将十进制数组（str）转换为二进制数组（bytes）

Question

Please provide some code for converting char[] array of decimal values to bytes array for big integer value in c.

How can I convert this below code for big array of decimal value, such as I get the result as array of long bytes?

static int dec2bin (char inbuf[], int num_convert)
{
  char ctemp;
  int result, power;

  num_convert--; /* index of LS char to convert */
  result = 0;
  power = 1;
  while (num_convert >= 0)
  {
    ctemp = inbuf[num_convert--]; /* working digit */
    if (isdigit(ctemp))
    {
      result += ((ctemp-'0') * power);
      power *= 10;
    }
    else
      return(0); /* error: non-numeric digit detected */
  }
  return (result);
}

No it is not just long value , it is really a biginteger value , can anyone give a simple dec to byte(binary conversion logic , i will replace the int with my bigint implementations and bigint operators(add, mult) etc.,

Samuel is correct!

Thanks in advance.

For Example i can replace the above as below

static int dec2bin (char inbuf[], bigint num_convert) 
{ 
  char ctemp; 
  bigint result, power; 

  num_convert--; /* index of LS char to convert */ 
  result = 0; 
  power = 1; 
  while (num_convert >= 0) 
  { 
    ctemp = inbuf[num_convert--]; /* working digit */ 
    if (isdigit(ctemp)) 
    { 
      result = bi_add(result ,(bi_mult((ctemp-'0'), power)); 
      power = bi_mult(power , 10); 
    } 
    else 
      return(0); /* error: non-numeric digit detected */ 
  } 
  return (result); 
} 

something like this will work ?

Answer 1

It sounds like you are looking for a solution where you do the low level arithmetic yourself.

Have you considered using an existing bignum package such as the GNU Multi Precision Arithmetic library ? It's pretty easy to convert your existing code once you have that. For example, this code:

result += ((ctemp-'0') * power);
power *= 10;

becomes:

mpz_t tmp;
mpz_init(tmp);
mpz_mul_ui(tmp, power, ctemp - '0');
mpz_add(result, result, tmp);
mpz_clear(tmp);

mpz_mul_ui(power, power, 10);