Converting string sequence of 1's and 0's into their Hexa sequences

ahzeeper · 04-27-2011, 08:44 AM

Hey all,
I have to write a code that converts a sequence of 1's and 0's(block) into their equivalent hexa number and copying to another array(byte). but this sequence is not always of length 8 and you are required to send strlen(sequence)%8 bits back into the string.
So i've written down this code and it works well with the sample input but fails in the real program.
Is there a smarter way of doing this?
thanks

Code:

int Convert_encode( char * block,unsigned char * byte)
{
 int len,iter,i,j,k,sum;
  char * temp=(char *)malloc(4*sizeof(char));
  len=strlen(block)/8;
  iter=strlen(block)/4;
  for (i=0;i<len;i++)
   {puts("a2");
   for(k=0;k<2;k++)    // this encodes first and last four bits seperatly    
   {
   sum=0;              //initializing sum
   for (j=0;j<4;j++)   //converting ASCII to 0-15 decimal
    {puts("a20");
    temp[j]=block[i*8+k*4+j]; puts("a21");
    if(temp[j]=='1')
    sum=sum+pow(2,3-j);
    } //puts("a3"); 
    printf("%d\n",sum);
   switch (sum)     // decimal to hexa
   {
   case 0:
   strcat(byte,"0");
   break;
   case 1:
   strcat(byte,"1");
   break;
   case 2:
   strcat(byte,"2");
   break;
   case 3:
   strcat(byte,"3");
   break;
   case 4:
   strcat(byte,"4");
   break;
   case 5:
   strcat(byte,"5");
   break;
   case 6:
   strcat(byte,"6");
   break;
   case 7:
   strcat(byte,"7");
   break;
   case 8:
   strcat(byte,"8");
   break;
   case 9:
   strcat(byte,"9");
   break;
   case 10:
   strcat(byte,"A");
   break;
   case 11:
   strcat(byte,"B");
   break;
   case 12:
   strcat(byte,"C");
   break;
   case 13:
   strcat(byte,"D");
   break;
   case 14:
   strcat(byte,"E");
   break;
   case 15:
   strcat(byte,"F");
   break;
   
   default :
   printf("unknown value in block\n");
   } 
    }
    } 
strcpy(block,&block[8*len]);
return(len);
free(temp);
}

grail · 04-27-2011, 08:54 AM

Please use [code][/code] tags as this is almost illegible.

Quote:

but fails in the real program.

You would need to elaborate on what you mean by 'fails'?

theNbomr · 04-27-2011, 09:02 AM

Your code is hard to read, because you failed to post in [CODE][/CODE] tags. In general, each char can be treated as a binary digit, and each binary digit can be multiplied by its binary weight, and summed with the other weighted values. The weight can be iteratively computed by multiplying by two on each loop, where the loop iterates once for each digit in the string.
Once the binary value represented by the string has been fully computed, you can use printf() to re-display it in a hexadecimal radix.

Code:

weight = 1
sumOfBits = 0
for each ASCII digit
   if ASCII digit = '1' then
      sumOfBits = sumOfBits + weight
   end if
   weight = weight * 2
end for

printf( "%04X", sumOfBits )

Any integer data type will automatically be an even number of bytes (8-bits).

--- rod.

dwhitney67 · 04-27-2011, 09:17 AM

First, could you please edit your original post so that it includes formatted code placed within CODE tags? It would make it much easier to read your code.

Second, could you please confirm if it is C code you are developing? Some developers wish to develop in C++, but end up writing C code.

Lastly, what does your input look like? Is it like: "1010000111101000"? Is the length of the input string a multiple of 8 or 4? Four bits (half-byte or nibble) can easily be translated to hex.

ahzeeper · 04-27-2011, 09:37 AM

Sorry for tags.

I am developing this code in C, no c++.
The input is a string of 1's and 0's but its length is not fixed. so it can be a multiple of 8 or not. but if it is not a multiple of eight then you take the remaining bits and copy them back into the original array ie block.
for instance lets say the block has '11110000101'. so now the "byte" should read F0 while the "block" output should be 101( these are the remainder of the bits that cant be converted).

This program is part of a JPEG encoder.
@dwhitney what is the easy method you are talking about?

thanks all

dwhitney67 · 04-27-2011, 09:52 AM

Quote:

Originally Posted by ahzeeper

This program is part of a JPEG encoder.
@dwhitney what is the easy method you are talking about?

See theNbomr's post for pseudocode that will help you with your algorithm.

Here's how I did the conversion in C++. However my code requires that the string have a length that is a multiple of 4. That's not a big deal; you can easily get the required nibbles using arithmetic; some number N div 4. Then copy that result multiplied by 4 to get the number of working digits; the remainder will be placed into your "output" buffer.

My code:

Code:

#include <iostream>
#include <string>
#include <cassert>
#include <cmath>

int main(int argc, char** argv)
{
   assert(argc == 2);

   std::string binstr = argv[1];

   assert(binstr.size() % 4 == 0);

   size_t pos = 0;

   while (pos != binstr.size())
   {
      std::string fourbits = binstr.substr(pos, 4);

      pos += 4;

      int sum = 0;

      for (size_t i = 0; i < 4; ++i)
      {
         if (int(fourbits[i] - '0') > 0)
         {
            sum += int(pow(double(2), double(3 - i)));
         }
      }

      std::cout << std::hex << sum;
   }
   std::cout << std::endl;
}

ahzeeper · 04-27-2011, 10:12 AM

thanks

ahzeeper · 04-27-2011, 10:22 AM

My code pretty much does the same thing, except after i convert it into a integer( between 0 and 15) i use switch case and strcat( ) to write equivalent hex values.
Anyways my code runs fine on the test samples but in the main program(which has been already written by someone else, and does'nt have any problems, that i know for sure) where i have to use this it gives me a segmentation fault. The error comes in a line where the HEX value of 4 is being written out by the strcat().
i'll post the error later
thanks again people