How to print wstring on Linux/OS X?

Question

How can I print a string like this: €áa¢cée£ on the console/screen? I tried this:

#include <iostream>    
#include <string>
using namespace std;

wstring wStr = L"€áa¢cée£";

int main (void)
{
    wcout << wStr << " : " << wStr.length() << endl;
    return 0;
}

which is not working. Even confusing, if I remove € from the string, the print out comes like this: ?a?c?e? : 7 but with € in the string, nothing gets printed after the € character.

If I write the same code in python:

#!/usr/bin/env python
# -*- coding: utf-8 -*-

wStr = u"€áa¢cée£"
print u"%s" % wStr

it prints out the string correctly on the very same console. What am I missing in c++ (well, I'm just a noob)? Cheers!!

---

Update 1: based on n.m.'s suggestion

#include <iostream>
#include <string>
using namespace std;

string wStr = "€áa¢cée£";
char *pStr = 0;

int main (void)
{
    cout << wStr << " : " << wStr.length() << endl;

    pStr = &wStr[0];
    for (unsigned int i = 0; i < wStr.length(); i++) {
        cout << "char "<< i+1 << " # " << *pStr << " => " << pStr << endl;
        pStr++;
    }
    return 0;
}

First of all, it reports 14 as the length of the string: €áa¢cée£ : 14 Is it because it's counting 2 byte per character?

And all I get this:

char 1 # ? => €áa¢cée£
char 2 # ? => ??áa¢cée£
char 3 # ? => ?áa¢cée£
char 4 # ? => áa¢cée£
char 5 # ? => ?a¢cée£
char 6 # a => a¢cée£
char 7 # ? => ¢cée£
char 8 # ? => ?cée£
char 9 # c => cée£
char 10 # ? => ée£
char 11 # ? => ?e£
char 12 # e => e£
char 13 # ? => £
char 14 # ? => ?

as the last cout output. So, actual problem still remains, I believe. Cheers!!

---

Update 2: based on n.m.'s second suggestion

#include <iostream>
#include <string>

using namespace std;

wchar_t wStr[] = L"€áa¢cée£";
int iStr = sizeof(wStr) / sizeof(wStr[0]);        // length of the string
wchar_t *pStr = 0;

int main (void)
{
    setlocale (LC_ALL,"");
    wcout << wStr << " : " << iStr << endl;

    pStr = &wStr[0];
    for (int i = 0; i < iStr; i++) {
       wcout << *pStr << " => " <<  static_cast<void*>(pStr) << " => " << pStr << endl;
       pStr++;
    }
    return 0;
}

And this is what I get as my result:

€áa¢cée£ : 9
€ => 0x1000010e8 => €áa¢cée£
á => 0x1000010ec => áa¢cée£
a => 0x1000010f0 => a¢cée£
¢ => 0x1000010f4 => ¢cée£
c => 0x1000010f8 => cée£
é => 0x1000010fc => ée£
e => 0x100001100 => e£
£ => 0x100001104 => £
 => 0x100001108 => 

Why there it's reported as 9 than 8? Or this is what I should expect? Cheers!!

Answer 1

Drop the L before the string literal. Use std::string, not std::wstring.

UPD: There's a better (correct) solution. keep wchar_t, wstring and the L, and call setlocale(LC_ALL,"") in the beginning of your program.

You should call setlocale(LC_ALL,"") in the beginning of your program anyway. This instructs your program to work with your environment's locale, instead of the default "C" locale. Your environment has a UTF-8 one so everything should work.

Without calling setlocale(LC_ALL,""), the program works with UTF-8 sequences without "realizing" that they are UTF-8. If a correct UTF-8 sequence is printed on the terminal, it will be interpreted as UTF-8 and everything will look fine. That's what happens if you use string and char: gcc uses UTF-8 as a default encoding for strings, and the ostream happily prints them without applying any conversion. It thinks it has a sequence of ASCII characters.

But when you use wchar_t, everything breaks: gcc uses UTF-32, the correct re-encoding is not applied (because the locale is "C") and the output is garbage.

When you call setlocale(LC_ALL,"") the program knows it should recode UTF-32 to UTF-8, and everything is fine and dandy again.

This all assumes that we only ever want to work with UTF-8. Using arbitrary locales and encodings is beyond the scope of this answer.