我的 C 程序粘贴在下面。在bash中,程序打印“char is”,不打印Ω。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);
}
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 说明符需要一个char
(即使在wprintf -version 中),因此您必须指定%lc
(因此%ls
对于字符串)。
其次,如果您像设置语言环境那样运行代码C
(它不会自动从环境中获取)。您必须setlocale
使用空字符串调用以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);
}
除了建议修复 LIBC 的答案之外,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;
}
输出前使用 {glib,libiconv,ICU} 将其转换为 UTF-8。