How to convert binary string to integer string

Question

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Converting string of 1s and 0s into binary value

Lets say that I have string which contains 1024 characters (represents 0 and 1). I want to present it as an number in decimal base (also as a string). The tricky part is that i have to do it in C/C++ without 3rd part libraries. Any clues?

Answer 1

There are probably more efficient ways, but I'd have an array of decimal digits, and implement a 'left-shift' function on it which runs from the least significant digit, doubling them and carrying over into the next.

It's then just a job of reading your binary data in one bit at a time and 'left-shifting' the decimal array and 'OR'ing in the binary digit as required.

Just iterate through the decimal digits to print out the answer.

void outputAsDecimal(char *binary)
{
   char digits[1000]; // arbitrary size for now

   for (int i=0; i< 1000; ++i)
     digits[i] = 0;

   while (*binary != 0)
   {

   // shift the digits, with carry
     int carry = 0;

     for (int i = 0; i< 1000; ++i)
     {
       int d = digits[i] *2 + carry;
       carry = d > 9;
       digits[i] = d % 10;
     }

   // or in the new bit
     if (*binary++ == '1') 
       digits[0] |= 1;
   }

    // output with leading zeroes!
    for (int i = 999; i >=0; --i)
    {
      putchar(digits[i] + '0'); // convert to ascii
    }
}

See it running here: http://ideone.com/CibAfw

Answer 2

Edit: Aha! I have just noticed the 1024 requirement. This makes it more complicated, but the idea remains the same. Instead of just having an int number, you need int number[32] (or long number[16], what have you.)

The math at the borders is annoying, but not impossible. Let me know if you can't figure it out.

This works for me. Decomposition and supporting values greater than offered in an (int) is left as an exercise to the reader...

#include <stdio.h> // only to print - not needed in computation
int main(int argc, char *argv[]) {
  printf("Converting: %s\n", argv[1]);
  int number = 0x0;
  char * binaryString = argv[1];
  int index = 0;
  int asciiZero = '0';
  char curr = binaryString[index];
  while(curr != '\0') {
    number = (number << 1) | (curr - asciiZero);
    index++;
    curr = binaryString[index];
  }

  printf("As number: %d\n", number);

  int MAX_DIGITS = 10; //adjust accordingly...
  char buffer[MAX_DIGITS];
  index = 0;
  while(number > 0) {
    buffer[index] = ((char) number % 10) + asciiZero;
    index++;
    number = number / 10;
  }
  buffer[index] = '\0';

  printf("As string: %s\n", buffer);
 }

If you want to support more than offered in the primitives available to you, you can make a struct containing multiple ints/longs/etc.

Answer 3

You can't represent a 1024 bit number as a decimal. The closest you could get is a floating point approximation.

EDIT:

A thought occurs. Put the numbers in a stack and count them that way. Still have to figure out how to add them together though. You're going to have to implement your own BigInt library if you want to do this, there's not a "trivial" way, as far as I can imagine.

Something like this is a good starting point:

http://gmplib.org/