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Концепции C++: Swappable

Материал из cppreference.com
< cpp‎ | concept


Any lvalue or rvalue of this type can be swapped with any lvalue or rvalue of the some other type, using unqualified function call swap() in the context where both std::swap and the user-defined swap()s are visible.

[править] Требования

Type U is swappable with type T if, for any object u of type U and any object t of type T,

Expression Requirements Semantics
#include <utility>

using std::swap;
swap(u, t);

After the call, the value of t is the value held by u before the call, and the value of u is the value held by t before the call. Calls the function named swap() found by overload resolution among all functions with that name that are found by argument-dependent lookup and the two std::swap templates defined in the header <utility>.
#include <utility>

using std::swap;
swap(t, u);

same same

Many standard library functions (for example, many algorithms) expect their arguments to satisfy Swappable, which means that any time the standard library performs a swap, it uses the equivalent of using std::swap; swap(t, u);.

[править] Заметки

It is unspecified if <utility> is actually included when the standard library functions perform the swap, so the user-provided swap() should not expect it to be included.

[править] Пример

#include <iostream>
#include <vector>
class IntVector {
    std::vector<int> v;
    IntVector& operator=(IntVector); // not assignable
    void swap(IntVector& other) {
void swap(IntVector& v1, IntVector& v2) {
int main()
    IntVector v1, v2;
//  std::swap(v1, v2); // compiler error! std::swap requires MoveAssignable
    std::iter_swap(&v1, &v2); // OK: library calls unqualified swap()