Question

我的软件在UTF8中得到的字符串比转换为ISO 8859要好。我知道UTF8域大于iso 8859.但是UTF8中的数据以前是从ISO上转换过来的，所以我不应该错过任何东西

我想知道是否有一种简单/直接的方法可以将UTF8转换为iso-8859-1。

由于

Answer 1

iconv - 执行字符集转换

size_t iconv(iconv_t cd, char **inbuf, size_t *inbytesleft, char **outbuf, size_t *outbytesleft);

iconv_t iconv_open(const char *tocode, const char *fromcode);

tocode为"ISO_8859-1"，fromcode为"UTF-8"。

工作示例：

#include <iconv.h>
#include <stdio.h>

int main (void) {
    iconv_t cd = iconv_open("ISO_8859-1", "UTF-8");
    if (cd == (iconv_t) -1) {
        perror("iconv_open failed!");
        return 1;
    }

    char input[] = "Test äöü";
    char *in_buf = &input[0];
    size_t in_left = sizeof(input) - 1;

    char output[32];
    char *out_buf = &output[0];
    size_t out_left = sizeof(output) - 1;

    do {
        if (iconv(cd, &in_buf, &in_left, &out_buf, &out_left) == (size_t) -1) {
            perror("iconv failed!");
            return 1;
        }
    } while (in_left > 0 && out_left > 0);
    *out_buf = 0;

    iconv_close(cd);

    printf("%s -> %s\n", input, output);
    return 0;
}

Answer 2

以下是您可能会觉得有用的功能：utf8_to_latin9()。它转换为ISO-8859-15（包括ISO-8859-1没有的{欧元}），但也适用于UTF-8的{{1}} - ＆gt; ISO-8859-1转换部分} - ＆gt; ISO-8859-1 - ＆gt; UTF-8往返。

该函数忽略了与iconv类似的ISO-8859-1标志的无效代码点，但不重新组合分解的UTF-8序列;也就是说，它不会将//IGNORE转变为U+006E U+0303。我不打算重新组合，因为iconv也没有。

该函数对字符串访问非常谨慎。它永远不会超出缓冲区。输出缓冲区必须比长度长一个字节，因为它总是附加字符串结尾的NUL字节。该函数返回输出中的字符数（字节），不包括字符串结尾的NUL字节。

U+00F1

请注意，您可以在/* 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); }功能中为特定代码点添加自定义音译，但您只能使用单字符替换。

正如当前编写的那样，该函数可以安全地进行就地转换：输入和输出指针可以是相同的。输出字符串永远不会长于输入字符串。如果您的输入字符串有额外字节的空间（例如，它有NUL终止字符串），您可以安全地使用上述函数将其从UTF-8转换为ISO-8859-1 / 15。我故意用这种方式写它，因为它应该在嵌入式环境中省去一些努力，尽管这种方法有点受限。定制和扩展。

编辑：

我为Latin-1/9和/或UTF-8转换（ISO-8859-1或-15到UTF-8）提供了一对转换函数in an edit to this answer;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。

Answer 3

以下示例也使用iconv库。即使您的文件（或输入流）包含UTF-8和ISO-8859-1混合字符（例如，如果您有UTF-8文件并在使用ISO的环境中对其进行编辑），它也可以工作-8859-1）。

int Utf8ToLatin1(char* input, char* output, size_t size)
{
    size_t in_left = size;
    size_t out_left = size;

    char *in_buf    = input;
    char *out_buf   = output;

    iconv_t cd = iconv_open("ISO_8859-1", "UTF-8");
    if (cd == (iconv_t)-1) {
        (void) fprintf(stderr, "iconv_open() failed, msg encoding will be kept!");
        strncpy(output, input, size);
        return -1;
    }

    do {
        if (iconv(cd, &in_buf, &in_left, &out_buf, &out_left) == (size_t) -1) {
            if (errno == EILSEQ) {
                /* Input conversion stopped due to an input byte that
                 * does not belong to the input codeset.
                 */
                printf("Input conversion stopped due to an input byte that does not belong to the input codeset.\n");
                *out_buf= *in_buf;
                out_buf++   ;out_left--;
                in_buf++    ;in_left--;

            } else if (errno == E2BIG) {
                /* Input conversion stopped due to lack of space in
                 * the output buffer.
                 */
                printf("Input conversion stopped due to lack of space in the output buffer.\n");
                perror("iconv failed!, propably the encoding is already Latin, msg encoding will be kept!\n");
                strncpy(output, input, size);
                return -1;
            } else if (errno == EINVAL) {
                /* Input conversion stopped due to an incomplete
                 * character or shift sequence at the end of the
                 * input buffer.
                 */
                printf("Input conversion stopped due to an incomplete character or shift sequence at the end of the input buffer.\n");
                *out_buf= *in_buf;
                out_buf++   ;out_left--;
                in_buf++    ;in_left--;
            }

        }
    } while (in_left > 0 && out_left > 0);
    *out_buf = 0;

    iconv_close(cd);

    printf("*********************************************************\n");
    printf("ISO-8859-1:\n %s\n", input, output);
    return 0;

}

有没有办法从UTF8转换为iso-8859-1？

3 个答案: