converting number characters in a char array to binary

Question

char *toBin(unsigned long long num)
{
  char s[9999];
  const int size= 9999;
  char *temp = new char[size];
  int i, j;
  int binBase = 2;

  for(i=0; num!= 0; i++)
  {
      s[i] = num % binBase + '0';
      num /= binBase;
  }//get backwards binary

  for(j=0; i>=0; j++)
  {
      i--;
      temp[j]=s[i];
  }//reverse binary sequence

  //increments one more so need to decrement
  temp[--j] = '\0';
  return temp;
  delete temp;
}

If I had "82" in a char array, I first converted the "82" to the decimal 82 and gave 82 to the function above to convert to binary. This only works within a certain limit of digits.

I think my problem is that unsigned long long can only hold a certain amount of digits so when it exceeds that limit, some random stuff sent to the function, which prints out the wrong binary.

The above function converts a unsigned long long num to binary. I realized that a better approach would be to look at each character in the char array, rather than converting it to a number first, and and use those values to convert to binary. However, I'm having a hard time trying to put it to code. Any suggestions or help would be appreciated.

Not really related to how to get a number to print in binary but you're allocating a lot of memory in this function that isn't needed. You need at most sizeof(unsigned long long)*8+1 characters (+1 for the null-terminator). Also, delete temp; is never being called because a function ends when return is reached. — jodag
– jodag, Commented Apr 11, 2014 at 0:17

jodag · Accepted Answer · 2014-04-11 00:52:55Z

1

the GNU Multiple Precision (GMP) library has the functionality to print any size number in any base. The following is an example.

#include <iostream>
#include <gmpxx>

int main()
{
    std::string numstr = "1234123412341234123412341234";

    // initialize mpz_class to have numstr in it (given in base 10)
    mpz_class bigint(numstr, 10);

    // convert to binary (base 2)
    std::string binstr = bigint.get_str(2);

    std::cout << numstr << " -> " << binstr << std::endl;
    return 0;
}

output of program

1234123412341234123412341234 -> 111111110011010111110011000011110100100000101001110100011000000101000011000000000111110010

answered Apr 11, 2014 at 0:52

jodag

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Deduplicator · Accepted Answer · 2014-04-11 00:17:19Z

0

Allocating an array of 9999 char on the stack is ill-advised because it might be too big for your platform. Anyway, you can easily calculate the maximum number of digits (+terminator) needed: CHAR_BIT*sizeof num+1.
Also, there is no need to allocate an equal amount on the heap.
Do not be afraid of pointers:
- Init a pointer to the last byte
- Set that to zero
- Add all digits beginning with the smallest (at least one, even if the number is 0).
- Return a copy of the finished number.

answered Apr 11, 2014 at 0:17

Deduplicator

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