Type of non-type parameter in template template class is non deducible in C++14 but deducible in C++172019 Community Moderator ElectionUse 'class' or 'typename' for template parameters?Clang variadic template specialization error: non-deducible template parameterCan't deduce template typewhy this variable isn't deduced as initializer_list in g++ in C++14?extern array as non-type template argument with clang c++1zIs default template template parameter value deduced context?Why is gcc failing when using lambda for non-type template parameter?Matching variadic non-type templatesTrailing class template arguments not deducedC++17 template constructor class template deduction
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Type of non-type parameter in template template class is non deducible in C++14 but deducible in C++17
2019 Community Moderator ElectionUse 'class' or 'typename' for template parameters?Clang variadic template specialization error: non-deducible template parameterCan't deduce template typewhy this variable isn't deduced as initializer_list in g++ in C++14?extern array as non-type template argument with clang c++1zIs default template template parameter value deduced context?Why is gcc failing when using lambda for non-type template parameter?Matching variadic non-type templatesTrailing class template arguments not deducedC++17 template constructor class template deduction
The title is a bit confusing but what I mean is this specific case:
template<class>
struct get_type_of_nontype;
template<class T, T Value, template<T> class Template>
struct get_type_of_nontype<Template<Value>>
using type = T;
;
So I can use it like this:
#include <type_traits>
template<int I>
class int_non_type ;
static_assert(
std::is_same<typename get_type_of_nontype<int_non_type<0>>::type, int>::value,
"T is deduced to be `int` as `template<T> class Template` is `template<int> class int_non_type`"
);
This works fine in C++17. In C++14 I get the following errors:
gcc 8:
<source>:5:8: error: template parameters not deducible in partial specialization:
struct get_type_of_nontype<Template<Value>> {
^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
<source>:5:8: note: 'T'
clang 7:
<source>:5:8: error: class template partial specialization contains a template parameter that cannot be deduced; this partial specialization will never be used [-Wunusable-partial-specialization]
struct get_type_of_nontype<Template<Value>> {
^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
<source>:4:16: note: non-deducible template parameter 'T'
template<class T, T Value, template<T> class Template>
^
And then they both complain that struct get_type_of_nontype<int_non_type<0>>
is incomplete, so typename get_type_of_non_type<int_non_type<0>>::type
can't compile.
Why is this different between C++14 and C++17? Is this just a compiler bug? If not, is there a way to do this in C++14?
c++ templates c++14 c++17 template-templates
add a comment |
The title is a bit confusing but what I mean is this specific case:
template<class>
struct get_type_of_nontype;
template<class T, T Value, template<T> class Template>
struct get_type_of_nontype<Template<Value>>
using type = T;
;
So I can use it like this:
#include <type_traits>
template<int I>
class int_non_type ;
static_assert(
std::is_same<typename get_type_of_nontype<int_non_type<0>>::type, int>::value,
"T is deduced to be `int` as `template<T> class Template` is `template<int> class int_non_type`"
);
This works fine in C++17. In C++14 I get the following errors:
gcc 8:
<source>:5:8: error: template parameters not deducible in partial specialization:
struct get_type_of_nontype<Template<Value>> {
^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
<source>:5:8: note: 'T'
clang 7:
<source>:5:8: error: class template partial specialization contains a template parameter that cannot be deduced; this partial specialization will never be used [-Wunusable-partial-specialization]
struct get_type_of_nontype<Template<Value>> {
^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
<source>:4:16: note: non-deducible template parameter 'T'
template<class T, T Value, template<T> class Template>
^
And then they both complain that struct get_type_of_nontype<int_non_type<0>>
is incomplete, so typename get_type_of_non_type<int_non_type<0>>::type
can't compile.
Why is this different between C++14 and C++17? Is this just a compiler bug? If not, is there a way to do this in C++14?
c++ templates c++14 c++17 template-templates
add a comment |
The title is a bit confusing but what I mean is this specific case:
template<class>
struct get_type_of_nontype;
template<class T, T Value, template<T> class Template>
struct get_type_of_nontype<Template<Value>>
using type = T;
;
So I can use it like this:
#include <type_traits>
template<int I>
class int_non_type ;
static_assert(
std::is_same<typename get_type_of_nontype<int_non_type<0>>::type, int>::value,
"T is deduced to be `int` as `template<T> class Template` is `template<int> class int_non_type`"
);
This works fine in C++17. In C++14 I get the following errors:
gcc 8:
<source>:5:8: error: template parameters not deducible in partial specialization:
struct get_type_of_nontype<Template<Value>> {
^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
<source>:5:8: note: 'T'
clang 7:
<source>:5:8: error: class template partial specialization contains a template parameter that cannot be deduced; this partial specialization will never be used [-Wunusable-partial-specialization]
struct get_type_of_nontype<Template<Value>> {
^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
<source>:4:16: note: non-deducible template parameter 'T'
template<class T, T Value, template<T> class Template>
^
And then they both complain that struct get_type_of_nontype<int_non_type<0>>
is incomplete, so typename get_type_of_non_type<int_non_type<0>>::type
can't compile.
Why is this different between C++14 and C++17? Is this just a compiler bug? If not, is there a way to do this in C++14?
c++ templates c++14 c++17 template-templates
The title is a bit confusing but what I mean is this specific case:
template<class>
struct get_type_of_nontype;
template<class T, T Value, template<T> class Template>
struct get_type_of_nontype<Template<Value>>
using type = T;
;
So I can use it like this:
#include <type_traits>
template<int I>
class int_non_type ;
static_assert(
std::is_same<typename get_type_of_nontype<int_non_type<0>>::type, int>::value,
"T is deduced to be `int` as `template<T> class Template` is `template<int> class int_non_type`"
);
This works fine in C++17. In C++14 I get the following errors:
gcc 8:
<source>:5:8: error: template parameters not deducible in partial specialization:
struct get_type_of_nontype<Template<Value>> {
^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
<source>:5:8: note: 'T'
clang 7:
<source>:5:8: error: class template partial specialization contains a template parameter that cannot be deduced; this partial specialization will never be used [-Wunusable-partial-specialization]
struct get_type_of_nontype<Template<Value>> {
^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
<source>:4:16: note: non-deducible template parameter 'T'
template<class T, T Value, template<T> class Template>
^
And then they both complain that struct get_type_of_nontype<int_non_type<0>>
is incomplete, so typename get_type_of_non_type<int_non_type<0>>::type
can't compile.
Why is this different between C++14 and C++17? Is this just a compiler bug? If not, is there a way to do this in C++14?
c++ templates c++14 c++17 template-templates
c++ templates c++14 c++17 template-templates
asked Mar 6 at 23:11
ArtyerArtyer
4,705728
4,705728
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
The Standard wording in [temp.deduct.type] paragraphs 13 and 14 changed. So yes, your example is invalid in C++14, but is allowed in C++17 thanks to a new language feature.
C++14:
A template type argument cannot be deduced from the type of a non-type template-argument.
[Example:
template<class T, T i> void f(double a[10][i]);
int v[10][20];
f(v); // error: argument for template-parameter T cannot be deduced
-- end example]
C++17:
When the value of the argument corresponding to a non-type template parameter
P
that is declared with a dependent type is deduced from an expression, the template parameters in the type ofP
are deduced from the type of the value. [Example:template<long n> struct A ;
template<typename T> struct C;
template<typename T, T n> struct C<A<n>>
using Q = T;
;
using R = long;
using R = C<A<2>>::Q; // OK; T was deduced to long from the
// template argument value in the type A<2>
-- end example] The type of
N
in the typeT[N]
isstd::size_t
. [Example:template<typename T> struct S;
template<typename T, T n> struct S<int[n]>
using Q = T;
;
using V = decltype(sizeof 0);
using V = S<int[42]>::Q; // OK; T was deduced to std::size_t from the type int[42]
-- end example]
[Example:
template<class T, T i> void f(int (&a)[i]);
int v[10];
void g()
f(v); // OK: T is std::size_t
-- end example]
This seems a bit related to another C++17 template change: C++17 is the first version to allow placeholder types in non-type template parameters, as in template <auto Value>
or template <auto* Ptr>
. I'd expect compiler implementations would need some similar logic for supporting both of the two language features.
Oh soT
is deduced fromT Value
and nottemplate <T> class
. This makes it sound like there is no way to do this at all in C++14.
– Artyer
Mar 6 at 23:44
Probably not. Unless maybe you only need to support a finite list of types like the arithmetic types, in which case you could write a specialization for each one.
– aschepler
Mar 6 at 23:46
add a comment |
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1 Answer
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1 Answer
1
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active
oldest
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active
oldest
votes
The Standard wording in [temp.deduct.type] paragraphs 13 and 14 changed. So yes, your example is invalid in C++14, but is allowed in C++17 thanks to a new language feature.
C++14:
A template type argument cannot be deduced from the type of a non-type template-argument.
[Example:
template<class T, T i> void f(double a[10][i]);
int v[10][20];
f(v); // error: argument for template-parameter T cannot be deduced
-- end example]
C++17:
When the value of the argument corresponding to a non-type template parameter
P
that is declared with a dependent type is deduced from an expression, the template parameters in the type ofP
are deduced from the type of the value. [Example:template<long n> struct A ;
template<typename T> struct C;
template<typename T, T n> struct C<A<n>>
using Q = T;
;
using R = long;
using R = C<A<2>>::Q; // OK; T was deduced to long from the
// template argument value in the type A<2>
-- end example] The type of
N
in the typeT[N]
isstd::size_t
. [Example:template<typename T> struct S;
template<typename T, T n> struct S<int[n]>
using Q = T;
;
using V = decltype(sizeof 0);
using V = S<int[42]>::Q; // OK; T was deduced to std::size_t from the type int[42]
-- end example]
[Example:
template<class T, T i> void f(int (&a)[i]);
int v[10];
void g()
f(v); // OK: T is std::size_t
-- end example]
This seems a bit related to another C++17 template change: C++17 is the first version to allow placeholder types in non-type template parameters, as in template <auto Value>
or template <auto* Ptr>
. I'd expect compiler implementations would need some similar logic for supporting both of the two language features.
Oh soT
is deduced fromT Value
and nottemplate <T> class
. This makes it sound like there is no way to do this at all in C++14.
– Artyer
Mar 6 at 23:44
Probably not. Unless maybe you only need to support a finite list of types like the arithmetic types, in which case you could write a specialization for each one.
– aschepler
Mar 6 at 23:46
add a comment |
The Standard wording in [temp.deduct.type] paragraphs 13 and 14 changed. So yes, your example is invalid in C++14, but is allowed in C++17 thanks to a new language feature.
C++14:
A template type argument cannot be deduced from the type of a non-type template-argument.
[Example:
template<class T, T i> void f(double a[10][i]);
int v[10][20];
f(v); // error: argument for template-parameter T cannot be deduced
-- end example]
C++17:
When the value of the argument corresponding to a non-type template parameter
P
that is declared with a dependent type is deduced from an expression, the template parameters in the type ofP
are deduced from the type of the value. [Example:template<long n> struct A ;
template<typename T> struct C;
template<typename T, T n> struct C<A<n>>
using Q = T;
;
using R = long;
using R = C<A<2>>::Q; // OK; T was deduced to long from the
// template argument value in the type A<2>
-- end example] The type of
N
in the typeT[N]
isstd::size_t
. [Example:template<typename T> struct S;
template<typename T, T n> struct S<int[n]>
using Q = T;
;
using V = decltype(sizeof 0);
using V = S<int[42]>::Q; // OK; T was deduced to std::size_t from the type int[42]
-- end example]
[Example:
template<class T, T i> void f(int (&a)[i]);
int v[10];
void g()
f(v); // OK: T is std::size_t
-- end example]
This seems a bit related to another C++17 template change: C++17 is the first version to allow placeholder types in non-type template parameters, as in template <auto Value>
or template <auto* Ptr>
. I'd expect compiler implementations would need some similar logic for supporting both of the two language features.
Oh soT
is deduced fromT Value
and nottemplate <T> class
. This makes it sound like there is no way to do this at all in C++14.
– Artyer
Mar 6 at 23:44
Probably not. Unless maybe you only need to support a finite list of types like the arithmetic types, in which case you could write a specialization for each one.
– aschepler
Mar 6 at 23:46
add a comment |
The Standard wording in [temp.deduct.type] paragraphs 13 and 14 changed. So yes, your example is invalid in C++14, but is allowed in C++17 thanks to a new language feature.
C++14:
A template type argument cannot be deduced from the type of a non-type template-argument.
[Example:
template<class T, T i> void f(double a[10][i]);
int v[10][20];
f(v); // error: argument for template-parameter T cannot be deduced
-- end example]
C++17:
When the value of the argument corresponding to a non-type template parameter
P
that is declared with a dependent type is deduced from an expression, the template parameters in the type ofP
are deduced from the type of the value. [Example:template<long n> struct A ;
template<typename T> struct C;
template<typename T, T n> struct C<A<n>>
using Q = T;
;
using R = long;
using R = C<A<2>>::Q; // OK; T was deduced to long from the
// template argument value in the type A<2>
-- end example] The type of
N
in the typeT[N]
isstd::size_t
. [Example:template<typename T> struct S;
template<typename T, T n> struct S<int[n]>
using Q = T;
;
using V = decltype(sizeof 0);
using V = S<int[42]>::Q; // OK; T was deduced to std::size_t from the type int[42]
-- end example]
[Example:
template<class T, T i> void f(int (&a)[i]);
int v[10];
void g()
f(v); // OK: T is std::size_t
-- end example]
This seems a bit related to another C++17 template change: C++17 is the first version to allow placeholder types in non-type template parameters, as in template <auto Value>
or template <auto* Ptr>
. I'd expect compiler implementations would need some similar logic for supporting both of the two language features.
The Standard wording in [temp.deduct.type] paragraphs 13 and 14 changed. So yes, your example is invalid in C++14, but is allowed in C++17 thanks to a new language feature.
C++14:
A template type argument cannot be deduced from the type of a non-type template-argument.
[Example:
template<class T, T i> void f(double a[10][i]);
int v[10][20];
f(v); // error: argument for template-parameter T cannot be deduced
-- end example]
C++17:
When the value of the argument corresponding to a non-type template parameter
P
that is declared with a dependent type is deduced from an expression, the template parameters in the type ofP
are deduced from the type of the value. [Example:template<long n> struct A ;
template<typename T> struct C;
template<typename T, T n> struct C<A<n>>
using Q = T;
;
using R = long;
using R = C<A<2>>::Q; // OK; T was deduced to long from the
// template argument value in the type A<2>
-- end example] The type of
N
in the typeT[N]
isstd::size_t
. [Example:template<typename T> struct S;
template<typename T, T n> struct S<int[n]>
using Q = T;
;
using V = decltype(sizeof 0);
using V = S<int[42]>::Q; // OK; T was deduced to std::size_t from the type int[42]
-- end example]
[Example:
template<class T, T i> void f(int (&a)[i]);
int v[10];
void g()
f(v); // OK: T is std::size_t
-- end example]
This seems a bit related to another C++17 template change: C++17 is the first version to allow placeholder types in non-type template parameters, as in template <auto Value>
or template <auto* Ptr>
. I'd expect compiler implementations would need some similar logic for supporting both of the two language features.
answered Mar 6 at 23:38
aschepleraschepler
53k580129
53k580129
Oh soT
is deduced fromT Value
and nottemplate <T> class
. This makes it sound like there is no way to do this at all in C++14.
– Artyer
Mar 6 at 23:44
Probably not. Unless maybe you only need to support a finite list of types like the arithmetic types, in which case you could write a specialization for each one.
– aschepler
Mar 6 at 23:46
add a comment |
Oh soT
is deduced fromT Value
and nottemplate <T> class
. This makes it sound like there is no way to do this at all in C++14.
– Artyer
Mar 6 at 23:44
Probably not. Unless maybe you only need to support a finite list of types like the arithmetic types, in which case you could write a specialization for each one.
– aschepler
Mar 6 at 23:46
Oh so
T
is deduced from T Value
and not template <T> class
. This makes it sound like there is no way to do this at all in C++14.– Artyer
Mar 6 at 23:44
Oh so
T
is deduced from T Value
and not template <T> class
. This makes it sound like there is no way to do this at all in C++14.– Artyer
Mar 6 at 23:44
Probably not. Unless maybe you only need to support a finite list of types like the arithmetic types, in which case you could write a specialization for each one.
– aschepler
Mar 6 at 23:46
Probably not. Unless maybe you only need to support a finite list of types like the arithmetic types, in which case you could write a specialization for each one.
– aschepler
Mar 6 at 23:46
add a comment |
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