有没有办法将UTF8转换为iso-8859-1？ [英] Is there a way to convert from UTF8 to iso-8859-1?
问题描述
我的软件在UTF8中得到一些字符串比我需要转换为ISO 8859 1.我知道UTF8域大于iso 8859.但UTF8中的数据以前已经从ISO升级转换，所以我不应该错过任何内容
My software is getting some strings in UTF8 than I need to convert to ISO 8859 1. I know that UTF8 domain is bigger than iso 8859. But the data in UTF8 has been previously upconverted from ISO, so I should not miss anything.
我想知道是否有一个简单/直接的方式从UTF8转换为iso-8859-1。
I would like to know if there is an easy / direct way to convert from UTF8 to iso-8859-1.
谢谢
推荐答案
这是一个您可能会发现有用的功能： utf8_to_latin9（） / code>。它转换为 ISO-8859-15 （包括欧元，其中 ISO-8859-1 没有），但是也可以正常运行 UTF-8 - > ISO-8859-1 转换部分 ISO-8859-1 - > UTF-8 - > ISO-8859-1 往返。
Here is a function you might find useful: utf8_to_latin9(). It converts to ISO-8859-15 (including EURO, which ISO-8859-1 does not have), but also works correctly for the UTF-8->ISO-8859-1 conversion part of a ISO-8859-1->UTF-8->ISO-8859-1 round-trip.
该函数忽略与iconv类似的 // IGNORE 标志的无效代码点，但是不重新分解UTF-8序列;也就是说，它不会将 U + 006E U + 0303 转换为 U + 00F1 。我不打扰重组，因为iconv也没有。
The function ignores invalid code points similar to //IGNORE flag for iconv, but does not recompose decomposed UTF-8 sequences; that is, it won't turn U+006E U+0303 into U+00F1. I don't bother recomposing because iconv does not either.
该函数非常小心字符串访问。它不会扫描超出缓冲区。输出缓冲区的长度必须长于一个字节，因为它总是附加字符串末尾的NUL字节。该函数返回输出中的字符数（字节），不包括字符串末尾NUL字节。
The function is very careful about the string access. It will never scan beyond the buffer. The output buffer must be one byte longer than length, because it always appends the end-of-string NUL byte. The function returns the number of characters (bytes) in output, not including the end-of-string NUL byte.
/* 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)
{
/* 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);
}
请注意，您可以在<$ c中添加特定代码点的自定义音译$ c> to_latin9（）函数，但您仅限于单字符替换。
Note that you can add custom transliteration for specific code points in the to_latin9() function, but you are limited to one-character replacements.
正如当前写的那样，该函数可以就地转换安全：输入和输出指针可以相同。输出字符串永远不会超过输入字符串。如果您的输入字符串有额外的字节空间（例如，它有NUL终止字符串），您可以安全地使用上述功能将其从UTF-8转换为ISO-8859-1 / 15。我以这种方式刻意写了，因为它应该在嵌入式环境中节省一些努力，尽管这种方法有点有限。
As it is currently written, the function can do in-place conversion safely: input and output pointers can be the same. The output string will never be longer than the input string. If your input string has room for an extra byte (for example, it has the NUL terminating the string), you can safely use the above function to convert it from UTF-8 to ISO-8859-1/15. I deliberately wrote it this way, because it should save you some effort in an embedded environment, although this approach is a bit limited wrt. customization and extension.
编辑：
我包含一对转换函数对这个答案的编辑对于拉丁语-1 / 9转换为UTF-8（ISO-8859-1或-15至UTF-8）;主要区别在于这些函数返回动态分配的副本，并保持原始字符串不变。
I included a pair of conversion functions in an edit to this answer for both Latin-1/9 to/from UTF-8 conversion (ISO-8859-1 or -15 to/from UTF-8); the main difference is that those functions return a dynamically allocated copy, and keep the original string intact.
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