Meta programming: Declare a new struct on the fly
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
add a comment |
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
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
add a comment |
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
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
c++ templates metaprogramming stateful compile-time-constant
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
add a comment |
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
add a comment |
3 Answers
3
active
oldest
votes
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);
6
You rely only on unicity of__LINE__
(so care with multiple TU, or several types on same line), sotemplate <size_t> struct unique_tag {};
would be enough ->#define make_new_type unique_tag<__LINE__>
. andusing 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
|
show 2 more comments
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.
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
Inusing discr = std::integral_type<int, Prev::value+1>;
do you meanusing discr = std::integral_type<int, Prev::discr+1>;
? Also? How doesusing 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
add a comment |
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, "");
}
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
add a comment |
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3 Answers
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active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
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);
6
You rely only on unicity of__LINE__
(so care with multiple TU, or several types on same line), sotemplate <size_t> struct unique_tag {};
would be enough ->#define make_new_type unique_tag<__LINE__>
. andusing 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
|
show 2 more comments
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);
6
You rely only on unicity of__LINE__
(so care with multiple TU, or several types on same line), sotemplate <size_t> struct unique_tag {};
would be enough ->#define make_new_type unique_tag<__LINE__>
. andusing 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
|
show 2 more comments
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);
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);
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), sotemplate <size_t> struct unique_tag {};
would be enough ->#define make_new_type unique_tag<__LINE__>
. andusing 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
|
show 2 more comments
6
You rely only on unicity of__LINE__
(so care with multiple TU, or several types on same line), sotemplate <size_t> struct unique_tag {};
would be enough ->#define make_new_type unique_tag<__LINE__>
. andusing 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
|
show 2 more comments
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.
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
Inusing discr = std::integral_type<int, Prev::value+1>;
do you meanusing discr = std::integral_type<int, Prev::discr+1>;
? Also? How doesusing 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
add a comment |
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.
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
Inusing discr = std::integral_type<int, Prev::value+1>;
do you meanusing discr = std::integral_type<int, Prev::discr+1>;
? Also? How doesusing 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
add a comment |
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.
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.
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
Inusing discr = std::integral_type<int, Prev::value+1>;
do you meanusing discr = std::integral_type<int, Prev::discr+1>;
? Also? How doesusing 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
add a comment |
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
Inusing discr = std::integral_type<int, Prev::value+1>;
do you meanusing discr = std::integral_type<int, Prev::discr+1>;
? Also? How doesusing 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
add a comment |
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, "");
}
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
add a comment |
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, "");
}
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
add a comment |
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, "");
}
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, "");
}
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
add a comment |
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
add a comment |
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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