An 'if constexpr branch' does not get discarded inside lambda that is inside a template functionC++0x error with constexpr and returning template functionPossible to instantiate templates using a for loop in a C++14 constexpr function?Calling `this` member function from generic lambda - clang vs gccInitializing a static constexpr data member of the base class by using a static constexpr data member of the derived classSFINAE constexpr with std::getStatic templated constexpr nested class memberShould decltype(foo(1)) instantiate the constexpr function template foo?Why can't lambda, when cast to function pointer, be used in constexpr context?False-branch of if constexpr not discarded in templated lambdaNested constexpr-if statement in discarded branch is still evaluated?
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An 'if constexpr branch' does not get discarded inside lambda that is inside a template function
C++0x error with constexpr and returning template functionPossible to instantiate templates using a for loop in a C++14 constexpr function?Calling `this` member function from generic lambda - clang vs gccInitializing a static constexpr data member of the base class by using a static constexpr data member of the derived classSFINAE constexpr with std::getStatic templated constexpr nested class memberShould decltype(foo(1)) instantiate the constexpr function template foo?Why can't lambda, when cast to function pointer, be used in constexpr context?False-branch of if constexpr not discarded in templated lambdaNested constexpr-if statement in discarded branch is still evaluated?
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty height:90px;width:728px;box-sizing:border-box;
The following code:
#include <type_traits>
struct X
static constexpr void x()
;
template <class T1, class T2>
constexpr bool makeFalse() return false;
template <class T>
void foo()
T tmp;
auto f = [](auto type)
if constexpr (makeFalse<T, decltype(type)>())
T::x(); // <- clang does not discard
else
// noop
;
int main()
foo<int>();
does not compile with Clang, but compiles with GCC. I can't see anything wrong with this code, but I'm not sure. Is Clang right not compiling it?
c++ c++17 if-constexpr
add a comment |
The following code:
#include <type_traits>
struct X
static constexpr void x()
;
template <class T1, class T2>
constexpr bool makeFalse() return false;
template <class T>
void foo()
T tmp;
auto f = [](auto type)
if constexpr (makeFalse<T, decltype(type)>())
T::x(); // <- clang does not discard
else
// noop
;
int main()
foo<int>();
does not compile with Clang, but compiles with GCC. I can't see anything wrong with this code, but I'm not sure. Is Clang right not compiling it?
c++ c++17 if-constexpr
worth mentioning thatT
is not dependant on the lambda template parameter. Don't know however howif constexpr
should handle that.
– bolov
4 hours ago
(somewhat) equivalent example without lambda compiles fine , so I suspect it's a clang bug godbolt.org/z/Xok1wC
– bolov
4 hours ago
1
@bolov if you remove the generic lambda, it compiles too: godbolt.org/z/xoTBT6
– Amadeus
4 hours ago
add a comment |
The following code:
#include <type_traits>
struct X
static constexpr void x()
;
template <class T1, class T2>
constexpr bool makeFalse() return false;
template <class T>
void foo()
T tmp;
auto f = [](auto type)
if constexpr (makeFalse<T, decltype(type)>())
T::x(); // <- clang does not discard
else
// noop
;
int main()
foo<int>();
does not compile with Clang, but compiles with GCC. I can't see anything wrong with this code, but I'm not sure. Is Clang right not compiling it?
c++ c++17 if-constexpr
The following code:
#include <type_traits>
struct X
static constexpr void x()
;
template <class T1, class T2>
constexpr bool makeFalse() return false;
template <class T>
void foo()
T tmp;
auto f = [](auto type)
if constexpr (makeFalse<T, decltype(type)>())
T::x(); // <- clang does not discard
else
// noop
;
int main()
foo<int>();
does not compile with Clang, but compiles with GCC. I can't see anything wrong with this code, but I'm not sure. Is Clang right not compiling it?
c++ c++17 if-constexpr
c++ c++17 if-constexpr
edited 18 mins ago
Peter Mortensen
14k1987114
14k1987114
asked 5 hours ago
nicolainicolai
331211
331211
worth mentioning thatT
is not dependant on the lambda template parameter. Don't know however howif constexpr
should handle that.
– bolov
4 hours ago
(somewhat) equivalent example without lambda compiles fine , so I suspect it's a clang bug godbolt.org/z/Xok1wC
– bolov
4 hours ago
1
@bolov if you remove the generic lambda, it compiles too: godbolt.org/z/xoTBT6
– Amadeus
4 hours ago
add a comment |
worth mentioning thatT
is not dependant on the lambda template parameter. Don't know however howif constexpr
should handle that.
– bolov
4 hours ago
(somewhat) equivalent example without lambda compiles fine , so I suspect it's a clang bug godbolt.org/z/Xok1wC
– bolov
4 hours ago
1
@bolov if you remove the generic lambda, it compiles too: godbolt.org/z/xoTBT6
– Amadeus
4 hours ago
worth mentioning that
T
is not dependant on the lambda template parameter. Don't know however how if constexpr
should handle that.– bolov
4 hours ago
worth mentioning that
T
is not dependant on the lambda template parameter. Don't know however how if constexpr
should handle that.– bolov
4 hours ago
(somewhat) equivalent example without lambda compiles fine , so I suspect it's a clang bug godbolt.org/z/Xok1wC
– bolov
4 hours ago
(somewhat) equivalent example without lambda compiles fine , so I suspect it's a clang bug godbolt.org/z/Xok1wC
– bolov
4 hours ago
1
1
@bolov if you remove the generic lambda, it compiles too: godbolt.org/z/xoTBT6
– Amadeus
4 hours ago
@bolov if you remove the generic lambda, it compiles too: godbolt.org/z/xoTBT6
– Amadeus
4 hours ago
add a comment |
1 Answer
1
active
oldest
votes
[stmt.if]/2:
During the instantiation of an enclosing templated entity, if the condition is not value-dependent after its instantiation, the discarded substatement (if any) is not instantiated.
Since makeFalse<T, decltype(type)>()
is value-dependent after the instantiation of foo<int>
, it appears that T::x()
should be instantiated per the standard, and since T::x
is ill-formed when T
is int
, Clang is right not compiling it.
Wouldn't this reasoning imply that a hypotheticalif constexpr (makeFalse<decltype(type)>) type.x();
would not be discarded either?
– Barry
3 hours ago
@Barry Yes. Buttype.x()
is a dependent and possibly valid expression after the instantiation.
– cpplearner
2 hours ago
I think "possibly valid" is muddling things, I don't think that's relevant necessarily. Are you saying it won't be discarded even ifmakeFalse<decltype(type)>
isfalse
? Assume it's actually an interesting check... more likeif constexpr (can_x<decltype(type)>) type.x();
– Barry
2 hours ago
I am not convinced. Why then are my example and Amadeus compiling?
– bolov
2 hours ago
3
@Barry Let me try to clarify. There are two instantiations that can be involved: (1) the instantiation offoo
(2) the instantiation off
's function call operator template (which does not happen in OP's example). (1) does not discard either branch, because the condition is still value-dependent after it. If either branch is ill-formed after (1), an error will occur (but [temp.res]/8 may kick in when an expression in a branch is dependent). (2) does discard one branch because the condition is no longer dependent.
– cpplearner
2 hours ago
|
show 1 more comment
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1 Answer
1
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active
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active
oldest
votes
[stmt.if]/2:
During the instantiation of an enclosing templated entity, if the condition is not value-dependent after its instantiation, the discarded substatement (if any) is not instantiated.
Since makeFalse<T, decltype(type)>()
is value-dependent after the instantiation of foo<int>
, it appears that T::x()
should be instantiated per the standard, and since T::x
is ill-formed when T
is int
, Clang is right not compiling it.
Wouldn't this reasoning imply that a hypotheticalif constexpr (makeFalse<decltype(type)>) type.x();
would not be discarded either?
– Barry
3 hours ago
@Barry Yes. Buttype.x()
is a dependent and possibly valid expression after the instantiation.
– cpplearner
2 hours ago
I think "possibly valid" is muddling things, I don't think that's relevant necessarily. Are you saying it won't be discarded even ifmakeFalse<decltype(type)>
isfalse
? Assume it's actually an interesting check... more likeif constexpr (can_x<decltype(type)>) type.x();
– Barry
2 hours ago
I am not convinced. Why then are my example and Amadeus compiling?
– bolov
2 hours ago
3
@Barry Let me try to clarify. There are two instantiations that can be involved: (1) the instantiation offoo
(2) the instantiation off
's function call operator template (which does not happen in OP's example). (1) does not discard either branch, because the condition is still value-dependent after it. If either branch is ill-formed after (1), an error will occur (but [temp.res]/8 may kick in when an expression in a branch is dependent). (2) does discard one branch because the condition is no longer dependent.
– cpplearner
2 hours ago
|
show 1 more comment
[stmt.if]/2:
During the instantiation of an enclosing templated entity, if the condition is not value-dependent after its instantiation, the discarded substatement (if any) is not instantiated.
Since makeFalse<T, decltype(type)>()
is value-dependent after the instantiation of foo<int>
, it appears that T::x()
should be instantiated per the standard, and since T::x
is ill-formed when T
is int
, Clang is right not compiling it.
Wouldn't this reasoning imply that a hypotheticalif constexpr (makeFalse<decltype(type)>) type.x();
would not be discarded either?
– Barry
3 hours ago
@Barry Yes. Buttype.x()
is a dependent and possibly valid expression after the instantiation.
– cpplearner
2 hours ago
I think "possibly valid" is muddling things, I don't think that's relevant necessarily. Are you saying it won't be discarded even ifmakeFalse<decltype(type)>
isfalse
? Assume it's actually an interesting check... more likeif constexpr (can_x<decltype(type)>) type.x();
– Barry
2 hours ago
I am not convinced. Why then are my example and Amadeus compiling?
– bolov
2 hours ago
3
@Barry Let me try to clarify. There are two instantiations that can be involved: (1) the instantiation offoo
(2) the instantiation off
's function call operator template (which does not happen in OP's example). (1) does not discard either branch, because the condition is still value-dependent after it. If either branch is ill-formed after (1), an error will occur (but [temp.res]/8 may kick in when an expression in a branch is dependent). (2) does discard one branch because the condition is no longer dependent.
– cpplearner
2 hours ago
|
show 1 more comment
[stmt.if]/2:
During the instantiation of an enclosing templated entity, if the condition is not value-dependent after its instantiation, the discarded substatement (if any) is not instantiated.
Since makeFalse<T, decltype(type)>()
is value-dependent after the instantiation of foo<int>
, it appears that T::x()
should be instantiated per the standard, and since T::x
is ill-formed when T
is int
, Clang is right not compiling it.
[stmt.if]/2:
During the instantiation of an enclosing templated entity, if the condition is not value-dependent after its instantiation, the discarded substatement (if any) is not instantiated.
Since makeFalse<T, decltype(type)>()
is value-dependent after the instantiation of foo<int>
, it appears that T::x()
should be instantiated per the standard, and since T::x
is ill-formed when T
is int
, Clang is right not compiling it.
answered 4 hours ago
cpplearnercpplearner
6,22122543
6,22122543
Wouldn't this reasoning imply that a hypotheticalif constexpr (makeFalse<decltype(type)>) type.x();
would not be discarded either?
– Barry
3 hours ago
@Barry Yes. Buttype.x()
is a dependent and possibly valid expression after the instantiation.
– cpplearner
2 hours ago
I think "possibly valid" is muddling things, I don't think that's relevant necessarily. Are you saying it won't be discarded even ifmakeFalse<decltype(type)>
isfalse
? Assume it's actually an interesting check... more likeif constexpr (can_x<decltype(type)>) type.x();
– Barry
2 hours ago
I am not convinced. Why then are my example and Amadeus compiling?
– bolov
2 hours ago
3
@Barry Let me try to clarify. There are two instantiations that can be involved: (1) the instantiation offoo
(2) the instantiation off
's function call operator template (which does not happen in OP's example). (1) does not discard either branch, because the condition is still value-dependent after it. If either branch is ill-formed after (1), an error will occur (but [temp.res]/8 may kick in when an expression in a branch is dependent). (2) does discard one branch because the condition is no longer dependent.
– cpplearner
2 hours ago
|
show 1 more comment
Wouldn't this reasoning imply that a hypotheticalif constexpr (makeFalse<decltype(type)>) type.x();
would not be discarded either?
– Barry
3 hours ago
@Barry Yes. Buttype.x()
is a dependent and possibly valid expression after the instantiation.
– cpplearner
2 hours ago
I think "possibly valid" is muddling things, I don't think that's relevant necessarily. Are you saying it won't be discarded even ifmakeFalse<decltype(type)>
isfalse
? Assume it's actually an interesting check... more likeif constexpr (can_x<decltype(type)>) type.x();
– Barry
2 hours ago
I am not convinced. Why then are my example and Amadeus compiling?
– bolov
2 hours ago
3
@Barry Let me try to clarify. There are two instantiations that can be involved: (1) the instantiation offoo
(2) the instantiation off
's function call operator template (which does not happen in OP's example). (1) does not discard either branch, because the condition is still value-dependent after it. If either branch is ill-formed after (1), an error will occur (but [temp.res]/8 may kick in when an expression in a branch is dependent). (2) does discard one branch because the condition is no longer dependent.
– cpplearner
2 hours ago
Wouldn't this reasoning imply that a hypothetical
if constexpr (makeFalse<decltype(type)>) type.x();
would not be discarded either?– Barry
3 hours ago
Wouldn't this reasoning imply that a hypothetical
if constexpr (makeFalse<decltype(type)>) type.x();
would not be discarded either?– Barry
3 hours ago
@Barry Yes. But
type.x()
is a dependent and possibly valid expression after the instantiation.– cpplearner
2 hours ago
@Barry Yes. But
type.x()
is a dependent and possibly valid expression after the instantiation.– cpplearner
2 hours ago
I think "possibly valid" is muddling things, I don't think that's relevant necessarily. Are you saying it won't be discarded even if
makeFalse<decltype(type)>
is false
? Assume it's actually an interesting check... more like if constexpr (can_x<decltype(type)>) type.x();
– Barry
2 hours ago
I think "possibly valid" is muddling things, I don't think that's relevant necessarily. Are you saying it won't be discarded even if
makeFalse<decltype(type)>
is false
? Assume it's actually an interesting check... more like if constexpr (can_x<decltype(type)>) type.x();
– Barry
2 hours ago
I am not convinced. Why then are my example and Amadeus compiling?
– bolov
2 hours ago
I am not convinced. Why then are my example and Amadeus compiling?
– bolov
2 hours ago
3
3
@Barry Let me try to clarify. There are two instantiations that can be involved: (1) the instantiation of
foo
(2) the instantiation of f
's function call operator template (which does not happen in OP's example). (1) does not discard either branch, because the condition is still value-dependent after it. If either branch is ill-formed after (1), an error will occur (but [temp.res]/8 may kick in when an expression in a branch is dependent). (2) does discard one branch because the condition is no longer dependent.– cpplearner
2 hours ago
@Barry Let me try to clarify. There are two instantiations that can be involved: (1) the instantiation of
foo
(2) the instantiation of f
's function call operator template (which does not happen in OP's example). (1) does not discard either branch, because the condition is still value-dependent after it. If either branch is ill-formed after (1), an error will occur (but [temp.res]/8 may kick in when an expression in a branch is dependent). (2) does discard one branch because the condition is no longer dependent.– cpplearner
2 hours ago
|
show 1 more comment
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worth mentioning that
T
is not dependant on the lambda template parameter. Don't know however howif constexpr
should handle that.– bolov
4 hours ago
(somewhat) equivalent example without lambda compiles fine , so I suspect it's a clang bug godbolt.org/z/Xok1wC
– bolov
4 hours ago
1
@bolov if you remove the generic lambda, it compiles too: godbolt.org/z/xoTBT6
– Amadeus
4 hours ago