Question

我有一个像"3bdb098f-b8b0-4d1b-baa2-0d93eb7169c4"这样的iBeacon UUID，我需要它看起来非常像：

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，但是需要一种方法来转换这个＆＃34;手工＆＃34;也会很酷（加上arduino代码不会每次都处理转换）

Answer 1

一次红色两个字符，将它们转换为与十六进制值对应的数字。做一个循环。遇到'-'字符时跳过该字符。将数组中的“current”元素设置为值。将“current”设置为数组中的下一个元素。

像这样的伪代码

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

这应该适用于您的问题

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替代方法，您只需为字符'0'和0 - 9（或简称'a' - 10）减去'W'字符{ {1}}预先进行转化。例如：

a - f

示例输出

#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;
}

（注意>您可以在尝试转换其他验证之前添加进一步检查，$ ./bin/uuid_to_uint8_t original: 3bdb098f-b8b0-4d1b-baa2-0d93eb7169c4 unsigned: 3bdb098fb8b04d1bbaa20d93eb7169c4是有效的十六进制字符或uuid[ui] ）。

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

3 个答案: