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C++ named requirements: ValueSwappable (since C++11)

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Two objects of this type can be dereferenced and the resulting values can be swapped using unqualified function call swap() in the context where both std::swap and the user-defined swap()s are visible.

[edit] Requirements

A type T is ValueSwappable if

  1. T satisfies the LegacyIterator requirements.
  2. For any dereferenceable object x of type T (that is, any value other than the end iterator), *x satisfies the Swappable requirements.

Many standard library functions expect their arguments to satisfy ValueSwappable, which means that any time the standard library performs a swap, it uses the equivalent of using std::swap; swap(*iter1, *iter2);.

[edit] Example

#include <iostream>
#include <vector>
 
class IntVector
{
    std::vector<int> v;
//  IntVector& operator=(IntVector); // not assignable (C++98 way)
public:
    IntVector& operator=(IntVector) = delete; // not assignable
    void swap(IntVector& other)
    {
        v.swap(other.v);
    }
};
 
void swap(IntVector& v1, IntVector& v2)
{
    v1.swap(v2);
}
 
int main()
{
    IntVector v1, v2;    // IntVector is Swappable, but not MoveAssignable
    IntVector* p1 = &v1;
    IntVector* p2 = &v2; // IntVector* is ValueSwappable
    std::iter_swap(p1, p2); // OK: iter_swap requires ValueSwappable
//  std::swap(v1, v2); // compiler error! std::swap requires MoveAssignable
}

[edit] See also

specifies that the values referenced by two indirectly_readable types can be swapped
(concept) [edit]