is_assignable<Incomplete&, Incomplete&>
Section: 21.3.3 [meta.type.synop] Status: Open Submitter: Casey Carter Opened: 2018-04-10 Last modified: 2024-08-21
Priority: 2
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Discussion:
LWG 2939 suggests that the the preconditions of the type traits need reevaluation.
This issue focuses specifically on is_assignable
and, by extension, its variants:
is_copy_assignable<T>
, equivalent to is_assignable<T&, const T&>
,
is_move_assignable<T>
, equivalent to is_assignable<T&, T>
,
is_trivially_assignable<T, U>
, equivalent to is_assignable<T, U> &&
/* magic */
,
is_trivially_copy_assignable<T>
, equivalent to is_assignable<T&, const T&> &&
/* magic */
,
is_trivially_move_assignable<T>
, equivalent to is_assignable<T&, T> &&
/* magic */
,
is_nothrow_assignable<T, U>
, equivalent to is_assignable<T, U> &&
noexcept(declval<T>() = declval<U>())
,
is_nothrow_copy_assignable<T>
, equivalent to is_assignable<T&, const T&> &&
noexcept(declval<T&>() = declval<const T&>())
,
is_nothrow_move_assignable<T>
, equivalent to is_assignable<T&, T> &&
noexcept(declval<T&>() = declval<T>())
, and
We note a discrepancy: is_copy_assignable<T>
requires T
to be a complete type, but the
equivalent form is_assignable<T&, const T&>
does not. The requirement for
is_copy_assignable<T>
seems sensible, since there's no way to determine whether or not the assignment
declval<T&>() = declval<const T&>()
is well-formed when T
is incomplete.
It seems that the same argument should apply to all of the above "assignable" traits, and that they must require that
the referent type is complete when given a reference type parameter to be implementable.
[2018-08 Batavia Monday issue discussion]
Issues 2797, 2939, 3022, and 3099 are all closely related. Walter to write a paper resolving them.
[2020-02-14, Prague]
LWG discussions. Set priority to 2.
[2023-06-12; Varna]
P1285R0 is related to this issue.
Previous resolution [SUPERSEDED]:
This wording is relative to N4741.
- In 21.3.5.4 [meta.unary.prop] Table 42, change the Precondition text for
is_assignable
,is_trivially_assignable
, andis_nothrow_assignable
as follows:remove_cvref_t<
T
>
andremove_cvref_t<
U
>
shall be complete types,cvvoid
, or arrays of unknown bound.- In 21.3.5.4 [meta.unary.prop] Table 42, change the Precondition text for
is_copy_assignable
,is_move_assignable
,is_trivially_copy_assignable
,is_trivially_move_assignable
,is_nothrow_copy_assignable
, andis_nothrow_move_assignable
as follows:remove_cvref_t<
T
>
shall be a complete type,cvvoid
, or an array of unknown bound.
[2024-08-21; Jonathan provides improved wording]
During LWG telecon review Tomasz pointed out that we don't always require
a complete type for the right operand of an assignment. Given
T::operator=(U&)
we should be able to give a correct answer for
is_assignable<T&, U&>
whether of not U
is complete.
This also affects e.g. is_constructible<T, U&>
if T::T(U&)
exists.
So for the examples above,
remove_cvref_t<U>
is not needed to give a correct answer.
However, if T::operator=(U&)
does not exist,
then we do need U
to be complete so we can tell if there is
an implicit conversion sequence to T
or another type that can be
assigned to T
.
We do not know how to solve this problem, and it's broader than just
is_assignable
.
It was suggested to make an incremental improvement to is_assignable
and open a new issue for the broader issue.
Proposed resolution:
This wording is relative to N4988.
- In 21.3.5.4 [meta.unary.prop] Table 51, change the Precondition text for
is_assignable
,is_trivially_assignable
, andis_nothrow_assignable
as follows:remove_cvref_t<
T
>
andU
shall be complete types, cvvoid
, or arrays of unknown bound.- In 21.3.5.4 [meta.unary.prop] Table 51, change the Precondition text for
is_copy_assignable
,is_move_assignable
,is_trivially_copy_assignable
,is_trivially_move_assignable
,is_nothrow_copy_assignable
, andis_nothrow_move_assignable
as follows:remove_cvref_t<
T
>
shall be a complete type,cvvoid
, or an array of unknown bound.