Category: utilities | Component type: concept |
X | A type that is a model of Assignable |
x, y | Object of type X |
Name | Expression | Type requirements | Return type |
---|---|---|---|
Copy constructor | X(x) | X | |
Copy constructor |
X x(y); X x = y; |
||
Assignment | x = y [1] | X& | |
Swap | swap(x,y) | void |
Name | Expression | Precondition | Semantics | Postcondition |
---|---|---|---|---|
Copy constructor | X(x) | X(x) is a copy of x [2] | ||
Copy constructor | X(x) | X(x) is a copy of x [2] | ||
Copy constructor |
X x(y); X x = y; |
x is a copy of y [2] | ||
Assignment | x = y [1] | x is a copy of y [2] | ||
Swap | swap(x,y) |
Equivalent to
{ X tmp = x; x = y; y = tmp; } |
[1] One implication of this requirement is that a const type is not Assignable. For example, const int is not Assignable: if x is declared to be of type const int, then x = 7 is illegal. Similarly, the type pair<const int, int> is not Assignable.
[2] The reason this says "x is a copy of y", rather than "x == y", is that operator== is not necessarily defined: equality is not a requirement of Assignable. If the type X is EqualityComparable as well as Assignable, then a copy of x should compare equal to x.