如何将字符串（字符数组）中的 UUID 转换为静态 uint8_t adv_data[31]

Question

我有一个像这样的 iBeacon UUID

"3bdb098f-b8b0-4d1b-baa2-0d93eb7169c4"

，我需要它看起来非常像这样：

static uint8_t adv_data[31] = { 0x02,0x01,0x06, 0x1A,0xFF,0x4C,0x00,0x02,0x15,0x71,0x3d,0x00,0x00,0x50,0x3e,0x4c,0x75,0xba,0x94,0x31,0x48,0xf1,0x8d,0x94,0x1e,0x00,0x00,0x00,0x00,0xC5 };

我需要一种方法来在代码中转换它，但是“手动”转换这个的一次性方法也很酷（而且arduino代码不必每次都处理转换）

Answer 1

一次读取两个字符，将其转换为十六进制值对应的数字。循环进行。遇到

'-'

字符时跳过它。将数组中的“当前”元素设置为该值。将“当前”设置为数组中的下一个元素。

类似这样的伪代码

while (not at end of string)
{
    char1 = get next character from string;
    char2 = get next character from string;
    value = make int from hex characters(char1, char2);
    array[current++] = value;
}

Answer 2

这应该可以解决你的问题

char *uuid = "3bdb098f-b8b0-4d1b-baa2-0d93eb7169c4";
static uint8_t adv_data[32];  //your uuid has 32 byte of data

// This algo will work for any size of uuid regardless where the hypen is placed
// However, you have to make sure that the destination have enough space.

int strCounter=0;      // need two counters: one for uuid string (size=38) and
int hexCounter=0;      // another one for destination adv_data (size=32)
while (i<strlen(uuid))
{
     if (uuid[strCounter] == '-') 
     {
         strCounter++;     //go to the next element
         continue;
     }

     // convert the character to string
     char str[2] = "\0";
     str[0] = uuid[strCounter];

     // convert string to int base 16
     adv_data[hexCounter]= (uint8_t)atoi(str,16);

     strCounter++;
     hexCounter++;
}

Answer 3

使用库函数转换为数字形式的另一种方法是，您可以简单地对字符

'0'

减去

0 - 9

，对字符

'a' - 10

减去

'W'

（或简单地

a - f

）来执行转换。例如：

#include <stdio.h>
#include <string.h>
#include <stdint.h>

#define MAXC 32

int main (int argc, char **argv) {

    char *uuid = argc > 1 ? argv[1] : "3bdb098f-b8b0-4d1b-baa2-0d93eb7169c4";
    uint8_t data[MAXC] = {0};

    size_t i, ui = 0, di = 0, ulen = strlen (uuid);

    for (;di < MAXC && ui < ulen; ui++, di++) {
        if (uuid[ui] == '-') ui++;    /* advance past any '-'  */
                                      /* convert to lower-case */
        if ('A' <= uuid[ui] && uuid[ui] <= 'Z') uuid[ui] |= (1 << 5);
        data[di] = ('0' <= uuid[ui] && uuid[ui] <= '9') ? uuid[ui] - '0' :
                    uuid[ui] - 'W';   /* convert to uint8_t */
    }

    if (di == MAXC && ui != ulen) {  /* validate all chars fit in data */
        fprintf (stderr, "error: uuid string exceeds storage.\n");
        return 1;
    }

    printf ("\n original: %s\n unsigned: ", uuid);
    for (i = 0; i < di; i++)
        putchar (data[i] + (data[i] < 10 ? '0' : 'W'));
    putchar ('\n');

    return 0;
}

示例输出

$ ./bin/uuid_to_uint8_t

 original: 3bdb098f-b8b0-4d1b-baa2-0d93eb7169c4
 unsigned: 3bdb098fb8b04d1bbaa20d93eb7169c4

（注意： 在尝试转换以进行额外验证之前，您可以添加进一步检查以确保

uuid[ui]

是有效的 十六进制 字符或

'-'

）。