将 wchar 打印到 Linux 控制台？

Question

我的 C 程序粘贴在下面。在bash中，程序打印“char is”，Ω 不打印。我的语言环境都是 en_US.utf8。

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

int main() {
   int r;
   wchar_t myChar1 = L'Ω';
   r = wprintf(L"char is %c\n", myChar1);
}

Answer 1

使用 {glib,libiconv,ICU} 将其转换为 UTF-8 后再输出。

Answer 2

这很有趣。显然编译器将欧米茄从 UTF-8 转换为 UNICODE，但不知何故 libc 把它搞砸了。

首先：%c-format 说明符需要 char（即使在 wprintf-版本中），因此您必须指定 %lc（因此对于字符串指定 %ls）。

其次，如果您像这样运行代码，则语言环境设置为C（它不会自动从环境中获取）。您必须使用空字符串调用 setlocale 才能从环境中获取语言环境，因此 libc 再次感到高兴。

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

int main() {
    int r;
    wchar_t myChar1 = L'Ω';
    setlocale(LC_CTYPE, "");
    r = wprintf(L"char is %lc (%x)\n", myChar1, myChar1);
}

Answer 3

除了建议修复 LIBC 的答案，您可以这样做：

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

// NOTE: *NOT* thread safe, not re-entrant
const char* unicode_to_utf8(wchar_t c)
{
    static unsigned char b_static[5];
    unsigned char* b = b_static; 

    if (c<(1<<7))// 7 bit Unicode encoded as plain ascii
    {
        *b++ = (unsigned char)(c);
    }
    else if (c<(1<<11))// 11 bit Unicode encoded in 2 UTF-8 bytes
    {
        *b++ = (unsigned char)((c>>6)|0xC0);
        *b++ = (unsigned char)((c&0x3F)|0x80);
    }
    else if (c<(1<<16))// 16 bit Unicode encoded in 3 UTF-8 bytes
        {
        *b++ = (unsigned char)(((c>>12))|0xE0);
        *b++ =  (unsigned char)(((c>>6)&0x3F)|0x80);
        *b++ =  (unsigned char)((c&0x3F)|0x80);
    }

    else if (c<(1<<21))// 21 bit Unicode encoded in 4 UTF-8 bytes
    {
        *b++ = (unsigned char)(((c>>18))|0xF0);
        *b++ = (unsigned char)(((c>>12)&0x3F)|0x80);
        *b++ = (unsigned char)(((c>>6)&0x3F)|0x80);
        *b++ = (unsigned char)((c&0x3F)|0x80);
    }
    *b = '\0';
    return b_static;
}


int main() {
    int r;
    wchar_t myChar1 = L'Ω';
    r = printf("char is %s\n", unicode_to_utf8(myChar1));
    return 0;
}