std::experimental::ranges::Assignable
Defined in header <experimental/ranges/concepts>
|
||
template
<
class T, class U >
concept bool Assignable =
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(ranges TS) | |
The concept Assignable<T, U>
specifies that an expression of the type and value category specified by U
can be assigned to an lvalue expression whose type is specified by T
.
Given
-
t
, an lvalue of type std::remove_reference_t<T> that refers to an objecto
-
u
, an expression such that decltype((u)) isU
u2
, a distinct object that is equal tou
,
Assignable<T, U>
is satisfied only if
- std::addressof (t = u) == std::addressof (o)
- After evaluating t = u:
t
is equal tou2
, unlessu
is a non-const xvalue that refers too
(i.e., the assignment is a self-move-assignment),- if
u
is a glvalue:- If it is a non-const xvalue, the object to which it refers is in a valid but unspecified state;
- Otherwise, the object it refers to is not modified;
There need not be any subsumption relationship between Assignable<T, U>
and
std::is_lvalue_reference
<T>
::
value
Equality preservation
An expression is equality preserving if it results in equal outputs given equal inputs.
- The inputs to an expression consist of its operands.
- The outputs of an expression consist of its result and all operands modified by the expression (if any).
Every expression required to be equality preserving is further required to be stable
Unless noted otherwise, every expression used in a requires-expression
Notes
A deduction constraint of the form
{ expression }
-
> Same<T>
&&
effectively requires
decltype((expression))&&
to be the exact same type as T&&
Assignment need not be a total function. In particular, if assigning to some object x
can cause some other object y
to be modified, then x = y is likely not in the domain of =
. This typically happens if the right operand is owned directly or indirectly by the left operand (e.g., with smart pointers to nodes in a node-based data structure, or with something like
std::vector
<
std::any
>