Converting hex char array to ascii char array

Question

Is there any function that can convert hex char array to an ASCII char array, something like this:

char array_input[] = "48656C6C6F3"; //Input hex sequence
char array_output[] = "Hello";      // Output ASCII

and that can accept input sequences that contain more than 300 letters.

something like:

char array_input[] = "08010000013feb55f000000000000f74000f0ea850209a690000940000120000001e09010002000300040016014703f0001504c8000c0900730a00460b00501300464306d7440000b5000bb60007422e9f180000cd0386ce000107c700000000f10000601a46000001344800000bb84900000bb84a00000bb84c00000000024e0000000000000000cf000001";

Use any way you know to separate the hex values and convert each of them into a char. Since this is apparently homework, show some effort how to solve this. — the busybee
– the busybee, Commented Dec 3, 2021 at 10:47

3 revs · Accepted Answer · 2021-12-03 11:09:43Z

2

Parse 2 hex digits at a time

Walk the input string looking for 2 hex digits. Form the ASCII value and then append to the output array. Append a final null character.

void foo(char *array_output, const char *array_input) {
  unsigned value;
  while (sscanf(array_input, "%2x", &value) == 1) {
    *array_output++ = value;
    array_input += 2;
  } 
  *array_output = '\0';
}

Pathological strings with white-space or signs like " A12-3" need additional code to handle.

Alternative: use is...() to test for hex digits. Then form a compound literal s to pass to strtol().

#include <ctype.h>
#include <stdio.h>
#include <stdlib.h>

void foo2(char *array_output, const char *array_input) {
  const unsigned char *h = (const unsigned char *) array_input;
  while (isxdigit(h[0]) && isxdigit(h[1])) {
    char s[3] = { h[0], h[1], '\0' };
    *array_output++ = (char) strtol(s, 0, 16);
    h += 2;
  }
  *array_output = '\0';
}

edited Dec 3, 2021 at 11:09

community wiki

3 revs
chux - Reinstate Monica

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3 Comments

chqrlie Over a year ago

You might show how to use %n to handle input strings with embedded spaces gracefully such as "48 65 6C 6C 6F 3", but spaces inside hex digit pairs and embedded + signs would also be ignored, which could be surprising. The input string could also be validated with array_input[strcspn(array_input, "0123456789abcdefABCDEF")] == '\0'

chux Over a year ago

@chqrlie Yes - good idea, yet sscanf() does have more shortcomings (too many). Alternate, more robust isxdigit() posted.

chqrlie Over a year ago

Indeed both sscanf and strtol feel like using a steam hammer to squash a fly :)

chqrlie · Accepted Answer · 2021-12-09 11:04:17Z

2

Here is an efficient alternative solution without any library calls:

void hextostr(char *dest, const char *src) {
    static unsigned char const val[(unsigned char)-1+1] = {
        ['0'] =  0+1, ['1'] =  1+1, ['2'] =  2+1, ['3'] =  3+1, ['4'] =  4+1,
        ['5'] =  5+1, ['6'] =  6+1, ['7'] =  7+1, ['8'] =  8+1, ['9'] =  9+1,
        ['a'] = 10+1, ['b'] = 11+1, ['c'] = 12+1, ['d'] = 13+1, ['e'] = 14+1, ['f'] = 15+1,
        ['A'] = 10+1, ['B'] = 11+1, ['C'] = 12+1, ['D'] = 13+1, ['E'] = 14+1, ['F'] = 15+1,
    };
    const unsigned char *p = (const unsigned char *)src;
    while (val[p[0]] && val[p[1]]) {
        *dest++ = (char)(val[p[0]] * 16 + val[p[1]] - 17);
        p += 2;
    }
    *dest = '\0';
}

edited Dec 9, 2021 at 11:04

answered Dec 3, 2021 at 11:15

chqrlie

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