May I change the held type in a std::variant from within a call to std::visit
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 Mar 15 at 17:23
Barry
185k21326601
asked Mar 15 at 17:05
burnpanckburnpanck
1,151622
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 Mar 15 at 17:23
Barry
185k21326601
asked Mar 15 at 17:05
burnpanckburnpanck
1,151622
5
Do you want quotes from the standard to back up the answer(s) you get?
– NathanOliver
Mar 15 at 17:08
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
Mar 15 at 17:27
@NathanOliver: I don't need actual quotes from the standard, as long as the experts here can agree on the answer:-).
– burnpanck
Mar 15 at 17:31
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 Mar 15 at 17:23
Barry
185k21326601
asked Mar 15 at 17:05
burnpanckburnpanck
1,151622
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 Mar 15 at 17:23
Barry
185k21326601
asked Mar 15 at 17:05
burnpanckburnpanck
1,151622
edited Mar 15 at 17:23
Barry
185k21326601
asked Mar 15 at 17:05
burnpanckburnpanck
1,151622
edited Mar 15 at 17:23
Barry
185k21326601
edited Mar 15 at 17:23
Barry
185k21326601
edited Mar 15 at 17:23
Barry
185k21326601
185k21326601
asked Mar 15 at 17:05
burnpanckburnpanck
1,151622
asked Mar 15 at 17:05
burnpanckburnpanck
1,151622
asked Mar 15 at 17:05
burnpanckburnpanck
1,151622
1,151622
5
Do you want quotes from the standard to back up the answer(s) you get?
– NathanOliver
Mar 15 at 17:08
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
Mar 15 at 17:27
@NathanOliver: I don't need actual quotes from the standard, as long as the experts here can agree on the answer:-).
– burnpanck
Mar 15 at 17:31
add a comment |
5
Do you want quotes from the standard to back up the answer(s) you get?
– NathanOliver
Mar 15 at 17:08
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
Mar 15 at 17:27
@NathanOliver: I don't need actual quotes from the standard, as long as the experts here can agree on the answer:-).
– burnpanck
Mar 15 at 17:31
5
5
Do you want quotes from the standard to back up the answer(s) you get?
– NathanOliver
Mar 15 at 17:08
Do you want quotes from the standard to back up the answer(s) you get?
– NathanOliver
Mar 15 at 17:08
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
Mar 15 at 17:27
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
Mar 15 at 17:27
@NathanOliver: I don't need actual quotes from the standard, as long as the experts here can agree on the answer:-).
– burnpanck
Mar 15 at 17:31
@NathanOliver: I don't need actual quotes from the standard, as long as the experts here can agree on the answer:-).
– burnpanck
Mar 15 at 17:31
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 Mar 15 at 17:28
BarryBarry
185k21326601
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 Mar 15 at 17:28
BarryBarry
185k21326601
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 Mar 15 at 17:28
BarryBarry
185k21326601
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 Mar 15 at 17:28
BarryBarry
185k21326601
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 Mar 15 at 17:28
BarryBarry
185k21326601
answered Mar 15 at 17:28
BarryBarry
185k21326601
answered Mar 15 at 17:28
BarryBarry
185k21326601
answered Mar 15 at 17:28
BarryBarry
185k21326601
185k21326601
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
Mar 15 at 17:08
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
Mar 15 at 17:27
@NathanOliver: I don't need actual quotes from the standard, as long as the experts here can agree on the answer:-).
– burnpanck
Mar 15 at 17:31