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variant
default constructor has vague constexpr
requirementsSection: 22.6.3.2 [variant.ctor] Status: New Submitter: Louis Dionne Opened: 2019-06-04 Last modified: 2020-09-06
Priority: 2
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Discussion:
In 22.6.3.2 [variant.ctor] p5, we say:
Remarks: This function shall be
constexpr
if and only if the value-initialization of the alternative typeT0
would satisfy the requirements for a constexpr function. […]
First of all, I find it confusing that we say "This function shall be constexpr
if […]",
when the declaration of the function clearly has the constexpr
keyword on it
unconditionally. Instead, I would use the wording "This function shall be usable in a constexpr context
if […]".
constexpr
-friendly as an extension. Instead, it seems better to just say "if".
Finally, I think the condition under which the function must be constexpr
-friendly is not
something we can test for because it says "value-initialization of the alternative type
T0
would satisfy the requirements for a constexpr function", which doesn't imply
the value initialization can actually be be performed inside a constexpr context (for example the
default constructor could be constexpr
friendly but not marked with the
constexpr
keyword).
[2017-06-17, Tim Song comments]
This issue is related to LWG 2833(i).
[2019-07 Issue Prioritization]
Priority to 2 after discussion on the reflector.
Previous resolution from Daniel [SUPERSEDED]:This wording is relative to N4810.
Modify 22.6.3.2 [variant.ctor] as indicated:
constexpr variant() noexcept(see below);-2- Effects: […]
-3- Ensures: […] -4- Throws: […] -5- Remarks: This function shall be usable in a context that requires constant evaluation if the alternative typeT0
can be value-initialized in a context that requires constant evaluation. […]constexpr
if and only if the value-initialization of the alternative typeT0
would satisfy the requirements for a constexpr function
[2020-06-08 Nina Dinka Ranns comments]
The revised wording provided by LWG 2833(i) should resolve this issue as well.
Proposed resolution: