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May I change the held type in a std::variant from within a call to std::visit

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May I change the held type in a std::variant from within a call to std::visit

16

2

Does the following code invoke undefined behaviour?

std::variant<A,B> v = ...;

std::visit([&v](auto& e)
 if constexpr (std::is_same_v<std::remove_reference_t<decltype(e)>,A>)
 e.some_modifying_operation_on_A();
 else 
 int i = e.some_accessor_of_B();
 v = some_function_returning_A(i); 
 
, v);

In particular, when the variant does not contain an A,
this code re-assigns an A while still holding a reference to the previously held object of type B.
However, because the reference is not used anymore after the assignment,
I feel the code is fine.
However, would a standard-library be free to implement std::visit
in a way such that the above is undefined behaviour?

c++ c++17 std-variant

share|improve this question

edited 4 hours ago

Barry

184k21323594

asked 4 hours ago

burnpanckburnpanck

1,131622

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  
  Do you want quotes from the standard to back up the answer(s) you get?
  
  – NathanOliver
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  From looking at [variant.visit], I'm 99% sure this code is compliant and guaranteed not to have UB, since std::visit(vis, variant) should be equivalent to vis(get</* active member */>(variant)), but I'm not confident enough in reading the standard to be certain
  
  – Justin
  3 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @NathanOliver: I don't need actual quotes from the standard, as long as the experts here can agree on the answer:-).
  
  – burnpanck
  3 hours ago
  

  
  
  
  

add a comment | 

16

2

Does the following code invoke undefined behaviour?

std::variant<A,B> v = ...;

std::visit([&v](auto& e)
 if constexpr (std::is_same_v<std::remove_reference_t<decltype(e)>,A>)
 e.some_modifying_operation_on_A();
 else 
 int i = e.some_accessor_of_B();
 v = some_function_returning_A(i); 
 
, v);

In particular, when the variant does not contain an A,
this code re-assigns an A while still holding a reference to the previously held object of type B.
However, because the reference is not used anymore after the assignment,
I feel the code is fine.
However, would a standard-library be free to implement std::visit
in a way such that the above is undefined behaviour?

c++ c++17 std-variant

share|improve this question

edited 4 hours ago

Barry

184k21323594

asked 4 hours ago

burnpanckburnpanck

1,131622

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  
  Do you want quotes from the standard to back up the answer(s) you get?
  
  – NathanOliver
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  From looking at [variant.visit], I'm 99% sure this code is compliant and guaranteed not to have UB, since std::visit(vis, variant) should be equivalent to vis(get</* active member */>(variant)), but I'm not confident enough in reading the standard to be certain
  
  – Justin
  3 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @NathanOliver: I don't need actual quotes from the standard, as long as the experts here can agree on the answer:-).
  
  – burnpanck
  3 hours ago
  

  
  
  
  

add a comment | 

Does the following code invoke undefined behaviour?

std::variant<A,B> v = ...;

std::visit([&v](auto& e)
 if constexpr (std::is_same_v<std::remove_reference_t<decltype(e)>,A>)
 e.some_modifying_operation_on_A();
 else 
 int i = e.some_accessor_of_B();
 v = some_function_returning_A(i); 
 
, v);

In particular, when the variant does not contain an A,
this code re-assigns an A while still holding a reference to the previously held object of type B.
However, because the reference is not used anymore after the assignment,
I feel the code is fine.
However, would a standard-library be free to implement std::visit
in a way such that the above is undefined behaviour?

c++ c++17 std-variant

share|improve this question

edited 4 hours ago

Barry

184k21323594

asked 4 hours ago

burnpanckburnpanck

1,131622

std::variant<A,B> v = ...;

std::visit([&v](auto& e)
 if constexpr (std::is_same_v<std::remove_reference_t<decltype(e)>,A>)
 e.some_modifying_operation_on_A();
 else 
 int i = e.some_accessor_of_B();
 v = some_function_returning_A(i); 
 
, v);

share|improve this question

edited 4 hours ago

Barry

184k21323594

asked 4 hours ago

burnpanckburnpanck

1,131622

share|improve this question

edited 4 hours ago

Barry

184k21323594

asked 4 hours ago

burnpanckburnpanck

1,131622

edited 4 hours ago

Barry

184k21323594

edited 4 hours ago

Barry

184k21323594

asked 4 hours ago

burnpanckburnpanck

1,131622

asked 4 hours ago

burnpanckburnpanck

1,131622

1 Answer
1

active

oldest

votes

9

The code is fine.

There is no requirement in the specification of std::visit that the visitor not change the alternative of any of the variants it is invoked on. The only requirement is:

Requires: For each valid pack m, e(m) shall be a valid expression. All such expressions shall be of the same type and value category; otherwise, the program is ill-formed.

Your visitor is a valid expression for each m and always returns void, so it satisfies the requirements and has well-defined behavior.

share|improve this answer

answered 3 hours ago

BarryBarry

184k21323594

add a comment | 

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9

The code is fine.

There is no requirement in the specification of std::visit that the visitor not change the alternative of any of the variants it is invoked on. The only requirement is:

Requires: For each valid pack m, e(m) shall be a valid expression. All such expressions shall be of the same type and value category; otherwise, the program is ill-formed.

Your visitor is a valid expression for each m and always returns void, so it satisfies the requirements and has well-defined behavior.

share|improve this answer

answered 3 hours ago

BarryBarry

184k21323594

add a comment | 

9

The code is fine.

There is no requirement in the specification of std::visit that the visitor not change the alternative of any of the variants it is invoked on. The only requirement is:

Requires: For each valid pack m, e(m) shall be a valid expression. All such expressions shall be of the same type and value category; otherwise, the program is ill-formed.

Your visitor is a valid expression for each m and always returns void, so it satisfies the requirements and has well-defined behavior.

share|improve this answer

answered 3 hours ago

BarryBarry

184k21323594

add a comment | 

The code is fine.

There is no requirement in the specification of std::visit that the visitor not change the alternative of any of the variants it is invoked on. The only requirement is:

Requires: For each valid pack m, e(m) shall be a valid expression. All such expressions shall be of the same type and value category; otherwise, the program is ill-formed.

Your visitor is a valid expression for each m and always returns void, so it satisfies the requirements and has well-defined behavior.

share|improve this answer

answered 3 hours ago

BarryBarry

184k21323594

share|improve this answer

answered 3 hours ago

BarryBarry

184k21323594

answered 3 hours ago

BarryBarry

184k21323594

answered 3 hours ago

BarryBarry

184k21323594