经过多次搜索，我找到了满足我需要的完美功能here
以下是代码：
/* UTF-8 to ISO-8859-1/ISO-8859-15 mapper.
* Return 0..255 for valid ISO-8859-15 code points, 256 otherwise.
*/
static inline unsigned int to_latin9(const unsigned int code)
{
//printf("\ncode = %d", code);
/* Code points 0 to U+00FF are the same in both. */
if (code < 256U) {
return code;
}
switch (code) {
case 0x0152U: return 188U; /* U+0152 = 0xBC: OE ligature */
case 0x0153U: return 189U; /* U+0153 = 0xBD: oe ligature */
case 0x0160U: return 166U; /* U+0160 = 0xA6: S with caron */
case 0x0161U: return 168U; /* U+0161 = 0xA8: s with caron */
case 0x0178U: return 190U; /* U+0178 = 0xBE: Y with diaresis */
case 0x017DU: return 180U; /* U+017D = 0xB4: Z with caron */
case 0x017EU: return 184U; /* U+017E = 0xB8: z with caron */
case 0x20ACU: return 164U; /* U+20AC = 0xA4: Euro */
default: return 256U;
}
}
/* Convert an UTF-8 string to ISO-8859-15.
* All invalid sequences are ignored.
* Note: output == input is allowed,
* but input < output < input + length
* is not.
* Output has to have room for (length+1) chars, including the trailing NUL byte.
*/
size_t utf8_to_latin9(char *const output, const char *const input, const size_t length)
{
unsigned char *out = (unsigned char *)output;
const unsigned char *in = (const unsigned char *)input;
const unsigned char *const end = (const unsigned char *)input + length;
unsigned int c;
while (in < end)
if (*in < 128)
*(out++) = *(in++); /* Valid codepoint */
else
if (*in < 192)
in++; /* 10000000 .. 10111111 are invalid */
else
if (*in < 224) { /* 110xxxxx 10xxxxxx */
if (in + 1 >= end)
break;
if ((in[1] & 192U) == 128U) {
c = to_latin9( (((unsigned int)(in[0] & 0x1FU)) << 6U)
| ((unsigned int)(in[1] & 0x3FU)) );
if (c < 256)
*(out++) = c;
}
in += 2;
} else
if (*in < 240) { /* 1110xxxx 10xxxxxx 10xxxxxx */
if (in + 2 >= end)
break;
if ((in[1] & 192U) == 128U &&
(in[2] & 192U) == 128U) {
c = to_latin9( (((unsigned int)(in[0] & 0x0FU)) << 12U)
| (((unsigned int)(in[1] & 0x3FU)) << 6U)
| ((unsigned int)(in[2] & 0x3FU)) );
if (c < 256)
*(out++) = c;
}
in += 3;
} else
if (*in < 248) { /* 11110xxx 10xxxxxx 10xxxxxx 10xxxxxx */
if (in + 3 >= end)
break;
if ((in[1] & 192U) == 128U &&
(in[2] & 192U) == 128U &&
(in[3] & 192U) == 128U) {
c = to_latin9( (((unsigned int)(in[0] & 0x07U)) << 18U)
| (((unsigned int)(in[1] & 0x3FU)) << 12U)
| (((unsigned int)(in[2] & 0x3FU)) << 6U)
| ((unsigned int)(in[3] & 0x3FU)) );
if (c < 256)
*(out++) = c;
}
in += 4;
} else
if (*in < 252) { /* 111110xx 10xxxxxx 10xxxxxx 10xxxxxx 10xxxxxx */
if (in + 4 >= end)
break;
if ((in[1] & 192U) == 128U &&
(in[2] & 192U) == 128U &&
(in[3] & 192U) == 128U &&
(in[4] & 192U) == 128U) {
c = to_latin9( (((unsigned int)(in[0] & 0x03U)) << 24U)
| (((unsigned int)(in[1] & 0x3FU)) << 18U)
| (((unsigned int)(in[2] & 0x3FU)) << 12U)
| (((unsigned int)(in[3] & 0x3FU)) << 6U)
| ((unsigned int)(in[4] & 0x3FU)) );
if (c < 256)
*(out++) = c;
}
in += 5;
} else
if (*in < 254) { /* 1111110x 10xxxxxx 10xxxxxx 10xxxxxx 10xxxxxx 10xxxxxx */
if (in + 5 >= end)
break;
if ((in[1] & 192U) == 128U &&
(in[2] & 192U) == 128U &&
(in[3] & 192U) == 128U &&
(in[4] & 192U) == 128U &&
(in[5] & 192U) == 128U) {
c = to_latin9( (((unsigned int)(in[0] & 0x01U)) << 30U)
| (((unsigned int)(in[1] & 0x3FU)) << 24U)
| (((unsigned int)(in[2] & 0x3FU)) << 18U)
| (((unsigned int)(in[3] & 0x3FU)) << 12U)
| (((unsigned int)(in[4] & 0x3FU)) << 6U)
| ((unsigned int)(in[5] & 0x3FU)) );
if (c < 256)
*(out++) = c;
}
in += 6;
} else
in++; /* 11111110 and 11111111 are invalid */
/* Terminate the output string. */
*out = '\0';
return (size_t)(out - (unsigned char *)output);
}
这项工作非常完美。
但是这个函数需要1 buffer_input并返回第二个buffer_output。
为了优化它，我只想使用一个buffer_input，函数返回buffer_input修改过。
我不知道是否可能。我自己没有足够的知识（我很难理解这段代码）。
很难根据我想要的内容编辑此代码？
目前我这样使用：
utf8_to_latin9(buffer_output, buffer_input, length)
编辑后我想使用这样：
utf8_to_latin9(buffer_input, buffer_input, length)
或
utf8_to_latin9(buffer_input, length)
答案 0 :(得分：1)
我已经不更正（或检查）了您的代码以解决任何问题，我只是将其修改为就地工作。请注意我对utf8_to_latin9()中错误转换的评论。
转换到就地操作非常简单：
切换到一个缓冲区参数，使out指向输入缓冲区的开头（而不是单独的输出缓冲区）：
< size_t utf8_to_latin9(char *const output, const char *const input, const size_t length)
< {
< unsigned char *out = (unsigned char *)output;
----
> size_t utf8_to_latin9(char *const input, const size_t length)
> {
> unsigned char *out = (unsigned char *)input;
在转换结束时，根据输入缓冲区（而不是单独的输出缓冲区）计算返回值：
< return (size_t)(out - (unsigned char *)output);
----
> return (size_t)(out - (unsigned char *)input);
由于UTF-8输入缓冲区无法以较慢速率消耗，而拉丁语输出缓冲区正在填充，因此非常安全。
我的测试main()：
int main()
{
char input[256] = {0};
/* ¤ *//* € *//* some kanji */
strcpy( input, "\xc2\xa4\xe2\x82\xac\xf0\xaf\xa4\xa9" );
utf8_to_latin9( input, 9 );
printf( "%hhx ", input[0] );
printf( "%hhx ", input[1] );
printf( "%hhx ", input[2] );
printf( "%hhx ", input[3] );
printf( "%hhx ", input[4] );
printf( "%hhx ", input[5] );
printf( "%hhx ", input[6] );
printf( "%hhx ", input[7] );
printf( "%hhx\n", input[8] );
return 0;
}
输出：
a4 a4 0 82 ac f0 af a4 a9
这就是我所期望的（展示我评论过的转换问题）。