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
    2 days ago






  • 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
    2 days ago








  • 1





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

    – HolyBlackCat
    2 days 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
    2 days ago






  • 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
    2 days ago








  • 1





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

    – HolyBlackCat
    2 days 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 2 days ago









kaykay

18.6k970118




18.6k970118








  • 7





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

    – Samer Tufail
    2 days ago






  • 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
    2 days ago








  • 1





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

    – HolyBlackCat
    2 days ago
















  • 7





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

    – Samer Tufail
    2 days ago






  • 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
    2 days ago








  • 1





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

    – HolyBlackCat
    2 days ago










7




7





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

– Samer Tufail
2 days ago





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

– Samer Tufail
2 days ago




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
2 days ago







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
2 days ago






1




1





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

– HolyBlackCat
2 days ago







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

– HolyBlackCat
2 days ago














3 Answers
3






active

oldest

votes


















21














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





















  • 6





    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
    2 days ago













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

    – NathanOliver
    2 days ago











  • Add it to the answer if you want..

    – Jarod42
    2 days ago






  • 1





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

    – Mooing Duck
    2 days ago






  • 1





    @MooingDuck: "error: lambda-expression in unevaluated context" before C++20 though.

    – Jarod42
    4 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
    2 days ago











  • 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
    2 days ago













  • 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
    2 days 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
    2 days ago





















2














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



















  • 2





    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
    2 days ago











  • @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
    2 days ago











Your Answer






StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "1"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f55341859%2fmeta-programming-declare-a-new-struct-on-the-fly%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









21














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





















  • 6





    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
    2 days ago













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

    – NathanOliver
    2 days ago











  • Add it to the answer if you want..

    – Jarod42
    2 days ago






  • 1





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

    – Mooing Duck
    2 days ago






  • 1





    @MooingDuck: "error: lambda-expression in unevaluated context" before C++20 though.

    – Jarod42
    4 hours ago
















21














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





















  • 6





    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
    2 days ago













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

    – NathanOliver
    2 days ago











  • Add it to the answer if you want..

    – Jarod42
    2 days ago






  • 1





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

    – Mooing Duck
    2 days ago






  • 1





    @MooingDuck: "error: lambda-expression in unevaluated context" before C++20 though.

    – Jarod42
    4 hours ago














21












21








21







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 2 days ago

























answered 2 days ago









NathanOliverNathanOliver

96.8k16137211




96.8k16137211








  • 6





    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
    2 days ago













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

    – NathanOliver
    2 days ago











  • Add it to the answer if you want..

    – Jarod42
    2 days ago






  • 1





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

    – Mooing Duck
    2 days ago






  • 1





    @MooingDuck: "error: lambda-expression in unevaluated context" before C++20 though.

    – Jarod42
    4 hours ago














  • 6





    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
    2 days ago













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

    – NathanOliver
    2 days ago











  • Add it to the answer if you want..

    – Jarod42
    2 days ago






  • 1





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

    – Mooing Duck
    2 days ago






  • 1





    @MooingDuck: "error: lambda-expression in unevaluated context" before C++20 though.

    – Jarod42
    4 hours ago








6




6





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
2 days ago







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
2 days ago















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

– NathanOliver
2 days ago





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

– NathanOliver
2 days ago













Add it to the answer if you want..

– Jarod42
2 days ago





Add it to the answer if you want..

– Jarod42
2 days ago




1




1





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

– Mooing Duck
2 days ago





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

– Mooing Duck
2 days ago




1




1





@MooingDuck: "error: lambda-expression in unevaluated context" before C++20 though.

– Jarod42
4 hours ago





@MooingDuck: "error: lambda-expression in unevaluated context" before C++20 though.

– Jarod42
4 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
    2 days ago











  • 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
    2 days ago













  • 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
    2 days 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
    2 days 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
    2 days ago











  • 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
    2 days ago













  • 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
    2 days 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
    2 days 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 2 days ago

























answered 2 days ago









Kuba OberKuba Ober

71k1083196




71k1083196













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

    – kay
    2 days ago











  • 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
    2 days ago













  • 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
    2 days 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
    2 days ago





















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

    – kay
    2 days ago











  • 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
    2 days ago













  • 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
    2 days 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
    2 days ago



















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

– kay
2 days ago





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

– kay
2 days ago













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
2 days ago







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
2 days ago















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
2 days ago





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
2 days 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
2 days 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
2 days ago













2














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



















  • 2





    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
    2 days ago











  • @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
    2 days ago
















2














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



















  • 2





    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
    2 days ago











  • @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
    2 days ago














2












2








2







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 2 days ago









max66max66

38.5k74473




38.5k74473








  • 2





    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
    2 days ago











  • @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
    2 days ago














  • 2





    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
    2 days ago











  • @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
    2 days ago








2




2





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
2 days ago





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
2 days ago













@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
2 days ago





@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
2 days ago


















draft saved

draft discarded




















































Thanks for contributing an answer to Stack Overflow!


  • Please be sure to answer the question. Provide details and share your research!

But avoid



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f55341859%2fmeta-programming-declare-a-new-struct-on-the-fly%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







Popular posts from this blog

How did Captain America manage to do this?

迪纳利

南乌拉尔铁路局