Question

我需要将包含十六进制值的字符串转换为字节数组。虽然第一个答案已经回答already here，但我收到以下错误：

warning: ISO C90 does not support the ‘hh’ gnu_scanf length modifier [-Wformat]

由于我不喜欢警告，而hh的遗漏只会产生另一个警告

warning: format ‘%x’ expects argument of type ‘unsigned int *’, but argument 3 has type ‘unsigned char *’ [-Wformat]

我的问题是：如何做到这一点？为了完成，我再次在这里发布示例代码：

#include <stdio.h>

int main(int argc, char **argv)
{
    const char hexstring[] = "deadbeef10203040b00b1e50", *pos = hexstring;
    unsigned char val[12];
    size_t count = 0;

     /* WARNING: no sanitization or error-checking whatsoever */
    for(count = 0; count < sizeof(val)/sizeof(val[0]); count++) {
        sscanf(pos, "%2hhx", &val[count]);
        pos += 2 * sizeof(char);
    }

    printf("0x");
    for(count = 0; count < sizeof(val)/sizeof(val[0]); count++)
        printf("%02x", val[count]);
    printf("\n");

    return(0);
}

Answer 1

您可以改为使用strtol()。

只需更换此行：

sscanf(pos, "%2hhx", &val[count]);

使用：

char buf[10];
sprintf(buf, "0x%c%c", pos[0], pos[1]);
val[count] = strtol(buf, NULL, 0);

更新：您可以使用此代码段避免使用sprintf()：

char buf[5] = {"0", "x", pos[0], pos[1], 0};
val[count] = strtol(buf, NULL, 0);

Answer 2

您可以将编译器切换到C99模式（hh长度修饰符在C99中标准化），或者您可以使用unsigned int临时变量：

unsigned int byteval;
if (sscanf(pos, "%2x", &byteval) != 1)
{
    /* format error */
}
val[count] = byteval;

Answer 3

为什么不在不使用sscanf，strol等的情况下这样做。下面是HexToBin和免费蜜蜂BinToHex。（注意最初通过错误记录系统返回枚举错误代码而不是简单的-1返回。）

unsigned char HexChar (char c)
{
    if ('0' <= c && c <= '9') return (unsigned char)(c - '0');
    if ('A' <= c && c <= 'F') return (unsigned char)(c - 'A' + 10);
    if ('a' <= c && c <= 'f') return (unsigned char)(c - 'a' + 10);
    return 0xFF;
}

int HexToBin (const char* s, unsigned char * buff, int length)
{
    int result;
    if (!s || !buff || length <= 0) return -1;

    for (result = 0; *s; ++result)
    {
        unsigned char msn = HexChar(*s++);
        if (msn == 0xFF) return -1;
        unsigned char lsn = HexChar(*s++);
        if (lsn == 0xFF) return -1;
        unsigned char bin = (msn << 4) + lsn;

        if (length-- <= 0) return -1;
        *buff++ = bin;
    }
    return result;
}

void BinToHex (const unsigned char * buff, int length, char * output, int outLength)
{
    char binHex[] = "0123456789ABCDEF";

    if (!output || outLength < 4) return;
    *output = '\0';

    if (!buff || length <= 0 || outLength <= 2 * length)
    {
        memcpy(output, "ERR", 4);
        return;
    }

    for (; length > 0; --length, outLength -= 2)
    {
        unsigned char byte = *buff++;

        *output++ = binHex[(byte >> 4) & 0x0F];
        *output++ = binHex[byte & 0x0F];
    }
    if (outLength-- <= 0) return;
    *output++ = '\0';
}

Answer 4

使用mvp的建议更改，我创建了此函数，其中包括错误检查（无效字符和不均匀长度）。

此函数将转换十六进制字符串 - 不以“0x”为前缀 - 使用偶数个字符转换为指定的字节数。如果遇到无效字符，或者十六进制字符串的长度为奇数，则返回-1，成功时返回0。

//convert hexstring to len bytes of data
//returns 0 on success, -1 on error
//data is a buffer of at least len bytes
//hexstring is upper or lower case hexadecimal, NOT prepended with "0x"
int hex2data(unsigned char *data, const unsigned char *hexstring, unsigned int len)
{
    unsigned const char *pos = hexstring;
    char *endptr;
    size_t count = 0;

    if ((hexstring[0] == '\0') || (strlen(hexstring) % 2)) {
        //hexstring contains no data
        //or hexstring has an odd length
        return -1;
    }

    for(count = 0; count < len; count++) {
        char buf[5] = {'0', 'x', pos[0], pos[1], 0};
        data[count] = strtol(buf, &endptr, 0);
        pos += 2 * sizeof(char);

        if (endptr[0] != '\0') {
            //non-hexadecimal character encountered
            return -1;
        }
    }

    return 0;
}

如何在C中正确地将十六进制字符串转换为字节数组？

4 个答案: