C++ named requirements: ValueSwappable (since C++11)
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C++ named requirements
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ValueSwappable
(C++11)
(C++11)
(C++11)
(C++17)
(C++11)
(C++11)
(C++11)
(C++11)
(C++11)
(C++11)
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(C++23)
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(C++11)
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(C++11)
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
-
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);.
[edit] Example
Run this code
#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
(C++20)
indirectly_readable types can be swapped (concept) [edit]