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24

6

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

share|improve this question

edited Apr 29 at 23:33

Peter Mortensen

14k1987114

asked Apr 29 at 18:34

nicolainicolai

366211

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  worth mentioning that T is not dependant on the lambda template parameter. Don't know however how if constexpr should handle that.
  
  – bolov
  Apr 29 at 19:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  (somewhat) equivalent example without lambda compiles fine , so I suspect it's a clang bug godbolt.org/z/Xok1wC
  
  – bolov
  Apr 29 at 19:03
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @bolov if you remove the generic lambda, it compiles too: godbolt.org/z/xoTBT6
  
  – Amadeus
  Apr 29 at 19:08
  

  
  
  
  

add a comment | 

24

6

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

share|improve this question

edited Apr 29 at 23:33

Peter Mortensen

14k1987114

asked Apr 29 at 18:34

nicolainicolai

366211

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  worth mentioning that T is not dependant on the lambda template parameter. Don't know however how if constexpr should handle that.
  
  – bolov
  Apr 29 at 19:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  (somewhat) equivalent example without lambda compiles fine , so I suspect it's a clang bug godbolt.org/z/Xok1wC
  
  – bolov
  Apr 29 at 19:03
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @bolov if you remove the generic lambda, it compiles too: godbolt.org/z/xoTBT6
  
  – Amadeus
  Apr 29 at 19:08
  

  
  
  
  

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

share|improve this question

edited Apr 29 at 23:33

Peter Mortensen

14k1987114

asked Apr 29 at 18:34

nicolainicolai

366211

#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>();

share|improve this question

edited Apr 29 at 23:33

Peter Mortensen

14k1987114

asked Apr 29 at 18:34

nicolainicolai

366211

share|improve this question

edited Apr 29 at 23:33

Peter Mortensen

14k1987114

asked Apr 29 at 18:34

nicolainicolai

366211

edited Apr 29 at 23:33

Peter Mortensen

14k1987114

edited Apr 29 at 23:33

Peter Mortensen

14k1987114

asked Apr 29 at 18:34

nicolainicolai

366211

asked Apr 29 at 18:34

nicolainicolai

366211

1 Answer
1

active

oldest

votes

17

[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.

share|improve this answer

answered Apr 29 at 19:12

cpplearnercpplearner

6,32122645

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Wouldn't this reasoning imply that a hypothetical if constexpr (makeFalse<decltype(type)>) type.x(); would not be discarded either?
  
  – Barry
  Apr 29 at 20:36
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Barry Yes. But type.x() is a dependent and possibly valid expression after the instantiation.
  
  – cpplearner
  Apr 29 at 20:52
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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
  Apr 29 at 20:56
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  
  @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
  Apr 29 at 21:09
  

  
  
  
  


• 
  
  
  

  
  6
  

  
  
  
  
  
  

  
  
  This is equivalent to the motivating example of P0588R0, except that there's not even an implicit capture that could complicate things under the old rules.
  
  – T.C.
  Apr 30 at 2:17
  

  
  
  
  

 | 
show 2 more comments

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17

[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.

share|improve this answer

answered Apr 29 at 19:12

cpplearnercpplearner

6,32122645

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Wouldn't this reasoning imply that a hypothetical if constexpr (makeFalse<decltype(type)>) type.x(); would not be discarded either?
  
  – Barry
  Apr 29 at 20:36
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Barry Yes. But type.x() is a dependent and possibly valid expression after the instantiation.
  
  – cpplearner
  Apr 29 at 20:52
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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
  Apr 29 at 20:56
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  
  @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
  Apr 29 at 21:09
  

  
  
  
  


• 
  
  
  

  
  6
  

  
  
  
  
  
  

  
  
  This is equivalent to the motivating example of P0588R0, except that there's not even an implicit capture that could complicate things under the old rules.
  
  – T.C.
  Apr 30 at 2:17
  

  
  
  
  

 | 
show 2 more comments

17

[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.

share|improve this answer

answered Apr 29 at 19:12

cpplearnercpplearner

6,32122645

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Wouldn't this reasoning imply that a hypothetical if constexpr (makeFalse<decltype(type)>) type.x(); would not be discarded either?
  
  – Barry
  Apr 29 at 20:36
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Barry Yes. But type.x() is a dependent and possibly valid expression after the instantiation.
  
  – cpplearner
  Apr 29 at 20:52
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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
  Apr 29 at 20:56
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  
  @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
  Apr 29 at 21:09
  

  
  
  
  


• 
  
  
  

  
  6
  

  
  
  
  
  
  

  
  
  This is equivalent to the motivating example of P0588R0, except that there's not even an implicit capture that could complicate things under the old rules.
  
  – T.C.
  Apr 30 at 2:17
  

  
  
  
  

 | 
show 2 more comments

[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.

share|improve this answer

answered Apr 29 at 19:12

cpplearnercpplearner

6,32122645

share|improve this answer

answered Apr 29 at 19:12

cpplearnercpplearner

6,32122645

answered Apr 29 at 19:12

cpplearnercpplearner

6,32122645

answered Apr 29 at 19:12

cpplearnercpplearner

6,32122645