How to convert an arbitrary large integer from base 10 to base 16?

Question

The program requires the input of an arbitrary large unsigned integer, which is expressed as one line in base 10. Outputs are another line that expresses an integer in base 16.

For example, the input is "1234567890987654321234567890987654321234567890987654321" and the output should be "CE3B5A137DD015278E09864703E4FF9952FF6B62C1CB1"

The faster the algorithm, the better.

It will be very simple if the input is limited to a 32-bit or 64-bit integer; for example, the following code can perform the conversion:

#define MAX_BUFFER 16
char hex[] = "0123456789ABCDEF";

char* dec2hex(unsigned input) {
    char buff[MAX_BUFFER];
    int i = 0, j = 0;
    char* output;

    if (input == 0) {
        buff[0] = hex[0];
        i = 1;
    } else {
        while (input) {
            buff[i++] = hex[input % 16];
            input = input / 16;
        }
    }

    output = malloc((i + 1) * sizeof(char));
    if (!output) 
        return NULL;

    while (i > 0) {
        output[j++] = buff[--i];        
    }
    output[j] = '\0';

    return output;
}

The real hard part is an "arbitrary large" unsigned integer. I have googled, but most of them talk about conversion in 32-bit or 64-bit. Bad luck.

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Answer 1

• , . 0s.

  16.

• . Mulitply 10 , , newvalue div 16.

  carryover = digit;
  for (i = (nElements-1); i >= 0; i--)
  {
      newVal = array[index] * 10) + carryover;
      array[index] = newval % 16;
      carryover = newval / 16;
  }
  

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#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include "sys/types.h"

char HexChar [16] = { '0', '1', '2', '3', '4', '5', '6', '7',
                      '8', '9', 'A', 'B', 'C', 'D', 'E', 'F' };

static int * initHexArray (char * pDecStr, int * pnElements);

static void addDecValue (int * pMyArray, int nElements, int value);
static void printHexArray (int * pHexArray, int nElements);

static void
addDecValue (int * pHexArray, int nElements, int value)
{
    int carryover = value;
    int tmp = 0;
    int i;

    /* start at the bottom of the array and work towards the top
     *
     * multiply the existing array value by 10, then add new value.
     * carry over remainder as you work back towards the top of the array
     */
    for (i = (nElements-1); (i >= 0); i--)
    {
        tmp = (pHexArray[i] * 10) + carryover;
        pHexArray[i] = tmp % 16;
        carryover = tmp / 16;
    }
}

static int *
initHexArray (char * pDecStr, int * pnElements)
{
    int * pArray = NULL;
    int lenDecStr = strlen (pDecStr);
    int i;

    /* allocate an array of integer values to store intermediate results
     * only need as many as the input string as going from base 10 to
     * base 16 will never result in a larger number of digits, but for values
     * less than "16" will use the same number
     */

    pArray = (int *) calloc (lenDecStr,  sizeof (int));

    for (i = 0; i < lenDecStr; i++)
    {
        addDecValue (pArray, lenDecStr, pDecStr[i] - '0');
    }

    *pnElements = lenDecStr;

    return (pArray);
}

static void
printHexArray (int * pHexArray, int nElements)
{
    int start = 0;
    int i;

    /* skip all the leading 0s */
    while ((pHexArray[start] == 0) && (start < (nElements-1)))
    {
        start++;
    }

    for (i = start; i < nElements; i++)
    {
        printf ("%c", HexChar[pHexArray[i]]);
    }

    printf ("\n");
}

int
main (int argc, char * argv[])
{
    int i;
    int * pMyArray = NULL;
    int nElements;

    if (argc < 2)
    {
        printf ("Usage: %s decimalString\n", argv[0]);
        return (-1);
    }

    pMyArray = initHexArray (argv[1], &nElements);

    printHexArray (pMyArray, nElements);

    if (pMyArray != NULL)
        free (pMyArray);

    return (0);
}

Answer 2

Python, , C, . , Python C , .

Python:

import math
import string

def incNumberByValue(digits, base, value):
   # The initial overflow is the 'value' to add to the number.
   overflow = value
   # Traverse list of digits in reverse order.
   for i in reversed(xrange(len(digits))):
      # If there is no overflow we can stop overflow propagation to next higher digit(s).
      if not overflow:
         return
      sum = digits[i] + overflow
      digits[i] = sum % base
      overflow = sum / base

def multNumberByValue(digits, base, value):
   overflow = 0
   # Traverse list of digits in reverse order.
   for i in reversed(xrange(len(digits))):
      tmp = (digits[i] * value) + overflow
      digits[i] = tmp % base
      overflow = tmp / base

def convertNumber(srcDigits, srcBase, destDigits, destBase):
   for srcDigit in srcDigits:
      multNumberByValue(destDigits, destBase, srcBase)
      incNumberByValue(destDigits, destBase, srcDigit)

def withoutLeadingZeros(digits):
   for i in xrange(len(digits)):
      if digits[i] != 0:
         break
   return digits[i:]

def convertNumberExt(srcDigits, srcBase, destBase):
   # Generate a list of zero which is long enough to hold the destination number.
   destDigits = [0] * int(math.ceil(len(srcDigits)*math.log(srcBase)/math.log(destBase)))
   # Do conversion.
   convertNumber(srcDigits, srcBase, destDigits, destBase)
   # Return result (without leading zeros).
   return withoutLeadingZeros(destDigits)


# Example: Convert base 10 to base 16
base10 = [int(c) for c in '1234567890987654321234567890987654321234567890987654321']
base16 = convertNumberExt(base10, 10, 16)
# Output list of base 16 digits as HEX string.
hexDigits = '0123456789ABCDEF'
string.join((hexDigits[n] for n in base16), '')

Answer 3

- " " .

GNU MP Bignum

Answer 4

BigInt:

http://www.codeproject.com/KB/cs/BigInt.aspx?msg=3038072#xx3038072xx

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Answer 5

Unix dc . BSD .

Answer 6

Python:

>>> from string import upper
>>> input = "1234567890987654321234567890987654321234567890987654321"
>>> output = upper(hex(int(input)))[2:-1]
>>> print output
CE3B5A137DD015278E09864703E4FF9952FF6B62C1CB1