Hexadecimal string to byte array in C

Question

Is there any standard C function that converts from hexadecimal string to byte array?
I do not want to write my own function.

Answer 1

As far as I know, there's no standard function to do so, but it's simple to achieve in the following manner:

#include <stdio.h>

int main(int argc, char **argv) {
    const char hexstring[] = "DEadbeef10203040b00b1e50", *pos = hexstring;
    unsigned char val[12];

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

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

    return 0;
}

---

Edit

As Al pointed out, in case of an odd number of hex digits in the string, you have to make sure you prefix it with a starting 0. For example, the string "f00f5" will be evaluated as {0xf0, 0x0f, 0x05} erroneously by the above example, instead of the proper {0x0f, 0x00, 0xf5}.

Amended the example a little bit to address the comment from @MassimoCallegari

Answer 2

I found this question by Googling for the same thing. I don't like the idea of calling sscanf() or strtol() since it feels like overkill. I wrote a quick function which does not validate that the text is indeed the hexadecimal presentation of a byte stream, but will handle odd number of hex digits:

uint8_t tallymarker_hextobin(const char * str, uint8_t * bytes, size_t blen)
{
   uint8_t  pos;
   uint8_t  idx0;
   uint8_t  idx1;

   // mapping of ASCII characters to hex values
   const uint8_t hashmap[] =
   {
     0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // ........
     0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // ........
     0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // ........
     0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // ........
     0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, //  !"#$%&'
     0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // ()*+,-./
     0x00, 0x01, 0x02, 0x03, 0x04, 0x05, 0x06, 0x07, // 01234567
     0x08, 0x09, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // 89:;<=>?
     0x00, 0x0a, 0x0b, 0x0c, 0x0d, 0x0e, 0x0f, 0x00, // @ABCDEFG
     0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // HIJKLMNO
     0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // PQRSTUVW
     0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // XYZ[\]^_
     0x00, 0x0a, 0x0b, 0x0c, 0x0d, 0x0e, 0x0f, 0x00, // `abcdefg
     0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // hijklmno
     0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // pqrstuvw
     0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // xyz{|}~.
     0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // ........
     0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // ........
     0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // ........
     0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // ........
     0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // ........
     0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // ........
     0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // ........
     0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // ........
     0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // ........
     0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // ........
     0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // ........
     0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // ........
     0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // ........
     0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // ........
     0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // ........
     0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00  // ........
   };

   bzero(bytes, blen);
   for (pos = 0; ((pos < (blen*2)) && (pos < strlen(str))); pos += 2)
   {
      idx0 = (uint8_t)str[pos+0];
      idx1 = (uint8_t)str[pos+1];
      bytes[pos/2] = (uint8_t)(hashmap[idx0] << 4) | hashmap[idx1];
   };

   return(0);
}

Answer 3

Apart from the excellent answers above I though I would write a C function that does not use any libraries and has some guards against bad strings.

uint8_t* datahex(char* string) {

    if(string == NULL) 
       return NULL;

    size_t slength = strlen(string);
    if((slength % 2) != 0) // must be even
       return NULL;

    size_t dlength = slength / 2;

    uint8_t* data = malloc(dlength);
    memset(data, 0, dlength);

    size_t index = 0;
    while (index < slength) {
        char c = string[index];
        int value = 0;
        if(c >= '0' && c <= '9')
          value = (c - '0');
        else if (c >= 'A' && c <= 'F') 
          value = (10 + (c - 'A'));
        else if (c >= 'a' && c <= 'f')
          value = (10 + (c - 'a'));
        else {
          free(data);
          return NULL;
        }

        data[(index/2)] += value << (((index + 1) % 2) * 4);

        index++;
    }

    return data;
}

Explanation:

a. index / 2 | Division between integers will round down the value, so 0/2 = 0, 1/2 = 0, 2/2 = 1, 3/2 = 1, 4/2 = 2, 5/2 = 2, etc. So, for every 2 string characters we add the value to 1 data byte.

b. (index + 1) % 2 | We want odd numbers to result to 1 and even to 0 since the first digit of a hex string is the most significant and needs to be multiplied by 16. so for index 0 => 0 + 1 % 2 = 1, index 1 => 1 + 1 % 2 = 0 etc.

c. << 4 | Shift by 4 is multiplying by 16. example: b00000001 << 4 = b00010000