std::ranges::fill_n() algorithm

since C++20

Simplified

// (1)
constexpr O fill_n( O first, std::iter_difference_t<O> n, const T& value );

The type of arguments are generic and have the following constraints:

• T - (none)
• O - std::output_iterator<const T&>

Detailed

// (1)
template<
  class T,
  std::output_iterator<const T&> O
>
constexpr O fill_n( O first, std::iter_difference_t<O> n, const T& value );

Assigns the given value to all elements in the range [first; first + n).

The function-like entities described on this page are niebloids.

Parameters

first	The beginning of the range of elements to modify.
n	Number of elements to modify.
value	The value to assign.

Return value

An output iterator that compares equal to first + n.

Complexity

Exactly n assignments.

Exceptions

(none)

Possible implementation

ranges::fill_n

struct fill_n_fn
{
    template<class T, std::output_iterator<const T&> O>
    constexpr O operator()(O first, std::iter_difference_t<O> n, const T& value) const
    {
        for (std::iter_difference_t<O> i {}; i != n; ++first, ++i)
            *first = value;
        return first;
    }
};

inline constexpr fill_n_fn fill_n {};

Examples

Main.cpp

#include <algorithm>
#include <iostream>
#include <string>
#include <vector>

void println(const auto& v)
{
    for (const auto& elem : v)
        std::cout << ' ' << elem;
    std::cout << '\n';
}

int main()
{
    constexpr auto n {010};

    std::vector<std::string> v(n, "▓▓░░");
    println(v);

    std::ranges::fill_n(v.begin(), n, "░░▓▓");
    println(v);
}

Output

 ▓▓░░ ▓▓░░ ▓▓░░ ▓▓░░ ▓▓░░ ▓▓░░ ▓▓░░ ▓▓░░
 ░░▓▓ ░░▓▓ ░░▓▓ ░░▓▓ ░░▓▓ ░░▓▓ ░░▓▓ ░░▓▓