C++ cout auto separator

Question

I am wondering if there is way that std::cout automatically will insert some predefined value between printed sequences.

For example:

st         


        

Answer 1

If you're just looking for a way to print a vector with its elements properly delimited (and without an extra delimiter at the end), try this:

#include <iostream>
#include <vector>

int main()
{
  std::vector<int> v{1, 2, 3};

  auto it = v.begin();
  if (it != v.end())
  {
    std::cout << *it;
    ++it;
  }

  for (; it != v.end(); ++it)
  {
    std::cout << ", " << *it;
  }
}

Answer 2

How about using the ostream_iterator

int main()
{
     std::vector<int> data {2,3,33,45};
     std::copy(std::begin(data), std::end(data),
               std::ostream_iterator(std::cout, " "));
     std::cout << "\n";
}

Answer 3

A C++17 fold expression with the comma operator can create a nice one-liner:

[](auto &&...xs){ ((std::cout << xs << ',') , ...); }(2,3,33,45);

Answer 4

Well, I got beaten to it. I'll post this anyway.

Edit : well, after reading Nim's answer, mine does achieve the exact syntax OP wished for.

#include <iostream>
#include <algorithm>

struct with_separator {
    with_separator(std::string sep)
    : sep(std::move(sep)) {}

    std::string sep;
};

struct separated_stream {
    separated_stream(std::ostream &stream, std::string sep)
    : _stream(stream), _sep(std::move(sep)), _first(true) {}

    template <class Rhs>
    separated_stream &operator << (Rhs &&rhs) {
        if(_first)
            _first = false;
        else
            _stream << _sep;

        _stream << std::forward<Rhs>(rhs);
        return *this;
    }

    separated_stream &operator << (std::ostream &(*manip)(std::ostream&)) {
        manip(_stream);
        return *this;
    }

    private:
    std::ostream &_stream;
    std::string _sep;
    bool _first;
};

separated_stream operator << (std::ostream &stream, with_separator wsep) {
    return separated_stream(stream, std::move(wsep.sep));
}

int main()
{
    std::cout << with_separator(", ") << 1 << 2 << 3 << std::endl;
}

Output :

1, 2, 3

Answer 5

Simple answer is No, however, you can roll your own...

#include <iostream>
#include <sstream>

using namespace std;

struct set_some_separator{
    set_some_separator(const char* sep) : _sep(sep)
    { };

    template <typename T>
    set_some_separator& operator<<(const T& v)
    {
        _str << v << _sep;
        return *this;
    }

    friend
    ostream& operator<<(ostream& os, const set_some_separator& s)
    { return os << s._str.str(); }

    const char* _sep;
    ostringstream _str;
};

int main()
{
    cout << (set_some_separator(" ") << 2 << 3 << 33 << 45) << endl;
}

Okay the format of the cout is slightly different, hey-ho...

Answer 6

Not quite the same thing, but:

#include <array>
#include <iostream>
#include <iterator>

int main() {
    std::array<int, 3> data = { 1, 2, 3 };
    std::ostream_iterator<int> out(std::cout, " ");
    std::copy(data.begin(), data.end(), out);
    std::cout << '\n';
    return 0;
}