ASCII to hex

Alsalamo Alaykom whoever following me :)

Today I am coming for you with a complete app in C, it is simple but yet complete and do a great job.

For some reason I wanted to convert from ASCII to hex, I know that many Linux app already exist for that (oc, hexdump), but may be I found them hard to use correctly :)

The following is a simple but complete app you can use for that purpose, have fun :) :

/*
 * @auther mhewedy
 * @email mohammed_a_hewedy@hotmail.com
 * @tested under gcc (GCC) 3.4.2 (mingw-special) (quite old :) )
 */

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

char* usage(char*);

int main(int argc, char* argv[])
{
    if (argc <= 1)
    {
        char* usage_string = usage(argv[0]);
        printf("%s\n", usage_string);
        free(usage_string);
        exit(1);
    }

    int s_option_present =0;
    char* string = NULL;

    if (strcmp("-s", argv[1]) != 0)
        string = argv[1];
    else
    {
        string = argv[2];
        s_option_present = 1;
    }

    while (*string != '\0')
    {
        printf("%X%s", *string++, (s_option_present? " " : ""));
    }
    printf("\n");

    return 0;
}

char* usage(char* app_name)
{
    char* usage = (char*) calloc(40, sizeof(char*));
    strcpy(usage, "Usage: ");
    strcat(usage, app_name);
    strcat(usage, " [-s] {ascii string}");
    return usage;
}

Example of usage:
atohex.exe -s ABC
atohex.exe -s 1
atohex.exe hello

Print unicode values of non-ascii characters

class GetUnicode {
 
 static final char digits[] = {'0', '1', '2', '3','4', '5', '6', '7', '8', '9', 'A', 'B', 'C', 'D', 'E', 'F'};
 
 public static void main(String[] args) {
  String name = "محمد هويدي";
  for (char ch : name.toCharArray())
   if (Character.isWhitespace(ch))
    System.out.print(ch);
   else
    System.out.print("\\u" + getHex(ch));
 }
 static String getHex(int ch) {
  char tmp[] = new char[32];
  int i, j, count =0;
  
  for (i=0; i<32; i++)
   tmp[i] = '0';
  while (ch>0)
  {
   int t = ch&0xf;
   ch>>=4;
   tmp[count++] = digits[t];
  }
  char ret[] = new char[count +=(4-count&3)]; // make ret size to be multiple of four
  
  for (i=count-1, j=0; i>=0; i--, j++)
   ret[j] = tmp[i];
  
  tmp = null;
  return new String(ret);
 }
}

Get bases for decimal nubmers

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

#define SIZE sizeof(unsigned)* CHAR_BIT
//#define DEBUG

int main(void){


    char digits[]={'0', '1', '2', '3', '4', '5', '6', '7', '8', 
                 '9', 'A', 'B', 'C', 'D', 'E', 'F'};
    char ret[SIZE] = {0};
    unsigned n, b;

    printf("Enter a Positive number? " );
    scanf("%u", &n);
    printf("Enter Base? " );
    scanf("%u", &b);
    
    int i=0;
    do{
        ret[i++] = n%b;
        n /=b;

    }while (n>0);

#ifdef DEBUG
    for (i=0; i<SIZE;i++)
        printf("%i", ret[i]);
    printf("\n");
#endif

    // SWAP:
    for (i=0; i<SIZE/2;i++){
        int tmp=ret[i];
        ret[i] = ret[SIZE-i-1];
        ret[SIZE-i-1] = tmp;
    }

    for (i=0; i<SIZE;i++)
        printf("%c", digits[ret[i]]);

    
    printf("\n");
    return 0;
}

Two ways to convert decimal numbers to binary strings

I'll show you two ways to convert decimal to binary.

First, the most basic way is apply the following pseudo-code:

int x
char[80] arr
do{
   reminder = x%2;
   arr[i] = reminder
}while (x/2 > 0);
reverse arr;

And the other is more advanced (and more simple), it use bitwise operations instead:

int x
while (x > 0){
        arr[i] = ((x & 1) ? '1' : '0');
        x >>=1;
}
reverse arr

and here's a C code for the last one:

 
//I am not the author of this function, I copied from the link below
void dec2binary(long i){
    
    char* str = malloc(sizeof(long)*8*sizeof(char));
    
    char* p = str;
    while (i > 0){
        *p++ = ((i & 1) ? '1' : '0');
        i >>=1;
    }

    while( p-- != str ) /* print out the result backwards */
        printf("%c",*p);

    free(str);
}

please see:
http://www.daniweb.com/code/snippet216349.html

Convert from Hex to Binary by eye

Here's a simple way to convert from Hexdecimals to Binary just by eye.
suppose you have the following hex expression:

int x = 0xb123f;

and you need to know the binary (or even decimal) representation of this integer.

first, divide this integer into 2 characters (hence, 2 characters in Hex = 1 byte) from right:
so, it will be:
b 12 3f
then write the binary representation for each symbol:
b (11 in decimal) = 1011
1 = 0001
2 = 0010
3 = 0011
f = (15 in decimal) 1111
so, you have the following bit pattern:
0000 1011 0001 0010 0011 1111 (= 725567 in decimal)

note: hence we have 5 hex symbols, we added 0 to the left, and this explains why we add 0000 at the left.


thats all.