Variadic template parameters from integerUse 'class' or 'typename' for template parameters?Returning multiple values from a C++ functionWhy can templates only be implemented in the header file?Partial specialization of variadic templatesWhy is reading lines from stdin much slower in C++ than Python?Partial specialization of class template for a type that appears in any position of a variadic template parameter packVariadic template class argument containers instantiationReplacing a 32-bit loop counter with 64-bit introduces crazy performance deviationsVariadic Template simulating “runtime” expansionC++ Variadic template method specialization
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Variadic template parameters from integer
Use 'class' or 'typename' for template parameters?Returning multiple values from a C++ functionWhy can templates only be implemented in the header file?Partial specialization of variadic templatesWhy is reading lines from stdin much slower in C++ than Python?Partial specialization of class template for a type that appears in any position of a variadic template parameter packVariadic template class argument containers instantiationReplacing a 32-bit loop counter with 64-bit introduces crazy performance deviationsVariadic Template simulating “runtime” expansionC++ Variadic template method specialization
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Given I have that type
template<int ...Is>
struct A ;
Can I "generate" the type A<0, 1, 2, 3, 4, 5,..., d> just from an integer d?
I thought about something like
template<int d>
struct B : A<std::index_sequence<d>...>
but it doesn't work.
Other option is to specialize manually:
template<int d>
struct B;
template<>
struct B<0>: A<> ;
template<>
struct B<1>: A<0> ;
template<>
struct B<2>: A<0, 1> ;
template<>
struct B<3>: A<0, 1, 2> ;
but obviously I don't be able to write B<3000> b;
[edit] my actual use-case is a "bit" more complex than that. I don't want to reimplement std::integer_sequence, but something more complex.
c++ variadic-templates
asked May 24 at 7:46
Regis PortalezRegis Portalez
3,20912033
|
show 1 more comment
Given I have that type
template<int ...Is>
struct A ;
Can I "generate" the type A<0, 1, 2, 3, 4, 5,..., d> just from an integer d?
I thought about something like
template<int d>
struct B : A<std::index_sequence<d>...>
but it doesn't work.
Other option is to specialize manually:
template<int d>
struct B;
template<>
struct B<0>: A<> ;
template<>
struct B<1>: A<0> ;
template<>
struct B<2>: A<0, 1> ;
template<>
struct B<3>: A<0, 1, 2> ;
but obviously I don't be able to write B<3000> b;
[edit] my actual use-case is a "bit" more complex than that. I don't want to reimplement std::integer_sequence, but something more complex.
c++ variadic-templates
asked May 24 at 7:46
Regis PortalezRegis Portalez
3,20912033
6
What you want isstd::make_integer_sequence.
– Evg
May 24 at 7:49
@Evg can you elaborate please?
– Regis Portalez
May 24 at 7:53
2
Your compiler will probably bail out at 3000 template parameters.
– n.m.
May 24 at 7:56
@n.m. If I'm right, there is a compiler setting for that :)
– Regis Portalez
May 24 at 7:56
As a stupid lowly Python programmer - why would one ever want to do this?
– Adam Barnes
May 24 at 19:36
|
show 1 more comment
Given I have that type
template<int ...Is>
struct A ;
Can I "generate" the type A<0, 1, 2, 3, 4, 5,..., d> just from an integer d?
I thought about something like
template<int d>
struct B : A<std::index_sequence<d>...>
but it doesn't work.
Other option is to specialize manually:
template<int d>
struct B;
template<>
struct B<0>: A<> ;
template<>
struct B<1>: A<0> ;
template<>
struct B<2>: A<0, 1> ;
template<>
struct B<3>: A<0, 1, 2> ;
but obviously I don't be able to write B<3000> b;
[edit] my actual use-case is a "bit" more complex than that. I don't want to reimplement std::integer_sequence, but something more complex.
c++ variadic-templates
asked May 24 at 7:46
Regis PortalezRegis Portalez
3,20912033
Given I have that type
template<int ...Is>
struct A ;
Can I "generate" the type A<0, 1, 2, 3, 4, 5,..., d> just from an integer d?
I thought about something like
template<int d>
struct B : A<std::index_sequence<d>...>
but it doesn't work.
Other option is to specialize manually:
template<int d>
struct B;
template<>
struct B<0>: A<> ;
template<>
struct B<1>: A<0> ;
template<>
struct B<2>: A<0, 1> ;
template<>
struct B<3>: A<0, 1, 2> ;
but obviously I don't be able to write B<3000> b;
[edit] my actual use-case is a "bit" more complex than that. I don't want to reimplement std::integer_sequence, but something more complex.
c++ variadic-templates
c++ variadic-templates
asked May 24 at 7:46
Regis PortalezRegis Portalez
3,20912033
asked May 24 at 7:46
Regis PortalezRegis Portalez
3,20912033
edited May 24 at 8:00
Regis Portalez
asked May 24 at 7:46
Regis PortalezRegis Portalez
3,20912033
asked May 24 at 7:46
Regis PortalezRegis Portalez
3,20912033
asked May 24 at 7:46
Regis PortalezRegis Portalez
3,20912033
3,20912033
6
What you want isstd::make_integer_sequence.
– Evg
May 24 at 7:49
@Evg can you elaborate please?
– Regis Portalez
May 24 at 7:53
2
Your compiler will probably bail out at 3000 template parameters.
– n.m.
May 24 at 7:56
@n.m. If I'm right, there is a compiler setting for that :)
– Regis Portalez
May 24 at 7:56
As a stupid lowly Python programmer - why would one ever want to do this?
– Adam Barnes
May 24 at 19:36
|
show 1 more comment
6
What you want isstd::make_integer_sequence.
– Evg
May 24 at 7:49
@Evg can you elaborate please?
– Regis Portalez
May 24 at 7:53
2
Your compiler will probably bail out at 3000 template parameters.
– n.m.
May 24 at 7:56
@n.m. If I'm right, there is a compiler setting for that :)
– Regis Portalez
May 24 at 7:56
As a stupid lowly Python programmer - why would one ever want to do this?
– Adam Barnes
May 24 at 19:36
6
6
What you want is
std::make_integer_sequence.– Evg
May 24 at 7:49
What you want is
std::make_integer_sequence.– Evg
May 24 at 7:49
@Evg can you elaborate please?
– Regis Portalez
May 24 at 7:53
@Evg can you elaborate please?
– Regis Portalez
May 24 at 7:53
2
2
Your compiler will probably bail out at 3000 template parameters.
– n.m.
May 24 at 7:56
Your compiler will probably bail out at 3000 template parameters.
– n.m.
May 24 at 7:56
@n.m. If I'm right, there is a compiler setting for that :)
– Regis Portalez
May 24 at 7:56
@n.m. If I'm right, there is a compiler setting for that :)
– Regis Portalez
May 24 at 7:56
As a stupid lowly Python programmer - why would one ever want to do this?
– Adam Barnes
May 24 at 19:36
As a stupid lowly Python programmer - why would one ever want to do this?
– Adam Barnes
May 24 at 19:36
|
show 1 more comment
2 Answers
2
active
oldest
votes
We already have what you want in the Standard library - std::make_integer_sequence. If you want to use your own type A<...> you can do this:
template<int... Is>
struct A ;
template<class>
struct make_A_impl;
template<int... Is>
struct make_A_impl<std::integer_sequence<int, Is...>>
using Type = A<Is...>;
;
template<int size>
using make_A = typename make_A_impl<std::make_integer_sequence<int, size>>::Type;
And then for A<0, ..., 2999> write
make_A<3000>
answered May 24 at 7:57
EvgEvg
4,99821739
So themake_A_impldeclaration is needed to workaround the fact that you can't have two parameter packs in one template argument list?
– Stack Danny
May 24 at 8:12
@StackDanny, I didn't fully get your question.make_A_implis used to "unpack"std::integer_sequenceintoint....
– Evg
May 24 at 8:28
add a comment |
A bit another way to do - use function signature to match the A<...> type:
#include <type_traits>
template<int ...Is>
struct A ;
namespace details
template <int ...Is>
auto GenrateAHelper(std::integer_sequence<int, Is...>) -> A<Is...>;
template<int I>
using GenerateA = decltype(details::GenrateAHelper(std::make_integer_sequence<int, I>()));
static_assert(std::is_same<GenerateA<3>, A<0, 1, 2>>::value, "");
answered May 24 at 7:59
Dmitry GordonDmitry Gordon
1,754616
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
We already have what you want in the Standard library - std::make_integer_sequence. If you want to use your own type A<...> you can do this:
template<int... Is>
struct A ;
template<class>
struct make_A_impl;
template<int... Is>
struct make_A_impl<std::integer_sequence<int, Is...>>
using Type = A<Is...>;
;
template<int size>
using make_A = typename make_A_impl<std::make_integer_sequence<int, size>>::Type;
And then for A<0, ..., 2999> write
make_A<3000>
answered May 24 at 7:57
EvgEvg
4,99821739
So themake_A_impldeclaration is needed to workaround the fact that you can't have two parameter packs in one template argument list?
– Stack Danny
May 24 at 8:12
@StackDanny, I didn't fully get your question.make_A_implis used to "unpack"std::integer_sequenceintoint....
– Evg
May 24 at 8:28
add a comment |
We already have what you want in the Standard library - std::make_integer_sequence. If you want to use your own type A<...> you can do this:
template<int... Is>
struct A ;
template<class>
struct make_A_impl;
template<int... Is>
struct make_A_impl<std::integer_sequence<int, Is...>>
using Type = A<Is...>;
;
template<int size>
using make_A = typename make_A_impl<std::make_integer_sequence<int, size>>::Type;
And then for A<0, ..., 2999> write
make_A<3000>
answered May 24 at 7:57
EvgEvg
4,99821739
So themake_A_impldeclaration is needed to workaround the fact that you can't have two parameter packs in one template argument list?
– Stack Danny
May 24 at 8:12
@StackDanny, I didn't fully get your question.make_A_implis used to "unpack"std::integer_sequenceintoint....
– Evg
May 24 at 8:28
add a comment |
We already have what you want in the Standard library - std::make_integer_sequence. If you want to use your own type A<...> you can do this:
template<int... Is>
struct A ;
template<class>
struct make_A_impl;
template<int... Is>
struct make_A_impl<std::integer_sequence<int, Is...>>
using Type = A<Is...>;
;
template<int size>
using make_A = typename make_A_impl<std::make_integer_sequence<int, size>>::Type;
And then for A<0, ..., 2999> write
make_A<3000>
answered May 24 at 7:57
EvgEvg
4,99821739
We already have what you want in the Standard library - std::make_integer_sequence. If you want to use your own type A<...> you can do this:
template<int... Is>
struct A ;
template<class>
struct make_A_impl;
template<int... Is>
struct make_A_impl<std::integer_sequence<int, Is...>>
using Type = A<Is...>;
;
template<int size>
using make_A = typename make_A_impl<std::make_integer_sequence<int, size>>::Type;
And then for A<0, ..., 2999> write
make_A<3000>
answered May 24 at 7:57
EvgEvg
4,99821739
answered May 24 at 7:57
EvgEvg
4,99821739
answered May 24 at 7:57
EvgEvg
4,99821739
answered May 24 at 7:57
EvgEvg
4,99821739
4,99821739
So themake_A_impldeclaration is needed to workaround the fact that you can't have two parameter packs in one template argument list?
– Stack Danny
May 24 at 8:12
@StackDanny, I didn't fully get your question.make_A_implis used to "unpack"std::integer_sequenceintoint....
– Evg
May 24 at 8:28
add a comment |
So themake_A_impldeclaration is needed to workaround the fact that you can't have two parameter packs in one template argument list?
– Stack Danny
May 24 at 8:12
@StackDanny, I didn't fully get your question.make_A_implis used to "unpack"std::integer_sequenceintoint....
– Evg
May 24 at 8:28
So the
make_A_impl declaration is needed to workaround the fact that you can't have two parameter packs in one template argument list?– Stack Danny
May 24 at 8:12
So the
make_A_impl declaration is needed to workaround the fact that you can't have two parameter packs in one template argument list?– Stack Danny
May 24 at 8:12
@StackDanny, I didn't fully get your question.
make_A_impl is used to "unpack" std::integer_sequence into int....– Evg
May 24 at 8:28
@StackDanny, I didn't fully get your question.
make_A_impl is used to "unpack" std::integer_sequence into int....– Evg
May 24 at 8:28
add a comment |
A bit another way to do - use function signature to match the A<...> type:
#include <type_traits>
template<int ...Is>
struct A ;
namespace details
template <int ...Is>
auto GenrateAHelper(std::integer_sequence<int, Is...>) -> A<Is...>;
template<int I>
using GenerateA = decltype(details::GenrateAHelper(std::make_integer_sequence<int, I>()));
static_assert(std::is_same<GenerateA<3>, A<0, 1, 2>>::value, "");
answered May 24 at 7:59
Dmitry GordonDmitry Gordon
1,754616
add a comment |
A bit another way to do - use function signature to match the A<...> type:
#include <type_traits>
template<int ...Is>
struct A ;
namespace details
template <int ...Is>
auto GenrateAHelper(std::integer_sequence<int, Is...>) -> A<Is...>;
template<int I>
using GenerateA = decltype(details::GenrateAHelper(std::make_integer_sequence<int, I>()));
static_assert(std::is_same<GenerateA<3>, A<0, 1, 2>>::value, "");
answered May 24 at 7:59
Dmitry GordonDmitry Gordon
1,754616
add a comment |
A bit another way to do - use function signature to match the A<...> type:
#include <type_traits>
template<int ...Is>
struct A ;
namespace details
template <int ...Is>
auto GenrateAHelper(std::integer_sequence<int, Is...>) -> A<Is...>;
template<int I>
using GenerateA = decltype(details::GenrateAHelper(std::make_integer_sequence<int, I>()));
static_assert(std::is_same<GenerateA<3>, A<0, 1, 2>>::value, "");
answered May 24 at 7:59
Dmitry GordonDmitry Gordon
1,754616
A bit another way to do - use function signature to match the A<...> type:
#include <type_traits>
template<int ...Is>
struct A ;
namespace details
template <int ...Is>
auto GenrateAHelper(std::integer_sequence<int, Is...>) -> A<Is...>;
template<int I>
using GenerateA = decltype(details::GenrateAHelper(std::make_integer_sequence<int, I>()));
static_assert(std::is_same<GenerateA<3>, A<0, 1, 2>>::value, "");
answered May 24 at 7:59
Dmitry GordonDmitry Gordon
1,754616
answered May 24 at 7:59
Dmitry GordonDmitry Gordon
1,754616
answered May 24 at 7:59
Dmitry GordonDmitry Gordon
1,754616
answered May 24 at 7:59
Dmitry GordonDmitry Gordon
1,754616
1,754616
add a comment |
add a comment |
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6
What you want is
std::make_integer_sequence.– Evg
May 24 at 7:49
@Evg can you elaborate please?
– Regis Portalez
May 24 at 7:53
2
Your compiler will probably bail out at 3000 template parameters.
– n.m.
May 24 at 7:56
@n.m. If I'm right, there is a compiler setting for that :)
– Regis Portalez
May 24 at 7:56
As a stupid lowly Python programmer - why would one ever want to do this?
– Adam Barnes
May 24 at 19:36