May I change the held type in a std::variant from within a call to std::visit2019 Community Moderator ElectionImplementing is_constexpr_copiableHow does std::visit work with std::variant?Visit and invoke a variant of std::functionVisit a variant with a monostateHow to simplify std::variant class typesis decltype required in trailing return using value of argument typeget currently held typeid of std::variant (like boost::variant type())noexcept visitation for std::variantHow to use std::visit with std::variant containing enumPredefined type list passed to a std::variant
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May I change the held type in a std::variant from within a call to std::visit
2019 Community Moderator ElectionImplementing is_constexpr_copiableHow does std::visit work with std::variant?Visit and invoke a variant of std::functionVisit a variant with a monostateHow to simplify std::variant class typesis decltype required in trailing return using value of argument typeget currently held typeid of std::variant (like boost::variant type())noexcept visitation for std::variantHow to use std::visit with std::variant containing enumPredefined type list passed to a std::variant
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
edited 10 hours ago
Barry
184k21324595
asked 10 hours ago
burnpanckburnpanck
1,141622
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
edited 10 hours ago
Barry
184k21324595
asked 10 hours ago
burnpanckburnpanck
1,141622
5
Do you want quotes from the standard to back up the answer(s) you get?
– NathanOliver
10 hours ago
1
From looking at [variant.visit], I'm 99% sure this code is compliant and guaranteed not to have UB, sincestd::visit(vis, variant)should be equivalent tovis(get</* active member */>(variant)), but I'm not confident enough in reading the standard to be certain
– Justin
9 hours ago
@NathanOliver: I don't need actual quotes from the standard, as long as the experts here can agree on the answer:-).
– burnpanck
9 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
edited 10 hours ago
Barry
184k21324595
asked 10 hours ago
burnpanckburnpanck
1,141622
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
c++ c++17 std-variant
edited 10 hours ago
Barry
184k21324595
asked 10 hours ago
burnpanckburnpanck
1,141622
edited 10 hours ago
Barry
184k21324595
asked 10 hours ago
burnpanckburnpanck
1,141622
edited 10 hours ago
Barry
184k21324595
edited 10 hours ago
Barry
184k21324595
edited 10 hours ago
Barry
184k21324595
184k21324595
asked 10 hours ago
burnpanckburnpanck
1,141622
asked 10 hours ago
burnpanckburnpanck
1,141622
asked 10 hours ago
burnpanckburnpanck
1,141622
1,141622
5
Do you want quotes from the standard to back up the answer(s) you get?
– NathanOliver
10 hours ago
1
From looking at [variant.visit], I'm 99% sure this code is compliant and guaranteed not to have UB, sincestd::visit(vis, variant)should be equivalent tovis(get</* active member */>(variant)), but I'm not confident enough in reading the standard to be certain
– Justin
9 hours ago
@NathanOliver: I don't need actual quotes from the standard, as long as the experts here can agree on the answer:-).
– burnpanck
9 hours ago
add a comment |
5
Do you want quotes from the standard to back up the answer(s) you get?
– NathanOliver
10 hours ago
1
From looking at [variant.visit], I'm 99% sure this code is compliant and guaranteed not to have UB, sincestd::visit(vis, variant)should be equivalent tovis(get</* active member */>(variant)), but I'm not confident enough in reading the standard to be certain
– Justin
9 hours ago
@NathanOliver: I don't need actual quotes from the standard, as long as the experts here can agree on the answer:-).
– burnpanck
9 hours ago
5
5
Do you want quotes from the standard to back up the answer(s) you get?
– NathanOliver
10 hours ago
Do you want quotes from the standard to back up the answer(s) you get?
– NathanOliver
10 hours ago
1
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
9 hours ago
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
9 hours ago
@NathanOliver: I don't need actual quotes from the standard, as long as the experts here can agree on the answer:-).
– burnpanck
9 hours ago
@NathanOliver: I don't need actual quotes from the standard, as long as the experts here can agree on the answer:-).
– burnpanck
9 hours ago
add a comment |
1 Answer
1
active
oldest
votes
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.
answered 9 hours ago
BarryBarry
184k21324595
add a comment |
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1 Answer
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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.
answered 9 hours ago
BarryBarry
184k21324595
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.
answered 9 hours ago
BarryBarry
184k21324595
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.
answered 9 hours ago
BarryBarry
184k21324595
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.
answered 9 hours ago
BarryBarry
184k21324595
answered 9 hours ago
BarryBarry
184k21324595
answered 9 hours ago
BarryBarry
184k21324595
answered 9 hours ago
BarryBarry
184k21324595
184k21324595
add a comment |
add a comment |
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5
Do you want quotes from the standard to back up the answer(s) you get?
– NathanOliver
10 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 tovis(get</* active member */>(variant)), but I'm not confident enough in reading the standard to be certain– Justin
9 hours ago
@NathanOliver: I don't need actual quotes from the standard, as long as the experts here can agree on the answer:-).
– burnpanck
9 hours ago