Why doesn't printf format unicode parameters?

Question

When using printf to format a double-byte string into a single-byte string:

printf(\"%ls\\n\", L\"s:\\\\яшертыHello\");   // %ls for a wide string (%s varies         


        

Answer 1

I expect your code to work -- and it works here on Linux -- but it is locale dependent. That means you have to set up the locale and your locale must support the character set used. Here is my test program:

#include <locale.h>
#include <stdio.h>

int main()
{
    int c;
    char* l = setlocale(LC_ALL, "");
    if (l == NULL) {
        printf("Locale not set\n");
    } else {
        printf("Locale set to %s\n", l);
    }
    printf("%ls\n", L"s:\\яшертыHello");
    return 0;
}

and here is an execution trace:

$ env LC_ALL=en_US.utf8 ./a.out
Locale set to en_US.utf8
s:\яшертыHello

If it says that the locale isn't set or is set to "C", it is normal that you don't get the result you expect.

Edit: see the answers to this question for the equivalent of en_US.utf8 for Windows.

Answer 2

In C++ I usually use std::stringstream to create formatted text. I also implemented an own operator to use Windows function to make the encoding:

ostream & operator << ( ostream &os, const wchar_t * str )
{
  if ( ( str == 0 ) || ( str[0] == L'\0' ) )
   return os;
  int new_size = WideCharToMultiByte( CP_UTF8, 0, str, -1, NULL, NULL, NULL, NULL );
  if ( new_size <= 0 )
    return os;
  std::vector<char> buffer(new_size);
  if ( WideCharToMultiByte( CP_UTF8, 0, str, -1, &buffer[0], new_size, NULL, NULL ) > 0 )
    os << &buffer[0];
  return os;
}

This code convert to UTF-8. For other possibilities check: WideCharToMultiByte.