Hex to Octal Conversion Program Without Using Decimal or Binary

Question

Today I was just playing around for basic conversions from one base to another. I goggled some code for converting from hex to octal, and I noticed that it mostly uses intermediate conversion to either decimal or binary and then back to octal.Is it possible write my own function for converting hex string to octal string without using any intermediate conversion.Also I do not want to use inbuilt printf option like %x or %o. Thanks for your inputs.

Answer 1

Of course it is possible. A number is a number no matter what numeric system it is in. The only problem is that people are used to decimal and that is why they understand it better. You may convert from any base to any other.

EDIT: more info on how to perform the conversion.

First note that 3 hexadecimal digits map to exactly 4 octal digits. So having the number of hexadecimal digits you may find the number of octal digits easily:

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
int get_val(char hex_digit) {
  if (hex_digit >= '0' && hex_digit <= '9') {
    return hex_digit - '0';
  } else {
    return hex_digit - 'A' + 10;
  }
}
void convert_to_oct(const char* hex, char** res) {
  int hex_len = strlen(hex);
  int oct_len = (hex_len/3) * 4;
  int i;

  // One hex digit left that is 4 bits or 2 oct digits.
  if (hex_len%3 == 1) {
    oct_len += 2;
  } else if (hex_len%3 == 2) { // 2 hex digits map to 3 oct digits
    oct_len += 3;
  }

  (*res) = malloc((oct_len+1) * sizeof(char));
  (*res)[oct_len] = 0; // don't forget the terminating char.

  int oct_index = oct_len - 1; // position we are changing in the oct representation.
  for (i = hex_len - 1; i - 3 >= 0; i -= 3) {
    (*res)[oct_index] = get_val(hex[i]) % 8 + '0';
    (*res)[oct_index - 1] = (get_val(hex[i])/8+ (get_val(hex[i-1])%4) * 2) + '0';
    (*res)[oct_index - 2] = get_val(hex[i-1])/4 + (get_val(hex[i-2])%2)*4 + '0';
    (*res)[oct_index - 3] = get_val(hex[i-2])/2 + '0'; 
    oct_index -= 4;
  }

  // if hex_len is not divisible by 4 we have to take care of the extra digits:
  if (hex_len%3 == 1) {
     (*res)[oct_index] = get_val(hex[0])%8 + '0';
     (*res)[oct_index - 1] = get_val(hex[0])/8 + '0';
  } else if (hex_len%3 == 2) {
     (*res)[oct_index] = get_val(hex[1])%8 + '0';
     (*res)[oct_index - 1] = get_val(hex[1])/8 + (get_val(hex[0])%4)*4 + '0';
     (*res)[oct_index - 2] = get_val(hex[0])/4 + '0';
  }
}

Also here is the example on ideone so that you can play with it: example.