How do I represent a Unicode character in a literal string ISO/ANSI C when the character set is ASCII?

Question

In Perl, I can say

my $s = "r\x{e9}sum\x{e9}";

to assign "résumé" to $s. I want to do something similar in C. Specifically, I want to say

sometype_that_can_hold_utf8 c = get_utf8_char();
if (c < '\x{e9}') {
    /* do something */
}

Answer 1

For UTF8, you have to generate the encoding yourself using rules found, for example, here. For example, the German sharp s (ß, code point 0xdf), has the UTF8 encoding 0xc3,0x9f. Your e-acute (é, code point 0xe9) has a UTF8 encoding of 0xc3,0xa9.

And you can put arbitrary hex characters in your strings with:

char *cv = "r\xc3\xa9sum\xc3\xa9";
char *sharpS = "\xc3\x9f";

Answer 2

If you have a C99 compiler you can use <wchar.h> (and <locale.h>) and enter the Unicode code points directly in the source.

$ cat wc.c

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

int main(void) {
  const wchar_t *name = L"r\u00e9sum\u00e9";
  setlocale(LC_CTYPE, "en_US.UTF-8");
  wprintf(L"name is %ls\n", name);
  return 0;
}

$ /usr/bin/gcc -std=c99 -pedantic -Wall wc.c

$ ./a.out

name is résumé

Answer 3

wchar_t is the type you are looking for: http://opengroup.org/onlinepubs/007908799/xsh/wchar.h.html