Sample 1 :

#include <stdio.h>
#include <stdlib.h>

int     main(void)
{
    wchar_t     *s;

    s = (wchar_t *)malloc(sizeof(wchar_t) * 2);
    s[0] = 42;
    s[1] = 0;
    printf("%ls\n", s);
    free(s);
    return (0);
}

output :

Everything is fine here : my character (*) is correctly displayed.

Sample 2 :

I wanted to display an other kind of character. On my system, wchar_t seem encoded on 4 bytes. So I tried to display the following character : É

#include <stdio.h>
#include <stdlib.h>

int     main(void)
{
    wchar_t     *s;

    s = (wchar_t *)malloc(sizeof(wchar_t) * 2);
    s[0] = 0xC389;
    s[1] = 0;
    printf("%ls\n", s);
    free(s);
    return (0);
}

But there is no output this time, I tried with many values from the "encoding" section (cf. previous link) for s[0] (0xC389, 201, 0xC9)... But I never get the É character displayed. I also tried with %S instead of %ls.

If I try to call printf like this : printf("<%ls>\n", s) the only character printed is '<', the display is truncated.

Why do I have this problem? How should I do?

回答1:

Why do I have this problem?

Make sure you check errno and the return value of printf!

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

int main(void)
{
    wchar_t *s;
    s = (wchar_t *) malloc(sizeof(wchar_t) * 2);
    s[0] = 0xC389;
    s[1] = 0;

    if (printf("%ls\n", s) < 0) {
        perror("printf");
    }

    free(s);
    return (0);
}

See the output:

$ gcc test.c && ./a.out
printf: Invalid or incomplete multibyte or wide character

How to fix

First of all, the default locale of a C program is C (also known as POSIX) which is ASCII-only. You will need to add a call to setlocale, specifically setlocale(LC_ALL,"").

If your LC_ALL, LC_CTYPE or LANG environment variables are not set to allow UTF-8 when blank, you'll have to explicitly select a locale. setlocale(LC_ALL, "C.UTF-8") works on most systems - C is standard, and the UTF-8 subset of C is generally implemented.

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

int main(void)
{
    wchar_t *s;
    s = (wchar_t *) malloc(sizeof(wchar_t) * 2);
    s[0] = 0xC389;
    s[1] = 0;

    setlocale(LC_ALL, "");

    if (printf("%ls\n", s) < 0) {
        perror("printf");
    }

    free(s);
    return (0);
}

See the output:

$ gcc test.c && ./a.out
쎉

The reason why the incorrect character printed out is because wchar_t represents a wide character (such as UTF-32), not a multibyte character (such as UTF-8). Note that wchar_t is always 32 bits wide in the GNU C Library, but the C standard doesn't require it to be. If you initialize the character using the UTF-32BE encoding (i.e. 0x000000C9), then it prints out correctly:

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

int main(void)
{
    wchar_t *s;
    s = (wchar_t *) malloc(sizeof(wchar_t) * 2);
    s[0] = 0xC9;
    s[1] = 0;

    setlocale(LC_ALL, "");

    if (printf("%ls\n", s) < 0) {
        perror("printf");
    }

    free(s);
    return (0);
}

Output:

$ gcc test.c && ./a.out
É

Note that you can also set the LC (locale) environment variables via command line:

$ LC_ALL=C.UTF-8
$ ./a.out
É

回答2:

One problem is that you are trying to encode UTF-8, which is a single-byte encoding scheme, as a multi-byte encoding. For UTF-8 you use plain char.

Also note that because you try to combine the UTF-8 sequence into a multi-byte type, you have endianness (byte-order) issues (in memory 0xC389 might be stored as 0x89 and 0xC3, in that order). And that the compiler will sign-extend your number as well (if sizeof(wchar_t) == 4 and you look at s[0] in a debugger it might be 0xFFFFC389).

Another problem is the terminal or console you use to print. Maybe it simply doesn't support UTF-8 or the other encodings you tried?

来源：https://stackoverflow.com/questions/40590207/displaying-wide-chars-with-printf

标签

encoding

printf

widechar

Displaying wide chars with printf