Implement is_destructible with Detected Idiom
3
Here is my implementation of is_destructible_v:
template<class T>
struct is_unknown_bound_array : std::false_type
{};
template<class T>
struct is_unknown_bound_array<T> : std::true_type
{};
template<typename T, typename U = std::remove_all_extents_t<T>>
using has_dtor = decltype(std::declval<U&>().~U());
template<typename T>
constexpr bool is_destructible_v
= (std::experimental::is_detected_v<has_dtor, T> or std::is_reference_v<T>)
and not is_unknown_bound_array<T>::value
and not std::is_function_v<T>;
template<typename T>
struct is_destructible : std::bool_constant<is_destructible_v<T>>
{};
clang compiled happily and passed all libstdcxx's testsuite, while gcc failed to compile:
prog.cc:177:47: error: 'std::declval<int&>()' is not of type 'int&'
177 | using has_dtor = decltype(std::declval<U&>().~U());
| ~~~~~~~~~~~~~~~~~~~~^
prog.cc: In substitution of 'template<class T, class U> using has_dtor = decltype (declval<U&>().~ U()) [with T = int&&; U = int&&]':
So, gcc cannot do SFINAE on using has_dtor = decltype(std::declval<U&>().~U());.
Question:
- Which compiler object to standard here?
- What's the most elegant solution/workaround here? The ways I can think of is a little ugly
c++ language-lawyer c++17 template-meta-programming typetraits
asked Nov 24 '18 at 9:28
陳 力陳 力
1,6001723
add a comment |
3
Here is my implementation of is_destructible_v:
template<class T>
struct is_unknown_bound_array : std::false_type
{};
template<class T>
struct is_unknown_bound_array<T> : std::true_type
{};
template<typename T, typename U = std::remove_all_extents_t<T>>
using has_dtor = decltype(std::declval<U&>().~U());
template<typename T>
constexpr bool is_destructible_v
= (std::experimental::is_detected_v<has_dtor, T> or std::is_reference_v<T>)
and not is_unknown_bound_array<T>::value
and not std::is_function_v<T>;
template<typename T>
struct is_destructible : std::bool_constant<is_destructible_v<T>>
{};
clang compiled happily and passed all libstdcxx's testsuite, while gcc failed to compile:
prog.cc:177:47: error: 'std::declval<int&>()' is not of type 'int&'
177 | using has_dtor = decltype(std::declval<U&>().~U());
| ~~~~~~~~~~~~~~~~~~~~^
prog.cc: In substitution of 'template<class T, class U> using has_dtor = decltype (declval<U&>().~ U()) [with T = int&&; U = int&&]':
So, gcc cannot do SFINAE on using has_dtor = decltype(std::declval<U&>().~U());.
Question:
- Which compiler object to standard here?
- What's the most elegant solution/workaround here? The ways I can think of is a little ugly
c++ language-lawyer c++17 template-meta-programming typetraits
asked Nov 24 '18 at 9:28
陳 力陳 力
1,6001723
It seems that GCC tries to make sure that the the type of expressionaina.~B()is the same asBwithout instantiating the function template, wandbox.org/permlink/4rhdNq4w7MMSIkcR
– Piotr Skotnicki
Nov 24 '18 at 11:05
msvc also can compile(though it doesn't supportand,or,not) godbolt.org/z/b9-Ii8
– 陳 力
Nov 25 '18 at 10:58
icc can compile: godbolt.org/z/ktZB3w
– 陳 力
Nov 25 '18 at 10:59
add a comment |
3
3
3
Here is my implementation of is_destructible_v:
template<class T>
struct is_unknown_bound_array : std::false_type
{};
template<class T>
struct is_unknown_bound_array<T> : std::true_type
{};
template<typename T, typename U = std::remove_all_extents_t<T>>
using has_dtor = decltype(std::declval<U&>().~U());
template<typename T>
constexpr bool is_destructible_v
= (std::experimental::is_detected_v<has_dtor, T> or std::is_reference_v<T>)
and not is_unknown_bound_array<T>::value
and not std::is_function_v<T>;
template<typename T>
struct is_destructible : std::bool_constant<is_destructible_v<T>>
{};
clang compiled happily and passed all libstdcxx's testsuite, while gcc failed to compile:
prog.cc:177:47: error: 'std::declval<int&>()' is not of type 'int&'
177 | using has_dtor = decltype(std::declval<U&>().~U());
| ~~~~~~~~~~~~~~~~~~~~^
prog.cc: In substitution of 'template<class T, class U> using has_dtor = decltype (declval<U&>().~ U()) [with T = int&&; U = int&&]':
So, gcc cannot do SFINAE on using has_dtor = decltype(std::declval<U&>().~U());.
Question:
- Which compiler object to standard here?
- What's the most elegant solution/workaround here? The ways I can think of is a little ugly
c++ language-lawyer c++17 template-meta-programming typetraits
asked Nov 24 '18 at 9:28
陳 力陳 力
1,6001723
Here is my implementation of is_destructible_v:
template<class T>
struct is_unknown_bound_array : std::false_type
{};
template<class T>
struct is_unknown_bound_array<T> : std::true_type
{};
template<typename T, typename U = std::remove_all_extents_t<T>>
using has_dtor = decltype(std::declval<U&>().~U());
template<typename T>
constexpr bool is_destructible_v
= (std::experimental::is_detected_v<has_dtor, T> or std::is_reference_v<T>)
and not is_unknown_bound_array<T>::value
and not std::is_function_v<T>;
template<typename T>
struct is_destructible : std::bool_constant<is_destructible_v<T>>
{};
clang compiled happily and passed all libstdcxx's testsuite, while gcc failed to compile:
prog.cc:177:47: error: 'std::declval<int&>()' is not of type 'int&'
177 | using has_dtor = decltype(std::declval<U&>().~U());
| ~~~~~~~~~~~~~~~~~~~~^
prog.cc: In substitution of 'template<class T, class U> using has_dtor = decltype (declval<U&>().~ U()) [with T = int&&; U = int&&]':
So, gcc cannot do SFINAE on using has_dtor = decltype(std::declval<U&>().~U());.
Question:
- Which compiler object to standard here?
- What's the most elegant solution/workaround here? The ways I can think of is a little ugly
c++ language-lawyer c++17 template-meta-programming typetraits
c++ language-lawyer c++17 template-meta-programming typetraits
asked Nov 24 '18 at 9:28
陳 力陳 力
1,6001723
asked Nov 24 '18 at 9:28
陳 力陳 力
1,6001723
edited Nov 28 '18 at 10:34
陳 力
asked Nov 24 '18 at 9:28
陳 力陳 力
1,6001723
asked Nov 24 '18 at 9:28
陳 力陳 力
1,6001723
asked Nov 24 '18 at 9:28
陳 力陳 力
1,6001723
1,6001723
It seems that GCC tries to make sure that the the type of expressionaina.~B()is the same asBwithout instantiating the function template, wandbox.org/permlink/4rhdNq4w7MMSIkcR
– Piotr Skotnicki
Nov 24 '18 at 11:05
msvc also can compile(though it doesn't supportand,or,not) godbolt.org/z/b9-Ii8
– 陳 力
Nov 25 '18 at 10:58
icc can compile: godbolt.org/z/ktZB3w
– 陳 力
Nov 25 '18 at 10:59
add a comment |
It seems that GCC tries to make sure that the the type of expressionaina.~B()is the same asBwithout instantiating the function template, wandbox.org/permlink/4rhdNq4w7MMSIkcR
– Piotr Skotnicki
Nov 24 '18 at 11:05
msvc also can compile(though it doesn't supportand,or,not) godbolt.org/z/b9-Ii8
– 陳 力
Nov 25 '18 at 10:58
icc can compile: godbolt.org/z/ktZB3w
– 陳 力
Nov 25 '18 at 10:59
It seems that GCC tries to make sure that the the type of expression
a in a.~B() is the same as B without instantiating the function template, wandbox.org/permlink/4rhdNq4w7MMSIkcR– Piotr Skotnicki
Nov 24 '18 at 11:05
It seems that GCC tries to make sure that the the type of expression
a in a.~B() is the same as B without instantiating the function template, wandbox.org/permlink/4rhdNq4w7MMSIkcR– Piotr Skotnicki
Nov 24 '18 at 11:05
msvc also can compile(though it doesn't support
and, or, not) godbolt.org/z/b9-Ii8– 陳 力
Nov 25 '18 at 10:58
msvc also can compile(though it doesn't support
and, or, not) godbolt.org/z/b9-Ii8– 陳 力
Nov 25 '18 at 10:58
icc can compile: godbolt.org/z/ktZB3w
– 陳 力
Nov 25 '18 at 10:59
icc can compile: godbolt.org/z/ktZB3w
– 陳 力
Nov 25 '18 at 10:59
add a comment |
1 Answer
1
active
oldest
votes
3
GCC seems to be broken when handling ~T() where T is a reference of scalar type.
It accepts the following code, which is clearly buggy per [expr.pseudo]/2:
template<typename T> using tester = decltype(int{}.~T(), char{});
tester<int&> ch;
int main() {}
I would use if constexpr to implement:
template<class T>
constexpr bool my_is_destructible() {
if constexpr (std::is_reference_v<T>) {
return true;
} else if constexpr (std::is_same_v<std::remove_cv_t<T>, void>
|| std::is_function_v<T>
|| is_unknown_bound_array<T>::value ) {
return false;
} else if constexpr (std::is_object_v<T>) {
return std::experimental::is_detected_v<has_dtor, T>;
} else {
return false;
}
}
It works with GCC too.
answered Nov 24 '18 at 12:33
llllllllllllllllllll
13.6k41742
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
3
GCC seems to be broken when handling ~T() where T is a reference of scalar type.
It accepts the following code, which is clearly buggy per [expr.pseudo]/2:
template<typename T> using tester = decltype(int{}.~T(), char{});
tester<int&> ch;
int main() {}
I would use if constexpr to implement:
template<class T>
constexpr bool my_is_destructible() {
if constexpr (std::is_reference_v<T>) {
return true;
} else if constexpr (std::is_same_v<std::remove_cv_t<T>, void>
|| std::is_function_v<T>
|| is_unknown_bound_array<T>::value ) {
return false;
} else if constexpr (std::is_object_v<T>) {
return std::experimental::is_detected_v<has_dtor, T>;
} else {
return false;
}
}
It works with GCC too.
answered Nov 24 '18 at 12:33
llllllllllllllllllll
13.6k41742
add a comment |
3
GCC seems to be broken when handling ~T() where T is a reference of scalar type.
It accepts the following code, which is clearly buggy per [expr.pseudo]/2:
template<typename T> using tester = decltype(int{}.~T(), char{});
tester<int&> ch;
int main() {}
I would use if constexpr to implement:
template<class T>
constexpr bool my_is_destructible() {
if constexpr (std::is_reference_v<T>) {
return true;
} else if constexpr (std::is_same_v<std::remove_cv_t<T>, void>
|| std::is_function_v<T>
|| is_unknown_bound_array<T>::value ) {
return false;
} else if constexpr (std::is_object_v<T>) {
return std::experimental::is_detected_v<has_dtor, T>;
} else {
return false;
}
}
It works with GCC too.
answered Nov 24 '18 at 12:33
llllllllllllllllllll
13.6k41742
add a comment |
3
3
3
GCC seems to be broken when handling ~T() where T is a reference of scalar type.
It accepts the following code, which is clearly buggy per [expr.pseudo]/2:
template<typename T> using tester = decltype(int{}.~T(), char{});
tester<int&> ch;
int main() {}
I would use if constexpr to implement:
template<class T>
constexpr bool my_is_destructible() {
if constexpr (std::is_reference_v<T>) {
return true;
} else if constexpr (std::is_same_v<std::remove_cv_t<T>, void>
|| std::is_function_v<T>
|| is_unknown_bound_array<T>::value ) {
return false;
} else if constexpr (std::is_object_v<T>) {
return std::experimental::is_detected_v<has_dtor, T>;
} else {
return false;
}
}
It works with GCC too.
answered Nov 24 '18 at 12:33
llllllllllllllllllll
13.6k41742
GCC seems to be broken when handling ~T() where T is a reference of scalar type.
It accepts the following code, which is clearly buggy per [expr.pseudo]/2:
template<typename T> using tester = decltype(int{}.~T(), char{});
tester<int&> ch;
int main() {}
I would use if constexpr to implement:
template<class T>
constexpr bool my_is_destructible() {
if constexpr (std::is_reference_v<T>) {
return true;
} else if constexpr (std::is_same_v<std::remove_cv_t<T>, void>
|| std::is_function_v<T>
|| is_unknown_bound_array<T>::value ) {
return false;
} else if constexpr (std::is_object_v<T>) {
return std::experimental::is_detected_v<has_dtor, T>;
} else {
return false;
}
}
It works with GCC too.
answered Nov 24 '18 at 12:33
llllllllllllllllllll
13.6k41742
edited Nov 24 '18 at 13:35
answered Nov 24 '18 at 12:33
llllllllllllllllllll
13.6k41742
answered Nov 24 '18 at 12:33
llllllllllllllllllll
13.6k41742
answered Nov 24 '18 at 12:33
llllllllllllllllllll
13.6k41742
13.6k41742
add a comment |
add a comment |
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It seems that GCC tries to make sure that the the type of expression
aina.~B()is the same asBwithout instantiating the function template, wandbox.org/permlink/4rhdNq4w7MMSIkcR– Piotr Skotnicki
Nov 24 '18 at 11:05
msvc also can compile(though it doesn't support
and,or,not) godbolt.org/z/b9-Ii8– 陳 力
Nov 25 '18 at 10:58
icc can compile: godbolt.org/z/ktZB3w
– 陳 力
Nov 25 '18 at 10:59