How to convert a hex string to a uint8_t array? (C language)

how can I convert a hex string to a uint8_t array? My string is 02012B1530A6E3958A98530031902003876940000000000CDF9844173BE512AFFFFFFE11DBBA1F00079387800E13012E11FC017FFFFFFFFE39C10F40 and I want to convert it into this array:

uint8_t array_uint[] = {0x02, 0x01, 0x2B, 0x15, 0x30, 0xA6, 0xE3, 0x95, 0x8A, 0x98, 0x53, 0x00, 0x31, 0x90, 0x20, 0x03, 0x87, 0x69, 0x40, 0x00, 0x00, 0x00, 0x00, 0x0C, 0xDF, 0x98, 0x44, 0x17, 0x3B, 0xE5, 0x12, 0xAF, 0xFF, 0xFF, 0xFE, 0x11, 0xDB, 0xBA, 0x1F, 0x00, 0x07, 0x93, 0x87, 0x80, 0x0E, 0x13, 0x01, 0x2E, 0x11, 0xFC, 0x01, 0x7F, 0xFF, 0xFF, 0xFF, 0xFE, 0x39, 0xC1, 0x0F, 0x40};

Thank you for your help!

Solution

You can use sscanf() to convert 2 bytes at a time from the source string into the destination array:

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

size_t convert_hex(uint8_t *dest, size_t count, const char *src) {
    size_t i;
    int value;
    for (i = 0; i < count && sscanf(src + i * 2, "%2x", &value) == 1; i++) {
        dest[i] = value;
    }
    return i;
}

Note however that this approach may be inefficient, with a quadratic time complexity, on architectures where the standard library computes the length of the source string for each call to sscanf(). Using an intermediary array solves this problem:

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

size_t convert_hex(uint8_t *dest, size_t count, const char *src) {
    char buf[3];
    size_t i;
    int value;
    for (i = 0; i < count && *src; i++) {
        buf[0] = *src++;
        buf[1] = '\0';
        if (*src) {
            buf[1] = *src++;
            buf[2] = '\0';
        }
        if (sscanf(buf, "%x", &value) != 1)
            break;
        dest[i] = value;
    }
    return i;
}

Storing the conversion result directly into the dest array is easy:

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

size_t convert_hex(uint8_t *dest, size_t count, const char *src) {
    char buf[3];
    size_t i;
    for (i = 0; i < count && *src; i++) {
        buf[0] = *src++;
        buf[1] = '\0';
        if (*src) {
            buf[1] = *src++;
            buf[2] = '\0';
        }
        if (sscanf(buf, "%hhx", &dest[i]) != 1)
            break;
    }
    return i;
}

Purists may argue that %hhx expects a pointer to an unsigned char instead of a uint8_t so the format should be "%"SCNx8 or "%2"SCNx8, defined in <inttypes.h>, but these alternatives are less readable and unnecessary as type uint8_t is always identical to type unsigned char on architectures where it is defined.