std::ranges::copy_n, std::ranges::copy_n_result

Defined in header <algorithm>
Call signature
template< std::input_iterator I, std::weakly_incrementable O >
  requires std::indirectly_copyable<I, O>
  constexpr copy_n_result<I, O>
            copy_n( I first, std::iter_difference_t<I> n, O result );
(1) (since C++20)
Helper type
template< class I, class O >
  using copy_n_result = ranges::in_out_result<I, O>;
(2) (since C++20)
1) Copies exactly n values from the range beginning at first to the range beginning at result by performing *(result + i) = *(first + i) for each integer in [0, n). The behavior is undefined if result is within the range [first, first + n) (ranges::copy_backward might be used instead in this case).

The function-like entities described on this page are niebloids, that is:

In practice, they may be implemented as function objects, or with special compiler extensions.


first - the beginning of the range of elements to copy from
n - number of the elements to copy
result - the beginning of the destination range

Return value

ranges::copy_n_result{first + n, result + n} or more formally, a value of type ranges::in_out_result that contains an std::input_iterator iterator equals to ranges::next(first, n) and a std::weakly_incrementable iterator equals to ranges::next(result, n).


Exactly n assignments.


In practice, implementations of std::ranges::copy_n may avoid multiple assignments and use bulk copy functions such as std::memmove if the value type is TriviallyCopyable and the iterator types satisfy contiguous_iterator. Alternativelly, such copy acceleration can be injected during an optimization phase of a compiler.

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

Possible implementation

struct copy_n_fn {
  template<std::input_iterator I, std::weakly_incrementable O>
    requires std::indirectly_copyable<I, O>
    constexpr ranges::copy_n_result<I, O>
    operator()(I first, std::iter_difference_t<I> n, O result) const {
        for (std::iter_difference_t<I> i{}; i != n; ++i, ++first, ++result)
            *result = *first;
        return {std::move(first), std::move(result)};
inline constexpr copy_n_fn copy_n{};


#include <algorithm>
#include <iterator>
#include <iostream>
#include <iomanip>
#include <string>
#include <string_view>
int main()
    const std::string_view in {"ABCDEFGH"};
    std::string out;
    std::ranges::copy_n(in.begin(), 4, std::back_inserter(out));
    std::cout << std::quoted(out) << '\n';
    out = "abcdefgh";
    const auto res = std::ranges::copy_n(in.begin(), 5, out.begin());
        << "*(res.in): '" << *(res.in) << "', distance: "
        << std::distance(std::begin(in), res.in) << '\n'
        << "*(res.out): '" << *(res.out) << "', distance: "
        << std::distance(std::begin(out), res.out) << '\n';


*(res.in): 'F', distance: 5
*(res.out): 'f', distance: 5

See also

copies a range of elements to a new location
copies a range of elements in backwards order
copies a range of elements omitting those that satisfy specific criteria
copies a range, replacing elements satisfying specific criteria with another value
creates a copy of a range that is reversed
copies and rotate a range of elements
creates a copy of some range of elements that contains no consecutive duplicates
moves a range of elements to a new location
moves a range of elements to a new location in backwards order
copies a number of elements to a new location
(function template)

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