> B;


Another approach would be to manually chain the types, and have the compiler statically validate that the chaining was done correctly, and bomb out with an error if you don’t. So it’d not be brittle (assuming the magic works out).



Something like:



namespace {
struct base_{
using discr = std::integral_type<int, 0>;
};

template <class Prev> class new_type {
[magic here]
using discr = std::integral_type<int, Prev::discr+1>;
};
}

using A = new_type<base_>;
using A2 = new_type<base_>;
using B = new_type<A>;
using C = new_type<B>;
using C2 = new_type<B>;


It takes only a small bit of magic to ensure that the lines with types A2 and C2 don’t compile. Whether that magic is possible in C++11 is another story.






share|improve this answer


























  • Is "lambda expression in an unevaluated operand" allowed in C++20?

    – kay
    yesterday











  • Yes – and it’s a big deal. It enables some constructs not previously possible at all, with no amount of template metaprogramming. Heck, it even makes them easy.

    – Kuba Ober
    yesterday













  • In using discr = std::integral_type<int, Prev::value+1>; do you mean using discr = std::integral_type<int, Prev::discr+1>;? Also? How does using A = new_type<base_>; using A2 = new_type<base_>; give two different types?

    – NathanOliver
    yesterday











  • It wouldn't give two different types because the second one would fail to compile. It's a bit of a pita to implement but that's the idea. It requires riding on the edges of the standard.

    – Kuba Ober
    23 hours ago





















3














I know... they are distilled evil... but seems to me that this is a works for an old C-style macro



#include <type_traits>

#define newType(x)
struct type_##x {};
using x = type_##x;

newType(A)
newType(B)
newType(C)

int main ()
{
static_assert(!std::is_same<A, B>::value, "");
static_assert(!std::is_same<B, C>::value, "");
static_assert(!std::is_same<A, C>::value, "");
}





share|improve this answer



















  • 1





    I don’t think this is any more “on the fly” than what OP already does. In particular, you still need to pass different identifiers to the macro.

    – Konrad Rudolph
    yesterday











  • @KonradRudolph - I don't know... the OP uses the identifier on the left of the = operator so I don't think is really different. But I have to admit that the lambda-based solution from NathanOliver is much better and really elegant.

    – max66
    yesterday











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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









19














You can almost get the syntax you want using



template <size_t>
constexpr auto make_new_type() { return (){}; }

using A = decltype(make_new_type<__LINE__>());
using B = decltype(make_new_type<__LINE__>());
using C = decltype(make_new_type<__LINE__>());


This works since every lambda expression results in a unique type. So for each unique value in <> you get a different function which returns a different closure.



If you introduce a macro you can get rid of having to type __LINE__ like



template <size_t>
constexpr auto new_type() { return (){}; }

#define make_new_type new_type<__LINE__>()

using A = decltype(make_new_type);
using B = decltype(make_new_type);
using C = decltype(make_new_type);





share|improve this answer





















  • 5





    You rely only on unicity of __LINE__ (so care with multiple TU, or several types on same line), so template <size_t> struct unique_tag {}; would be enough -> #define make_new_type unique_tag<__LINE__>. and using A = make_new_type;

    – Jarod42
    yesterday













  • @Jarod42 Good point. Mind if I add that to the answer as an alternative?

    – NathanOliver
    yesterday











  • Add it to the answer if you want..

    – Jarod42
    23 hours ago











  • Why not simply using A = (){};?

    – Mooing Duck
    19 hours ago






  • 1





    using A = decltype((){}); then.

    – Mooing Duck
    18 hours ago
















19














You can almost get the syntax you want using



template <size_t>
constexpr auto make_new_type() { return (){}; }

using A = decltype(make_new_type<__LINE__>());
using B = decltype(make_new_type<__LINE__>());
using C = decltype(make_new_type<__LINE__>());


This works since every lambda expression results in a unique type. So for each unique value in <> you get a different function which returns a different closure.



If you introduce a macro you can get rid of having to type __LINE__ like



template <size_t>
constexpr auto new_type() { return (){}; }

#define make_new_type new_type<__LINE__>()

using A = decltype(make_new_type);
using B = decltype(make_new_type);
using C = decltype(make_new_type);





share|improve this answer





















  • 5





    You rely only on unicity of __LINE__ (so care with multiple TU, or several types on same line), so template <size_t> struct unique_tag {}; would be enough -> #define make_new_type unique_tag<__LINE__>. and using A = make_new_type;

    – Jarod42
    yesterday













  • @Jarod42 Good point. Mind if I add that to the answer as an alternative?

    – NathanOliver
    yesterday











  • Add it to the answer if you want..

    – Jarod42
    23 hours ago











  • Why not simply using A = (){};?

    – Mooing Duck
    19 hours ago






  • 1





    using A = decltype((){}); then.

    – Mooing Duck
    18 hours ago














19












19








19







You can almost get the syntax you want using



template <size_t>
constexpr auto make_new_type() { return (){}; }

using A = decltype(make_new_type<__LINE__>());
using B = decltype(make_new_type<__LINE__>());
using C = decltype(make_new_type<__LINE__>());


This works since every lambda expression results in a unique type. So for each unique value in <> you get a different function which returns a different closure.



If you introduce a macro you can get rid of having to type __LINE__ like



template <size_t>
constexpr auto new_type() { return (){}; }

#define make_new_type new_type<__LINE__>()

using A = decltype(make_new_type);
using B = decltype(make_new_type);
using C = decltype(make_new_type);





share|improve this answer















You can almost get the syntax you want using



template <size_t>
constexpr auto make_new_type() { return (){}; }

using A = decltype(make_new_type<__LINE__>());
using B = decltype(make_new_type<__LINE__>());
using C = decltype(make_new_type<__LINE__>());


This works since every lambda expression results in a unique type. So for each unique value in <> you get a different function which returns a different closure.



If you introduce a macro you can get rid of having to type __LINE__ like



template <size_t>
constexpr auto new_type() { return (){}; }

#define make_new_type new_type<__LINE__>()

using A = decltype(make_new_type);
using B = decltype(make_new_type);
using C = decltype(make_new_type);






share|improve this answer














share|improve this answer



share|improve this answer








edited yesterday

























answered yesterday









NathanOliverNathanOliver

96.6k16137210




96.6k16137210








  • 5





    You rely only on unicity of __LINE__ (so care with multiple TU, or several types on same line), so template <size_t> struct unique_tag {}; would be enough -> #define make_new_type unique_tag<__LINE__>. and using A = make_new_type;

    – Jarod42
    yesterday













  • @Jarod42 Good point. Mind if I add that to the answer as an alternative?

    – NathanOliver
    yesterday











  • Add it to the answer if you want..

    – Jarod42
    23 hours ago











  • Why not simply using A = (){};?

    – Mooing Duck
    19 hours ago






  • 1





    using A = decltype((){}); then.

    – Mooing Duck
    18 hours ago














  • 5





    You rely only on unicity of __LINE__ (so care with multiple TU, or several types on same line), so template <size_t> struct unique_tag {}; would be enough -> #define make_new_type unique_tag<__LINE__>. and using A = make_new_type;

    – Jarod42
    yesterday













  • @Jarod42 Good point. Mind if I add that to the answer as an alternative?

    – NathanOliver
    yesterday











  • Add it to the answer if you want..

    – Jarod42
    23 hours ago











  • Why not simply using A = (){};?

    – Mooing Duck
    19 hours ago






  • 1





    using A = decltype((){}); then.

    – Mooing Duck
    18 hours ago








5




5





You rely only on unicity of __LINE__ (so care with multiple TU, or several types on same line), so template <size_t> struct unique_tag {}; would be enough -> #define make_new_type unique_tag<__LINE__>. and using A = make_new_type;

– Jarod42
yesterday







You rely only on unicity of __LINE__ (so care with multiple TU, or several types on same line), so template <size_t> struct unique_tag {}; would be enough -> #define make_new_type unique_tag<__LINE__>. and using A = make_new_type;

– Jarod42
yesterday















@Jarod42 Good point. Mind if I add that to the answer as an alternative?

– NathanOliver
yesterday





@Jarod42 Good point. Mind if I add that to the answer as an alternative?

– NathanOliver
yesterday













Add it to the answer if you want..

– Jarod42
23 hours ago





Add it to the answer if you want..

– Jarod42
23 hours ago













Why not simply using A = (){};?

– Mooing Duck
19 hours ago





Why not simply using A = (){};?

– Mooing Duck
19 hours ago




1




1





using A = decltype((){}); then.

– Mooing Duck
18 hours ago





using A = decltype((){}); then.

– Mooing Duck
18 hours ago













18














In C++20:



using A = decltype({}); // an idiom
using B = decltype({});
...


This is idiomatic code: that’s how one writes “give me a unique type” in C++20.



In C++11, the clearest and simplest approach uses __LINE__:



namespace {
template <int> class new_type {};
}

using A = new_type<__LINE__>; // an idiom - pretty much
using B = new_type<__LINE__>;


The anonymous namespace is the most important bit. It is a serious mistake not to put the new_type class in the anonymous namespace: the types then won't be unique anymore across translation units. All sorts of hilarity will ensue 15 minutes before you plan to ship :)



This extends to C++98:



namespace {
template <int> class new_type {};
}

typedef new_type<__LINE__> A; // an idiom - pretty much
typedef new_type<__LINE__> B;


Another approach would be to manually chain the types, and have the compiler statically validate that the chaining was done correctly, and bomb out with an error if you don’t. So it’d not be brittle (assuming the magic works out).



Something like:



namespace {
struct base_{
using discr = std::integral_type<int, 0>;
};

template <class Prev> class new_type {
[magic here]
using discr = std::integral_type<int, Prev::discr+1>;
};
}

using A = new_type<base_>;
using A2 = new_type<base_>;
using B = new_type<A>;
using C = new_type<B>;
using C2 = new_type<B>;


It takes only a small bit of magic to ensure that the lines with types A2 and C2 don’t compile. Whether that magic is possible in C++11 is another story.






share|improve this answer


























  • Is "lambda expression in an unevaluated operand" allowed in C++20?

    – kay
    yesterday











  • Yes – and it’s a big deal. It enables some constructs not previously possible at all, with no amount of template metaprogramming. Heck, it even makes them easy.

    – Kuba Ober
    yesterday













  • In using discr = std::integral_type<int, Prev::value+1>; do you mean using discr = std::integral_type<int, Prev::discr+1>;? Also? How does using A = new_type<base_>; using A2 = new_type<base_>; give two different types?

    – NathanOliver
    yesterday











  • It wouldn't give two different types because the second one would fail to compile. It's a bit of a pita to implement but that's the idea. It requires riding on the edges of the standard.

    – Kuba Ober
    23 hours ago


















18














In C++20:



using A = decltype({}); // an idiom
using B = decltype({});
...


This is idiomatic code: that’s how one writes “give me a unique type” in C++20.



In C++11, the clearest and simplest approach uses __LINE__:



namespace {
template <int> class new_type {};
}

using A = new_type<__LINE__>; // an idiom - pretty much
using B = new_type<__LINE__>;


The anonymous namespace is the most important bit. It is a serious mistake not to put the new_type class in the anonymous namespace: the types then won't be unique anymore across translation units. All sorts of hilarity will ensue 15 minutes before you plan to ship :)



This extends to C++98:



namespace {
template <int> class new_type {};
}

typedef new_type<__LINE__> A; // an idiom - pretty much
typedef new_type<__LINE__> B;


Another approach would be to manually chain the types, and have the compiler statically validate that the chaining was done correctly, and bomb out with an error if you don’t. So it’d not be brittle (assuming the magic works out).



Something like:



namespace {
struct base_{
using discr = std::integral_type<int, 0>;
};

template <class Prev> class new_type {
[magic here]
using discr = std::integral_type<int, Prev::discr+1>;
};
}

using A = new_type<base_>;
using A2 = new_type<base_>;
using B = new_type<A>;
using C = new_type<B>;
using C2 = new_type<B>;


It takes only a small bit of magic to ensure that the lines with types A2 and C2 don’t compile. Whether that magic is possible in C++11 is another story.






share|improve this answer


























  • Is "lambda expression in an unevaluated operand" allowed in C++20?

    – kay
    yesterday











  • Yes – and it’s a big deal. It enables some constructs not previously possible at all, with no amount of template metaprogramming. Heck, it even makes them easy.

    – Kuba Ober
    yesterday













  • In using discr = std::integral_type<int, Prev::value+1>; do you mean using discr = std::integral_type<int, Prev::discr+1>;? Also? How does using A = new_type<base_>; using A2 = new_type<base_>; give two different types?

    – NathanOliver
    yesterday











  • It wouldn't give two different types because the second one would fail to compile. It's a bit of a pita to implement but that's the idea. It requires riding on the edges of the standard.

    – Kuba Ober
    23 hours ago
















18












18








18







In C++20:



using A = decltype({}); // an idiom
using B = decltype({});
...


This is idiomatic code: that’s how one writes “give me a unique type” in C++20.



In C++11, the clearest and simplest approach uses __LINE__:



namespace {
template <int> class new_type {};
}

using A = new_type<__LINE__>; // an idiom - pretty much
using B = new_type<__LINE__>;


The anonymous namespace is the most important bit. It is a serious mistake not to put the new_type class in the anonymous namespace: the types then won't be unique anymore across translation units. All sorts of hilarity will ensue 15 minutes before you plan to ship :)



This extends to C++98:



namespace {
template <int> class new_type {};
}

typedef new_type<__LINE__> A; // an idiom - pretty much
typedef new_type<__LINE__> B;


Another approach would be to manually chain the types, and have the compiler statically validate that the chaining was done correctly, and bomb out with an error if you don’t. So it’d not be brittle (assuming the magic works out).



Something like:



namespace {
struct base_{
using discr = std::integral_type<int, 0>;
};

template <class Prev> class new_type {
[magic here]
using discr = std::integral_type<int, Prev::discr+1>;
};
}

using A = new_type<base_>;
using A2 = new_type<base_>;
using B = new_type<A>;
using C = new_type<B>;
using C2 = new_type<B>;


It takes only a small bit of magic to ensure that the lines with types A2 and C2 don’t compile. Whether that magic is possible in C++11 is another story.






share|improve this answer















In C++20:



using A = decltype({}); // an idiom
using B = decltype({});
...


This is idiomatic code: that’s how one writes “give me a unique type” in C++20.



In C++11, the clearest and simplest approach uses __LINE__:



namespace {
template <int> class new_type {};
}

using A = new_type<__LINE__>; // an idiom - pretty much
using B = new_type<__LINE__>;


The anonymous namespace is the most important bit. It is a serious mistake not to put the new_type class in the anonymous namespace: the types then won't be unique anymore across translation units. All sorts of hilarity will ensue 15 minutes before you plan to ship :)



This extends to C++98:



namespace {
template <int> class new_type {};
}

typedef new_type<__LINE__> A; // an idiom - pretty much
typedef new_type<__LINE__> B;


Another approach would be to manually chain the types, and have the compiler statically validate that the chaining was done correctly, and bomb out with an error if you don’t. So it’d not be brittle (assuming the magic works out).



Something like:



namespace {
struct base_{
using discr = std::integral_type<int, 0>;
};

template <class Prev> class new_type {
[magic here]
using discr = std::integral_type<int, Prev::discr+1>;
};
}

using A = new_type<base_>;
using A2 = new_type<base_>;
using B = new_type<A>;
using C = new_type<B>;
using C2 = new_type<B>;


It takes only a small bit of magic to ensure that the lines with types A2 and C2 don’t compile. Whether that magic is possible in C++11 is another story.







share|improve this answer














share|improve this answer



share|improve this answer








edited 20 hours ago

























answered yesterday









Kuba OberKuba Ober

71k1083196




71k1083196













  • Is "lambda expression in an unevaluated operand" allowed in C++20?

    – kay
    yesterday











  • Yes – and it’s a big deal. It enables some constructs not previously possible at all, with no amount of template metaprogramming. Heck, it even makes them easy.

    – Kuba Ober
    yesterday













  • In using discr = std::integral_type<int, Prev::value+1>; do you mean using discr = std::integral_type<int, Prev::discr+1>;? Also? How does using A = new_type<base_>; using A2 = new_type<base_>; give two different types?

    – NathanOliver
    yesterday











  • It wouldn't give two different types because the second one would fail to compile. It's a bit of a pita to implement but that's the idea. It requires riding on the edges of the standard.

    – Kuba Ober
    23 hours ago





















  • Is "lambda expression in an unevaluated operand" allowed in C++20?

    – kay
    yesterday











  • Yes – and it’s a big deal. It enables some constructs not previously possible at all, with no amount of template metaprogramming. Heck, it even makes them easy.

    – Kuba Ober
    yesterday













  • In using discr = std::integral_type<int, Prev::value+1>; do you mean using discr = std::integral_type<int, Prev::discr+1>;? Also? How does using A = new_type<base_>; using A2 = new_type<base_>; give two different types?

    – NathanOliver
    yesterday











  • It wouldn't give two different types because the second one would fail to compile. It's a bit of a pita to implement but that's the idea. It requires riding on the edges of the standard.

    – Kuba Ober
    23 hours ago



















Is "lambda expression in an unevaluated operand" allowed in C++20?

– kay
yesterday





Is "lambda expression in an unevaluated operand" allowed in C++20?

– kay
yesterday













Yes – and it’s a big deal. It enables some constructs not previously possible at all, with no amount of template metaprogramming. Heck, it even makes them easy.

– Kuba Ober
yesterday







Yes – and it’s a big deal. It enables some constructs not previously possible at all, with no amount of template metaprogramming. Heck, it even makes them easy.

– Kuba Ober
yesterday















In using discr = std::integral_type<int, Prev::value+1>; do you mean using discr = std::integral_type<int, Prev::discr+1>;? Also? How does using A = new_type<base_>; using A2 = new_type<base_>; give two different types?

– NathanOliver
yesterday





In using discr = std::integral_type<int, Prev::value+1>; do you mean using discr = std::integral_type<int, Prev::discr+1>;? Also? How does using A = new_type<base_>; using A2 = new_type<base_>; give two different types?

– NathanOliver
yesterday













It wouldn't give two different types because the second one would fail to compile. It's a bit of a pita to implement but that's the idea. It requires riding on the edges of the standard.

– Kuba Ober
23 hours ago







It wouldn't give two different types because the second one would fail to compile. It's a bit of a pita to implement but that's the idea. It requires riding on the edges of the standard.

– Kuba Ober
23 hours ago













3














I know... they are distilled evil... but seems to me that this is a works for an old C-style macro



#include <type_traits>

#define newType(x)
struct type_##x {};
using x = type_##x;

newType(A)
newType(B)
newType(C)

int main ()
{
static_assert(!std::is_same<A, B>::value, "");
static_assert(!std::is_same<B, C>::value, "");
static_assert(!std::is_same<A, C>::value, "");
}





share|improve this answer



















  • 1





    I don’t think this is any more “on the fly” than what OP already does. In particular, you still need to pass different identifiers to the macro.

    – Konrad Rudolph
    yesterday











  • @KonradRudolph - I don't know... the OP uses the identifier on the left of the = operator so I don't think is really different. But I have to admit that the lambda-based solution from NathanOliver is much better and really elegant.

    – max66
    yesterday
















3














I know... they are distilled evil... but seems to me that this is a works for an old C-style macro



#include <type_traits>

#define newType(x)
struct type_##x {};
using x = type_##x;

newType(A)
newType(B)
newType(C)

int main ()
{
static_assert(!std::is_same<A, B>::value, "");
static_assert(!std::is_same<B, C>::value, "");
static_assert(!std::is_same<A, C>::value, "");
}





share|improve this answer



















  • 1





    I don’t think this is any more “on the fly” than what OP already does. In particular, you still need to pass different identifiers to the macro.

    – Konrad Rudolph
    yesterday











  • @KonradRudolph - I don't know... the OP uses the identifier on the left of the = operator so I don't think is really different. But I have to admit that the lambda-based solution from NathanOliver is much better and really elegant.

    – max66
    yesterday














3












3








3







I know... they are distilled evil... but seems to me that this is a works for an old C-style macro



#include <type_traits>

#define newType(x)
struct type_##x {};
using x = type_##x;

newType(A)
newType(B)
newType(C)

int main ()
{
static_assert(!std::is_same<A, B>::value, "");
static_assert(!std::is_same<B, C>::value, "");
static_assert(!std::is_same<A, C>::value, "");
}





share|improve this answer













I know... they are distilled evil... but seems to me that this is a works for an old C-style macro



#include <type_traits>

#define newType(x)
struct type_##x {};
using x = type_##x;

newType(A)
newType(B)
newType(C)

int main ()
{
static_assert(!std::is_same<A, B>::value, "");
static_assert(!std::is_same<B, C>::value, "");
static_assert(!std::is_same<A, C>::value, "");
}






share|improve this answer












share|improve this answer



share|improve this answer










answered yesterday









max66max66

38.4k74473




38.4k74473








  • 1





    I don’t think this is any more “on the fly” than what OP already does. In particular, you still need to pass different identifiers to the macro.

    – Konrad Rudolph
    yesterday











  • @KonradRudolph - I don't know... the OP uses the identifier on the left of the = operator so I don't think is really different. But I have to admit that the lambda-based solution from NathanOliver is much better and really elegant.

    – max66
    yesterday














  • 1





    I don’t think this is any more “on the fly” than what OP already does. In particular, you still need to pass different identifiers to the macro.

    – Konrad Rudolph
    yesterday











  • @KonradRudolph - I don't know... the OP uses the identifier on the left of the = operator so I don't think is really different. But I have to admit that the lambda-based solution from NathanOliver is much better and really elegant.

    – max66
    yesterday








1




1





I don’t think this is any more “on the fly” than what OP already does. In particular, you still need to pass different identifiers to the macro.

– Konrad Rudolph
yesterday





I don’t think this is any more “on the fly” than what OP already does. In particular, you still need to pass different identifiers to the macro.

– Konrad Rudolph
yesterday













@KonradRudolph - I don't know... the OP uses the identifier on the left of the = operator so I don't think is really different. But I have to admit that the lambda-based solution from NathanOliver is much better and really elegant.

– max66
yesterday





@KonradRudolph - I don't know... the OP uses the identifier on the left of the = operator so I don't think is really different. But I have to admit that the lambda-based solution from NathanOliver is much better and really elegant.

– max66
yesterday


















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