How to initialize a wchar_t variable?

Question

I am reading the book: C: In a Nutshell, and after reading the section Character Sets, which talks about wide characters, I wrote this program:

#include <stdio.h>
#include <stddef.h>
#include <wchar.h>

int main() {
  wchar_t wc = '\x3b1';
  wprintf(L"%lc\n", wc);
  return 0;
}

I then compiled it using gcc, but gcc gave me this warning:

main.c:7:15: warning: hex escape sequence out of range [enabled by default]

And the program does not output the character α (whose unicode is U+03B1), which is what I wanted it to do.

How do I change the program to print the character α?

Answer 1

This works for me

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

int main(void) {
  wchar_t wc = L'\x3b1';

  setlocale(LC_ALL, "en_US.UTF-8");
  wprintf(L"%lc\n", wc);
  return 0;
}

Answer 2

wchar_t wc = L'\x3b1';

is the correct way to initialise a wchar_t variable to U+03B1. The L prefix is used to specify a wchar_t literal. Your code defines a char literal and that's why the compiler is warning.

The fact that you don't see the desired character when printing is down to your local environment's console settings.

Answer 3

try L'\x03B1' It might just solve your problem. IF you're in doubt you can try :

'\u03b1' to initialize.