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An 'if constexpr branch' does not get discarded inside lambda that is inside a template function

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}

17

5

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 24 mins ago

Peter Mortensen

14k1987114

asked 5 hours ago

nicolainicolai

331211

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @bolov if you remove the generic lambda, it compiles too: godbolt.org/z/xoTBT6
  
  – Amadeus
  4 hours ago
  

  
  
  
  

add a comment | 

17

5

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 24 mins ago

Peter Mortensen

14k1987114

asked 5 hours ago

nicolainicolai

331211

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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
  
  
  

  
  
  
  


• 
  
  
  

  
  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

share|improve this question

edited 24 mins ago

Peter Mortensen

14k1987114

asked 5 hours ago

nicolainicolai

331211

#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 24 mins ago

Peter Mortensen

14k1987114

asked 5 hours ago

nicolainicolai

331211

share|improve this question

edited 24 mins ago

Peter Mortensen

14k1987114

asked 5 hours ago

nicolainicolai

331211

edited 24 mins ago

Peter Mortensen

14k1987114

edited 24 mins ago

Peter Mortensen

14k1987114

asked 5 hours ago

nicolainicolai

331211

asked 5 hours ago

nicolainicolai

331211

1 Answer
1

active

oldest

votes

10

[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 4 hours ago

cpplearnercpplearner

6,23122543

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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
  3 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
  3 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 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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10

[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 4 hours ago

cpplearnercpplearner

6,23122543

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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
  3 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
  3 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 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

10

[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 4 hours ago

cpplearnercpplearner

6,23122543

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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
  3 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
  3 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 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

[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 4 hours ago

cpplearnercpplearner

6,23122543

share|improve this answer

answered 4 hours ago

cpplearnercpplearner

6,23122543

answered 4 hours ago

cpplearnercpplearner

6,23122543

answered 4 hours ago

cpplearnercpplearner

6,23122543