std::copy_backward - cppreference.com (original) (raw)

| | | | | ---------------------------------------------------------------------------------------------------------------------- | | ----------------------- | | template< class BidirIt1, class BidirIt2 >BidirIt2 copy_backward( BidirIt1 first, BidirIt1 last, BidirIt2 d_last ); | | (constexpr since C++20) |

Copies the elements from the range [first, last) to another range ending at d_last. The elements are copied in reverse order (the last element is copied first), but their relative order is preserved.

The behavior is undefined if d_last is within (first, last]. std::copy must be used instead of std::copy_backward in that case.

Contents

• 1 Parameters
• 2 Return value
• 3 Complexity
• 4 Notes
• 5 Possible implementation
• 6 Example
• 7 Defect reports
• 8 See also

[edit] Parameters

[edit] Return value

Iterator to the last element copied.

[edit] Complexity

Exactly std::distance(first, last) assignments.

[edit] Notes

When copying overlapping ranges, std::copy is appropriate when copying to the left (beginning of the destination range is outside the source range) while std::copy_backward is appropriate when copying to the right (end of the destination range is outside the source range).

[edit] Possible implementation

template<class BidirIt1, class BidirIt2> BidirIt2 copy_backward(BidirIt1 first, BidirIt1 last, BidirIt2 d_last) { while (first != last) *(--d_last) = *(--last); return d_last; }

[edit] Example

#include #include #include #include   int main() { std::vector source(4); std::iota(source.begin(), source.end(), 1); // fills with 1, 2, 3, 4   std::vector destination(6);   std::copy_backward(source.begin(), source.end(), destination.end());   std::cout << "destination contains: "; for (auto i: destination) std::cout << i << ' '; std::cout << '\n'; }

Output:

destination contains: 0 0 1 2 3 4

[edit] Defect reports

The following behavior-changing defect reports were applied retroactively to previously published C++ standards.

[edit] See also