Hexadecimal string to byte array in C
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
For short strings, strtol, strtoll, and strtoimax will work just fine (note that the third argument is the base to use in processing the string...set it to 16). If your input is longer than number-of-bits-in-the-longest-integer-type/4 then you'll need one of the more flexible methods suggested by other answers.
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);
}
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