Join elements with delimiter (tools/join.hpp)

template <std::ranges::range R, typename T>
std::string join(R e, T d);

Given $n$ elements $e_0, e_1, \cdots, e_{n - 1}$, it returns $e_0 + d + e_1 + d + \cdots + d + e_{n - 1}$ where $+$ is string concatenation. Note that it returns an empty string if $n = 0$.

Constraints

operator<<(std::ostream, std::ranges::range_value_t<R>) is defined.
operator<<(std::ostream, T) is defined.

Time Complexity

$O\left( \sum_{i = 0}^{n - 1} |e_i| + n |d| \right)$

License

Author

anqooqie

Verified with

Code

#ifndef TOOLS_JOIN_HPP
#define TOOLS_JOIN_HPP

#include <ranges>
#include <sstream>

namespace tools {
  template <::std::ranges::range R, typename T>
  ::std::string join(R&& e, const T& d) {
    ::std::ostringstream ss;
    auto it = ::std::ranges::begin(e);
    const auto end = ::std::ranges::end(e);
    if (it != end) {
      ss << *it;
      for (++it; it != end; ++it) {
        ss << d << *it;
      }
    }
    return ss.str();
  }
}

#endif

#line 1 "tools/join.hpp"



#include <ranges>
#include <sstream>

namespace tools {
  template <::std::ranges::range R, typename T>
  ::std::string join(R&& e, const T& d) {
    ::std::ostringstream ss;
    auto it = ::std::ranges::begin(e);
    const auto end = ::std::ranges::end(e);
    if (it != end) {
      ss << *it;
      for (++it; it != end; ++it) {
        ss << d << *it;
      }
    }
    return ss.str();
  }
}