From uint8_t to hex string array? - c

Question

I know it's a simple question, but I am completely lost.
I have in input a string:
input: 220209
I want:
output: uint8_t myarr[8]={0x22,0x02,0x09,0x00, 0x00, 0x00,0x00,0x00}
So myarr has always a dimension of 8, and if I have a lower input string I have to complete with 0x00. Can someone give me any idea?
Sorry for bothering you, any help will be appreciated.
Regards

Answer 1

Using strtol() to convert 2 characters at a time, explicitly using base 16 makes it easy:
#include <inttypes.h>
#include <stdio.h>
#include <stdlib.h>
// Assume s has an even number of hexdigit chars (at most 16) followed by a nul
void convert(const char *s, uint8_t out[static 8]) {
int i = 0;
// Handle 2 chars at a time
while (*s) {
char byte[3] = { *s, *(s + 1), 0 };
out[i++] = strtol(byte, NULL, 16);
s += 2;
}
// Fill the rest of the array with nuls
for (; i < 8; i += 1) {
out[i] = 0;
}
}
int main(void)
{
const char *input = "220209";
uint8_t myarr[8];
convert(input, myarr);
// Pretty-print the array using hex values
fputs("uint8_t myarr[8] = { ", stdout);
for (int i = 0; i < 8; i += 1) {
printf("0x%02" PRIx8, myarr[i]);
if (i < 7) {
fputs(", ", stdout);
}
}
puts(" };");
return 0;
}

Answer 2

Another option is to use sscanf() to parse two characters at a time:
void convert(const char *s, uint8_t out[8])
{
memset(out, 0, 8);
while (sscanf(s, "%2"SCNx8, out++) == 1) {
s += 2;
}
}
The first line sets the 8 bytes to 0 by taking advantage of 64 bits arithmetics. Then we scan two characters from string s, reading an integer of type uint8_t from hexdecimal ASCII representation. If the read was successful, we move to the next two characters.
It looks like a code golf submission, but it's nice, because nothing gets copied.
Another version could be:
void convert(const char *s, uint8_t out[8])
{
static char tbl[] = {
0, 0, 0, 0, 0, 0, 0,0,0,0,0,0,0,0,0,0,
0, 0, 0, 0, 0, 0, 0,0,0,0,0,0,0,0,0,0,
0, 0, 0, 0, 0, 0, 0,0,0,0,0,0,0,0,0,0,
0, 1, 2, 3, 4, 5, 6,7,8,9,0,0,0,0,0,0,
0,10,11,12,13,14,15,0,0,0,0,0,0,0,0,0,
0, 0, 0, 0, 0, 0, 0,0,0,0,0,0,0,0,0,0,
0,10,11,12,13,14,15,0,0,0,0,0,0,0,0,0,
0, 0, 0, 0, 0, 0, 0,0,0,0,0,0,0,0,0,0,
};
memset(out, 0, 8);
while (*s) {
*out++ = tbl[s[0]] * 16 + tbl[s[1]];
s += 2;
}
}

Answer 3

Hope it can work for you
memset(myarr, 0, 8);
int szLen = strlen(szInput);
for(int i = 0; i < szLen / 2; i++) {
uint8_t high = toupper(szInput[i*2]);
uint8_t low = toupper(szInput[i*2 + 1]);
if(high >= '0' && high <= '9')
high = high - '0';
if(low >= '0' && low <= '9')
low = low - '0';
if(high >= 'A' && high <= 'F')
high = high - 'A' + 10;
if(low >= 'A' && low <= 'F')
low = low - 'A' + 10;
myarr[i] = (high << 4) | low; // bit operation is faster than arithmetic
}