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An algorithm for converting an infinitely long base 2 ^ 32 numbers to a printable base 10

I represent an infinitely accurate integer as an array of unsigned ints to process on the GPU. For debugging purposes, I would like to print a basic representation of 10 of one of these numbers, but I have difficulty wrapping around it. Here is what I would like to do:

//the number 4*(2^32)^2+5*(2^32)^1+6*(2^32)^0
unsigned int aNumber[3] = {4,5,6};
char base10TextRepresentation[50];
convertBase2To32ToBase10Text(aNumber,base10TextRepresentation);

Any suggestions on how to approach this issue?

Edit: here is the full implementation thanks to drhirsch

#include <string.h>
#include <stdio.h>
#include <stdint.h>

#define SIZE 4

uint32_t divideBy10(uint32_t * number) {
  uint32_t r = 0;
  uint32_t d;
  for (int i=0; i<SIZE; ++i) {
    d = (number[i] + r*0x100000000) / 10;
    r = (number[i] + r*0x100000000) % 10;
    number[i] = d;
  }
  return r;
}

int zero(uint32_t* number) {
  for (int i=0; i<SIZE; ++i) {
    if (number[i] != 0) {
      return 0;
    }
  }
  return 1;
}

void swap(char *a, char *b) {
  char tmp = *a;
  *a = *b;
  *b = tmp;
}

void reverse(char *str) {
  int x = strlen(str);
  for (int y = 0; y < x/2; y++) {
    swap(&str[y],&str[x-y-1]);
  }
}

void convertTo10Text(uint32_t* number, char* buf) {
  int n = 0;
  do {
    int digit = divideBy10(number);
    buf[n++] = digit + '0';
  } while(!zero(number));
  buf[n] = '\0';
  reverse(buf);
}

int main(int argc, char** argv) {
  uint32_t aNumber[SIZE] = {0,0xFFFFFFFF,0xFFFFFFFF,0xFFFFFFFF};
  uint32_t bNumber[4] = {1,0,0,0};

  char base10TextRepresentation[50];

  convertTo10Text(aNumber, base10TextRepresentation);
  printf("%s\n",base10TextRepresentation);
  convertTo10Text(bNumber, base10TextRepresentation);
  printf("%s\n",base10TextRepresentation);
}
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4 answers

If you have access to 64-bit arithmetic, this is easier. I would do something along the line:

int32_t divideBy10(int32_t* number) {
    uint32_t r = 0;
    uint32_t d;
    for (int i=0; i<SIZE; ++i) {
        d = (number[i] + r*0x100000000) / 10;
        r = (number[i] + r*0x100000000) % 10;
        number[i] = d;
        number[i] = r;
}

void convertTo10Text(int32_t* number, char* buf) {
    do {
        digit = divideBy10(number);
        *buf++ = digit + '0';
    } while (!isEqual(number, zero));
    reverse(buf);
}

isEqual () and reverse (), which must be implemented. divideBy10 divides by 10 and returns the remainder.

+4

, ( 2 ^ 32) . (, ) 10 , , , .

C ++, GNU Bignum. .

, , . ; .

+4

.NET, BigInteger.

+3

How about using long doubles? Then you get 80 bits in the mantissa, but I think the accuracy is lost when using floating point numbers.

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