hex string to byte array in C/C++

Note: this repository has been replaced by the chex library of the same author.

tl;dr

/* Copyright (c) 2022 Tero 'stedo' Liukko, MIT License */
static unsigned char unhex(char h){
  return ((h & 0x0f) + (h >> 6) * 9);
}

static unsigned hex2bin(const char* hex, unsigned hlen, void* bin, unsigned blen){
  unsigned i, j;
  for(i = 0, j = 0; (i < blen) && (j+1 < hlen); ++i, j+=2){
    unsigned char hi = unhex(hex[j+0]);
    unsigned char lo = unhex(hex[j+1]);
    ((unsigned char*)bin)[i] = (hi << 4) | lo;
  }
  return i;
}

Features

• does exactly what it says on the tin
• single header file only
• no dependencies, not even the standard library
• written in ANSI C, can be used in C++
• cross-platform / platform-agnostic
• branchless implementation without look-up tables
• no input validation whatsoever
• suitable for embedded development and microcontrollers

Background

The Web is full of examples (in various languages) on how to decode a hex string to an array of bytes, but none has been as elegant as one would hope; they either

• use libraries (like sscanf from cstdio)
• use look-up tables that take space
• use multiple branches to evaluate whether input is in the hexadecimal character set and whether it's in uppercase or lowercase

A more elegant elegant solution exists to map the input space to output space; consider the ASCII character table and the characters we are interested in

ASCII	hex	bin	ASCII	hex	bin	ASCII	hex	bin
0	30	0011 0000
1	31	0011 0001	A	41	0100 0001	a	61	0110 0001
2	32	0011 0010	B	42	0100 0010	b	62	0110 0010
3	33	0011 0011	C	43	0100 0011	c	63	0110 0011
4	34	0011 0100	D	44	0100 0100	d	64	0110 0100
5	35	0011 0101	E	45	0100 0101	e	65	0110 0101
6	36	0011 0110	F	46	0100 0110	f	66	0110 0110
7	37	0011 0111
8	38	0011 1000
9	39	0011 1001

From the above one can observe that the second most significant bit can be used to determine whether the input is 0xA (10 in decimal) or above, and that the lower nibble can be used (almost) as-is in either case, and by almost I mean that if the input is indeed 10 or above, we can always add 9 to the lower nibble to get the actual value.

In (pseudo-)code this means

// char hex = ...
uint8_t dec = (hex & 0x0f);   // always use the lowest 4 bits
bool add_more = (hex & 0x40); // ??
if(add_more) dec += 9;        // profit

which we can improve to make it a neat (or ugly, YMMV) one-liner by replacing the if branch with some basic arithmetic, like

(hex & 0x0f) + (!!(hex & 0x40)) * 9

or, even better like

(hex & 0x0f) + (hex >> 6) * 9

which can be trivially used in a loop to decode a buffer of hex characters. The latter variant also seems to compile without resulting in actual branching instructions. The most significant bit is assumed to be always 0. See the implementation in hex2bin.h.

It should be noted that the single mask bit (0x40) works for both lowercase as well as uppercase hexadecimal characters, i.e. valid inputs are the characters 0123456789ABCDEFabcdef.

Test

Run the snippet below for a simple test

#include <stdio.h>
#include <string.h>
#include <assert.h>

#include "hex2bin.h"

int main(void){
  /* check all possible 16-bit values in network byte order */
  unsigned i;
  for(i = 0; i <= 0xffff; ++i){
    /* encode to hex string using a function from the standard library */
    char hex[5];
    snprintf(hex, sizeof(hex), "%02hhX%02hhx", i>>8, i);

    /* decode from hex string using our implementation */
    unsigned char bin[2];
    unsigned n = hex2bin(hex, sizeof(hex), bin, sizeof(bin));
    assert(n == sizeof(bin));

    /* decoded matches the original */
    const unsigned char org[2] = {i>>8, i};
    assert(!memcmp(org, bin, sizeof(bin)));
  }
  return 0;
}

Disclaimer

The algorithm assumes valid input, and no effort is made to detect invalid input.

Author

Tero 'stedo' Liukko

License

MIT