How can I take characters of a string with the data type of uint8_t Variable and use them as hexadecimal variables?

Question

i have an uint8_t Variable which contains a substring of 4 hexadecimal variables. Example:

uint8_t String[10] = "00AABBCC";

I would like to take these 4 hex Variables into different hex values:

uint8_t Data_Byte[4];
Data_Byte[0]=0x00;
Data_Byte[1]=0xAA;
Data_Byte[2]=0xBB;
Data_Byte[3]=0xCC;

How can I take these 4 substrings into 4 different uint8_t Variables?

Answer 1

You can use sscanf to parse each two-character pair in the string into a number:

uint8_t arr[strlen(String) / 2];
for (int i = 0; i < strlen(String); i += 2) {
    sscanf(String + i, "%2hhx", &arr[i / 2]);
}

If you're developing on a system with limited sscanf support, you can use something like this:

for (int i = 0; i < strlen(String); i += 2) {
    uint8_t val1 = isdigit(String[i]) ? (String[i] - '0') : (String[i] - 'A' + 10);
    uint8_t val2 = isdigit(String[i + 1]) ? (String[i + 1] - '0') : (String[i + 1] - 'A' + 10);

    arr[i / 2] = val1 << 4 | val2;
}

Answer 2

static char digits[] = "0123456789ABCDEF";

uint8_t *convert(uint8_t *chrs, uint8_t *buff)
{
    size_t len = strlen((char *)chrs);

    for(size_t i = 0; i < len; i++)
    {
        int is_first_digit = !(i & 1);
        int shift = is_first_digit << 2;
        buff[i / 2] += (strchr(digits, chrs[i]) - digits) << shift;
    }
    return buff;
}

int main(void)
{
    uint8_t String[] = "00AABBCC";
    uint8_t buff[4];

    convert(String, buff);
    for(size_t i = 0; i < sizeof(buff); i++)
    {
        printf("%hhx", buff[i]); // I know it is wrong format
    }
}

https://godbolt.org/z/9c8aexTvq

Or even faster solution:

int getDigit(uint8_t ch)
{
    switch(ch)
    {
        case '0':
        case '1':
        case '2':
        case '3':
        case '4':
        case '5':
        case '6':
        case '7':
        case '8':
        case '9':
            return ch - '0';
        case 'A':
        case 'B':
        case 'C':
        case 'D':
        case 'E':
        case 'F':
            return ch - 'A' + 10;
    }
    return 0;
}

uint8_t *convert(uint8_t *chrs, uint8_t *buff)
{
    size_t len = strlen((char *)chrs);
    
    for(size_t i = 0; i < len; i++)
    {
        int is_first_digit = !(i & 1);
        int shift = is_first_digit << 2;

        buff[i / 2] += (getDigit(chrs[i])) << shift;
    }
    return buff;
}

Remember: use functions for this kind of tasks. Do not program in main.

Answer 3

I took some inspiration from 0___________ and made my own:

static char digits[] = "0123456789ABCDEF";
void convert(uint8_t *chrs, uint8_t *buff)
{
    size_t len = strlen((char *)chrs);
    size_t i;

    for(i = 0; i < len; i+=2) {
        buff[i / 2] = (strchr(digits, chrs[i]) - digits);
        buff[i / 2] += (strchr(digits, chrs[i+1]) - digits) << 4;
    }

    if(i<len)
        buff[i / 2] = (strchr(digits, chrs[i]) - digits);
}

The changes are that I find it much more natural to do a complete element in every iteration. To account for odd length input strings, I just added an if statement in the end that takes care of it. This can be removed if input strings always have even length. And I skipped returning the buffer for simplicity. However, as 0___________ pointed out in comments, there are good reasons to return a pointer to the output buffer. Read about those reasons here: c++ memcpy return value

Answer 4

With your stipulation the strings will represent 4 bytes, this a far-easier-to-read-and-understand solution IMO. I have no comment on efficiency.

#include <stdio.h>
#include <stdbool.h>
#include <stdlib.h>
#include <stdint.h>
#include <inttypes.h>
#include <errno.h>
#include <limits.h>
#include <string.h>
#include <arpa/inet.h>

bool convert(const uint8_t* strValue, uint8_t* cvrtArray)
{
    // make 2nd parameter non-NULL for better error checking
    errno = 0;
    char* endptr = NULL;
    // convert to unsigned long
    unsigned long val = strtoul((const char*)strValue, &endptr, 16);
    // do some error checking, this probably needs some improvements
    if (errno == ERANGE && val == ULONG_MAX)
    {
        fprintf(stderr, "Overflow\n");

        return false;
    }
    else if ((strValue != NULL) && (*endptr != '\0'))
    {
        fprintf(stderr, "Cannot convert\n");
        return false;
    }

    // potential need to flip the bytes (your string is big endian, and the
    // test machine on godbolt is little endian)
    val = htonl(val);

    // copy to our array
    memcpy(cvrtArray, &val, 4);
    return true;
}

int main(void)
{
    uint8_t Data_Byte[4] = { 0 };
    uint8_t String[10] = "00AABBCC";

    if (convert(String, Data_Byte) == true)
    {
        for(size_t i=0; i<sizeof Data_Byte; i++)
        {
            printf("Data_Byte[%zu] = 0x%02" PRIX8 "\n", i, Data_Byte[i]);
        }
    }
    else
    {
        fprintf(stderr, "There was a problem converting %s to byte array\n", String);
    }

    return 0;
}

code in action