how to convert digits from a integer in a byte array in C++

Question

i tried to convert the digits from a number like 9140 to a char array of bytes, i finally did it, but for some reason one of the numbers is converted wrong.

The idea is separate each digit an convert it in a byte[4] and save it a global array of bytes, that means that array have a digit each 4 positions, i insert each digit at the end of array and finally i insert the amount of digits at the end of the array.

the problem is randomly with some values, for example for the value 25 it works but for 9140 it return me 9040, which could be the problem? this is the code:

void convertCantToByteArray4Digits(unsigned char *bufferDigits,int cant){
    //char bufferDigits[32];
    int bufferPos=20;
    double cantAux=cant;
    int digit=0,cantDigits=0;
    double subdigit=0;
    while(cantAux > 0){
        cout<<"VUELTA"<<endl;
        cantAux/=10;
        cout<<"cantAux/=10:"<<cantAux<<endl;
        cout<<"floor"<<floor(cantAux)<<endl;
        subdigit=cantAux-floor(cantAux);
        cout<<"subdigit"<<subdigit<<endl;
        digit=static_cast<int>(subdigit*10);
        cout<<"digit:"<<subdigit*10<<endl;
        cantAux=cantAux-subdigit;
        cout<<"cantAux=cantAux-subdigit:"<<cantAux<<endl;
        bufferDigits[bufferPos-4] = (digit >> 24) & 0xFF;
        std::cout<<static_cast<int>(bufferDigits[bufferPos-4])<<std::endl;
        bufferDigits[bufferPos-3] = (digit >> 16) & 0xFF;
        std::cout<<static_cast<int>(bufferDigits[bufferPos-3])<<std::endl;
        bufferDigits[bufferPos-2] = (digit >> 8) & 0xFF;
        std::cout<<static_cast<int>(bufferDigits[bufferPos-2])<<std::endl;
        bufferDigits[bufferPos-1] = (digit) & 0xFF;
        std::cout<<static_cast<int>(bufferDigits[bufferPos-1])<<std::endl;
        /*bufferDigits[0] = digit >> 24;
        std::cout<<bufferDigits[0]<<std::endl;
        bufferDigits[1] = digit >> 16;
        bufferDigits[2] = digit >> 8;
        bufferDigits[3] = digit;*/
        bufferPos-=4;
        cantDigits++;
    }
    cout<<"sizeof"<<sizeof(bufferDigits)<<endl;
    cout<<"cantDigits"<<cantDigits<<endl;
    bufferPos=24;
    bufferDigits[bufferPos-4] = (cantDigits) >> 24;
        //std::cout<<bufferDigits[bufferPos-4]<<std::endl;
    bufferDigits[bufferPos-3] = (cantDigits) >> 16;
    bufferDigits[bufferPos-2] = (cantDigits) >> 8;
    bufferDigits[bufferPos-1] = (cantDigits);

}

the bufferDigits have a size of 24 bytes, the cant parameter is the number to convert, i receive any question about my code.

I did it, and then but the result is wrong. Nice contradiction — sehe
– sehe, Commented Nov 12, 2011 at 20:07
You are mixing decimal and binary number shifts/divisions. What is the char[] array that would match the sample, 9140? — sehe
– sehe, Commented Nov 12, 2011 at 20:09
jaja @sehe yeah i was confused when i write this :P, when i say "i did it" is when i can convert the digits to bytes and when i say "but the result is wrong" is when some of the digits was wrong converted :P — Diego Fernando Murillo Valenci
– Diego Fernando Murillo Valenci, Commented Nov 12, 2011 at 20:39
when i convert each digit of 9140 the output in the char array is: (0,0,0,9)(for 9),(0,0,0,0)(for 1 ?), (0,0,0,4)(for 4) and (0,0,0,0)(for 0) so i write each group of 4 bytes into the char array from right to left (...some empty bytes...0,0,0,9,0,0,0,0,0,0,0,4,0,0,0,0,...bytes of amount of digits). — Diego Fernando Murillo Valenci
– Diego Fernando Murillo Valenci, Commented Nov 12, 2011 at 20:45
@DiegoFernandoMurilloValenci, did you try my solution below? — Jim Rhodes
– Jim Rhodes, Commented Nov 12, 2011 at 20:49

sehe · Accepted Answer · 2011-11-12 20:19:06Z

2

I feel this is the most c++ way that probably answers your question, if I understood correctly:

#include <string>
#include <iterator>
#include <iostream>
#include <algorithm>

template <typename It>
It tochars(unsigned int i, It out)
{
    It save = out;

    do    *out++ = '0' + i%10;
    while (i/=10);

    std::reverse(save, out);
    return out;
}

int main()
{
    char buf[10];

    char* end = tochars(9140, buf);
    *end = 0; // null terminate

    std::cout << buf << std::endl;
}

answered Nov 12, 2011 at 20:19

sehe

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2 Comments

Diego Fernando Murillo Valenci Over a year ago

well my friend, but that code convert the int to a char array, is like convert int to string, but i need is convert each int digit to a byte[4] and save it the array of char, because i send this array to java and convertit again to an integer.

sehe Over a year ago

You do ... what? Good luck with that.

Jim Rhodes · Accepted Answer · 2011-11-12 20:47:31Z

Instead of using a double and the floor function, just use an int and the modulus operator instead.

void convertCantToByteArray4Digits(unsigned char *bufferDigits,int cant)
{
  int bufferPos=20;
  int cantAux=cant;
  int digit=0,cantDigits=0;
  while(cantAux > 0)
  {
    cout<<"VUELTA"<<endl;
    digit = cantAux % 10;
    cout<<"digit:"<<digit<<endl;
    cantAux /= 10;
    cout<<"cantAux/=10:"<<cantAux<<endl;
    bufferDigits[bufferPos-4] = (digit >> 24) & 0xFF;
    std::cout<<static_cast<int>(bufferDigits[bufferPos-4])<<std::endl;
    bufferDigits[bufferPos-3] = (digit >> 16) & 0xFF;
    std::cout<<static_cast<int>(bufferDigits[bufferPos-3])<<std::endl;
    bufferDigits[bufferPos-2] = (digit >> 8) & 0xFF;
    std::cout<<static_cast<int>(bufferDigits[bufferPos-2])<<std::endl;
    bufferDigits[bufferPos-1] = (digit) & 0xFF;
    std::cout<<static_cast<int>(bufferDigits[bufferPos-1])<<std::endl;
    bufferPos-=4;
    cantDigits++;
  }

Pubby · Accepted Answer · 2011-11-12 20:04:14Z

0

Why not use a union?

union {
  int i;
  char c[4];
};

i = 2530;
// now c is set appropriately

Or memcpy?

memcpy(bufferDigits, &cant, sizeof(int));

answered Nov 12, 2011 at 20:04

Pubby

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3 Comments

Diego Fernando Murillo Valenci Over a year ago

hello thanks for you reply, the union convert each digit of i into a byte[4]? the same occurs for memcpy?

Jim Rhodes Over a year ago

-1 because the problem states: "The idea is separate each digit an convert it in a byte[4]". He wants five, byte[4] entities, one for each of the possible 4 digits and one for the count of digits.

Diego Fernando Murillo Valenci Over a year ago

i use union but for each digit not for the entire number and get the same result, it convert 9140 to a byte array that contain 9040, i need to convert each digit because i send this to JAVA for convert it again to integer.

Kerrek SB · Accepted Answer · 2011-11-12 20:18:48Z

0

Why so complicated? Just divide and take remainders. Here's a reentrant example to which you provide a buffer, and you get back a pointer to the beginning of the converted string:

char * to_string(unsigned int n, char * buf, unsigned int len)
{
  if (len < 1) return buf;

  buf[--len] = 0;

  if (n == 0 && len > 0) { buf[--len] = '0'; }

  while (n != 0 && len > 0) { buf[--len] = '0' + (n % 10); n /= 10; }

  return &buf[len];
}

Usage: char buf[100]; char * s = to_string(4160, buf, 100);

answered Nov 12, 2011 at 20:18

Kerrek SB

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