binary to hex

[loaded_tk]

hey,

first of all am new to C programming ! am trying to write a program that changes a binary 8 or 16 bit number into hex, i used to the select case in my program, my question is how can i make it recognise the whole 8/16 digit binary number i enter and not only the 4 bits that are identified in the look up table.. thank you

#include <stdio.h>

int main ();
void bintohex (int);


int main ()s
{
  int digit;

  scanf ("%d", &digit);
  bintohex (digit);
  return 0;
}


void bintohex (int digit)        /* print out bintohex code */
{
  switch (digit)
  {
    case 0000 : printf ("0");
    break; 
    case 0001 : printf ("1");
    break;  
    case 0010 : printf ("2");
    break; 
    case 0011 : printf ("3");
    break; 
    case 0100 : printf ("4");
    break;
    case 0101 : printf ("5");
    break; 
    case 0110 : printf ("6");
    break; 
    case 0111 : printf ("7");
    break; 
    case 1000 : printf ("8");
    break;
    case 1001 : printf ("9");
    break;
    case 1010 : printf ("A");
    break; 
    case 1011 : printf ("B");
    break; 
    case 1100 : printf ("C");
    break; 
    case 1101 : printf ("D");
    break;
    case 1110 : printf ("E");
    break;
    case 1111 : printf ("F");
    break;


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

[MadCow257]

scroll down to my post
http://cboard.cprogramming.com/showt...hlight=lecture
it's c++, but conversion is trivial

[hk_mp5kpdw]

    case 0000 : printf ("0");
    break; 
    case 0001 : printf ("1");
    break;  
    case 0010 : printf ("2");
    break; 
    case 0011 : printf ("3");
    break; 
    case 0100 : printf ("4");
    break;
    case 0101 : printf ("5");
    break; 
    case 0110 : printf ("6");
    break; 
    case 0111 : printf ("7");
    break;

There is a problem with these as they will be interpreted by the compiler as octal values. You are effectively saying this:

    case 0 : printf ("0");
    break; 
    case 1 : printf ("1");
    break;  
    case 8 : printf ("2");
    break; 
    case 9 : printf ("3");
    break; 
    case 64 : printf ("4");
    break;
    case 65 : printf ("5");
    break; 
    case 72 : printf ("6");
    break; 
    case 73 : printf ("7");
    break;

Values with leading zeroes are interpreted as octal in your compiled code, i.e. 0100 is seen as 100 (octal) which is 64 (decimal).

[loaded_tk]

oh thanks i didn't know that, where can i start form then writing a code?

[OnionKnight]

Also, %d wants an integer in decimal, that is base 10. So if you enter 101 you will get 101 in decimal, not 5.
You should retrieve a string and then process it four bytes a time (four bits = one hex character) and convert it to decimal then print out a number if it's in the interval 0 to 9 or print out a letter if it's in the interval 10 to 15.