C - Convert Integer to Binary Array

Question

I am very new to the C language. I will need a small program to convert int to binary and the binary preferably stored in an array so that I can further break them apart for decoding purpose. I have following:

#include <stdio.h>
#include <stdlib.h>

int main()
{
    int arr[20];
    int dec = 40;
    int i = 0, ArrLen;

    if(dec > 0)
    {
        while(dec > 0)
        {
            arr[i] = dec % 2;
            i++;
            dec = dec / 2;
        }
    }
    else
    {
        printf("Invalid Number");
    }
}

From the code above, I can store the binary value in arr. But instead of getting the binary equivalent: 101000, the array now is like {0, 0, 0, 1, 0, 1}, which is the reversed of the correct answer. So the question is, how to get an array in the correct order or possibly flip it? I have one thing for sure, and that is the maximum array length will not exceed 8 elements.

This conversion will be used repeatedly. So I plan to put it in a function so I'd be able to call the function, pass in an integer, and then get the array as the return value. So another question is, is it feasible to get an array as the return value?

Answer 1

You can parameterize the array by using a pointer to int. It may be useful to parameterize the number of digits as well.

void int_to_bin_digit(unsigned int in, int count, int* out)
{
    /* assert: count <= sizeof(int)*CHAR_BIT */
    unsigned int mask = 1U << (count-1);
    int i;
    for (i = 0; i < count; i++) {
        out[i] = (in & mask) ? 1 : 0;
        in <<= 1;
    }
}

int main(int argc, char* argv[])
{
    int digit[8];
    int_to_bin_digit(40, 8, digit);
    return 0;
}

Answer 2

or recursive V2.0:

#include <stdio.h>

char *binaryToAbits(unsigned int answer, char *result) {
  if(answer==0) return result;
  else {
    result=binaryToAbits(answer>>1,result);
    *result='0'+(answer & 0x01);
    return result+1;
  }
}

int main(void) {
    unsigned int numToConvert=0x1234ABCD;
    char ascResult[64];
    *binaryToAbits(numToConvert,ascResult)='\0';
    printf("%s",ascResult);
    return 0;
}

Note, thanks to @chux, here is a better recursive function that handles the case of converting 0 - it outputs "0" instead of "":

char *binaryToAbits(unsigned int answer, char *result) {
  if(answer>1) {
    result=binaryToAbits(answer>>1,result);
  }
  *result='0'+(answer & 0x01);
  return result+1;
};

Answer 3

Same answer as Johnny Cage with the addtion of a function to get the length of a digit

#include <math.h>

int bit_len(unsigned int n){
   return floor(log(n)/log(2))+1;
}
void int_to_bin_digit(unsigned int in, int len_digitis,int* out_digit){

  unsigned int mask = 1U << (len_digitis-1);
  int i;
  for (i = 0; i < len_digitis; i++) {
    out_digit[i] = (in & mask) ? 1 : 0;
    in <<= 1;
  }
}

int main(int argc, char* argv[]){
   int number = 30;
   int len = bit_len(number);
   int digits[len];
   int_to_bin_digit(number,len, digits);
   for( int i =0;i<len;i++){
       printf("%d",digits[i]);
   }
  return 0;
 }

Answer 4

Try something like this:

uint8_t * intToBin(int x) {
    uint8_t *bin = (int *) malloc(8);
    uint8_t i = 0;
    int mask = 0x80;
    for (i = 0; i < 8; i++) {
        bin[i] = (x & mask) >> (7-i);
        mask >>= 1;
    }
    return bin;
}

Include <stdint.h> for the declaration of uint8_t. Remember to free the malloc-ed memory if you don't want memory leaks.

Answer 5

Using bitwise logic :

for(int i = 0 ; i < 8 ; i++)
{
    bytearray[i] = inputint & pow(2,7-i);
}

Answer 6

This may be helpful:

void binary(unsigned n)
{
    unsigned i;
    for (i = 1 << 31; i > 0; i = i / 2)
        (n & i)?`/*STORE 1*/` : `/*STORE 0*/` ;
}

Answer 7

This should work.

#include <stdio.h>

void intToBin(int dec, int bin[], int numBits){
    for(int i = 0; i < numBits; i++){
        bin[i] = 1 & (dec >> i);
    }
}

void printArr(int arr[], int arrSize){
    for(int i = 0; i < arrSize; i++) {
        printf("%d ", arr[i]);
    }
}

int main(int argc, char* argv[]){
    int bin[32];
    intToBin(-15, bin, 32); 
    printArr(bin, 32);
}

Answer 8

Recursive Implementation:

(Since you cannot have the prior idea of number of digits(0/1) in the binary format of the given number)

int arr[200]; //for storing the binary representation of num
int i=0; // to keep the count of the no of digits in the binary representation

void calBinary(int n) // function to recalculate
{
   if(n>1)
      calBinary(n/2);
   arr[i++]=n%2;
}