std::wostream_iterator

Question

Why there are no std::wostream_iterator in C++?

Is there any good reason for this?

#include <fstream>
#include <iterator>
#include <string>
#include <vector>

int main()
{
    std::vector<std::wstring> myvec = { L"first", L"second" };

    std::wofstream f("New.txt");
//  std::copy(myvec.begin(), myvec.end(), std::wostream_iterator<std::wstring>(f)); // Error
//  std::copy(myvec.begin(), myvec.end(), std::ostream_iterator<std::wstring>(f)); // Error
    std::copy(myvec.begin(), myvec.end(), std::ostream_iterator<std::wstring, wchar_t>(f)); // Ok
    std::copy(myvec.begin(), myvec.end(), std::ostream_iterator<std::wstring, wchar_t, std::char_traits<wchar_t>>(f)); // Ok
}

Because ostream_iterator does the job? – VoidStar Aug 26 '12 at 13:06 — VoidStar, Aug 26 '12 at 13:06
wondering why your 2nd `copy` function is error – Mr.Anubis Aug 26 '12 at 13:13 — Mr.Anubis, Aug 26 '12 at 13:13
@Mr.Anubis: Because by default char_type is char – awesoon Aug 26 '12 at 13:24 — awesoon, Aug 26 '12 at 13:24

ForEveR · Accepted Answer · 2012-08-26T13:43:30.893

Because std::ostream_iterator is not typedef (and std::wstring is typedef on std::basic_string with charT = wchar_t).

template <class T, class charT = char, class traits = char_traits<charT> >
class ostream_iterator:
public iterator<output_iterator_tag, void, void, void, void>

Second copy operation

std::copy(myvec.begin(), myvec.end(), std::ostream_iterator<std::wstring>(f));

is incorrect, since

typedef basic_ostream<charT,traits> ostream_type;

so, in this case ostream_type is basic_ostream<char, std::char_traits<char> >

ostream_iterator(ostream_type& s);

Constructor can receive only this type, but we pass basic_ofstream<wchar_t, std::char_traits<wchar_t> >, it's incorrect.

still wondering why that 2nd copy function is error since `T` and `charT` parameters can be different — Mr.Anubis, Aug 26 '12 at 13:26

std::wostream_iterator

1 Answers1