C++ named requirements: ValueSwappable
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.
Requirements
A type T is ValueSwappable if.
-
Tsatisfies the LegacyIterator requirements - For any dereferenceable object
xof typeT(that is, any value other than the end iterator),*xsatisfies 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);.
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
}
See also
|
(C++20)
|
specifies that the values referenced by two indirectly_readable types can be swapped (concept) |
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https://en.cppreference.com/w/cpp/named_req/ValueSwappable
