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?

Start with thinking about how you would do this by hand. Then put that into code. The value in an int is already binary by the way. There are many different ways to do this, especially if you levy what you know about how computers store data. But I'm guessing this is assignment for a class since this is something that most schools have you do early. — ydobonebi
– ydobonebi, Commented Jul 23, 2015 at 4:00
HI Quinn, this is not an assignment anyway. I am far after school. Just I am actually doing more on C#, PHP & JAVA. Not any C language.Printing out as an BINARY is easy from an int, but to store them, I can't catch any ball yet. — Fei Hap Lee
– Fei Hap Lee, Commented Jul 23, 2015 at 4:02
You can't pass an array in C. You have to malloc in the function and return a pointer to the address. — M. Shaw
– M. Shaw, Commented Jul 23, 2015 at 4:04
@M.Shaw struct R { unsigned char ar[8]; }; struct R func(int value) { struct R r; .... return r; } - If you bury a fixed array in a struct, you can return it by-value. The compiler will generate the appropriate code to do it if asked. Whether that is needed, or desirable, for this task is a different issue. — WhozCraig
– WhozCraig, Commented Jul 23, 2015 at 5:54

Johnny Cage · Accepted Answer · 2015-07-23 05:27:51Z

8

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;
}

answered Jul 23, 2015 at 5:27

Johnny Cage

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3 Comments

chux Over a year ago

Minor: /* assert: count <= sizeof(unsigned)*CHAR_BIT && count > 0 */

Johnny Cage Over a year ago

@chux I was on the fence about sizeof(?), since it's evaluating a signed type and only casting to unsigned to use a logical shift. A count of zero shouldn't be flagged as an error, but less than zero is probably a mistake. In practice count should be unsigned.

chux Over a year ago

Asserting count>0 helps with as count == 0 is a problem with 1U << (count-1);. Note count is not cast to unsigned in ` 1U << (count-1)`.

chux · Accepted Answer · 2015-07-23 14:58:04Z

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;
};

edited Jul 23, 2015 at 14:58

chux

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answered Jul 23, 2015 at 11:20

Martin James

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5 Comments

chux Over a year ago

1) numToConvert=0 --> "" rather than "0". 2) Certainly 64 enough for 0x1234ABCD, but I'd expect 65 or 33.

Martin James Over a year ago

@chux - sure, but that would require another function to check for 0. Easy enough, I just didn't bother with it. Buffer size: I thought 32, then 33 for the nul, then rounded up to 64:)

chux Over a year ago

"require another function to check for 0". Hmmm, maybe a small re-write instead: if (answer > 1) {result=binaryToAbits(answer>>1,result); } *result='0'+(answer & 1); return result+1; Somewhat simple and yet handles 0 well.

Martin James Over a year ago

@chux good idea! With your permission, I would edit it in and add an attribution.

Martin James Over a year ago

In fact, sod it, I'm gonna add it anyway. If you object, I'll remove it, or you can:)

Marcelo de Mattos Nascimento · Accepted Answer · 2018-08-22 17:29:03Z

2

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;
 }

edited Aug 22, 2018 at 17:29

answered Aug 22, 2018 at 15:16

Marcelo de Mattos Nascimento

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Comments

ydobonebi · Accepted Answer · 2015-07-23 04:15:17Z

0

Using bitwise logic :

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

edited Jul 23, 2015 at 4:15

answered Jul 23, 2015 at 4:07

ydobonebi

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10 Comments

ydobonebi Over a year ago

You could do it faster with exploiting C unions and structs btw.

M. Shaw Over a year ago

sizeof(int) is 4, not 8.

M. Shaw Over a year ago

And it should be pow(2, 7-i) so the array isn't reversed.

ydobonebi Over a year ago

@M.Shaw Oops, I was excited about my bits I forgot sizeof is more into bytes

M. Shaw Over a year ago

sizeof(int) * 8 is 32. OP said the maximum array length will not exceed 8 elements.

|

TryinHard · Accepted Answer · 2015-07-23 04:37:42Z

0

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*/` ;
}

answered Jul 23, 2015 at 4:37

TryinHard

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1 Comment

chux Over a year ago

Note: Better to use 1u << 31 as 1 << 31 is undefined behavior with 32-bit int.

M. Shaw · Accepted Answer · 2015-07-23 13:18:16Z

0

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.

edited Jul 23, 2015 at 13:18

answered Jul 23, 2015 at 4:01

M. Shaw

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3 Comments

chux Over a year ago

Not enough memory int *bin = (int * ) malloc(8); --> int *bin = malloc(8 * sizeof *bin);

M. Shaw Over a year ago

@chux The OP needs 8 bits stored in an array. I should probably change the type of bin to unsigned char or uint8_t instead. The OP needs 8 bits and I'm already allocating 64. There is no reason at all I would need 256 bits of memory to store 8 bits of data.

chux Over a year ago

"The OP needs 8 bits and I'm already allocating 64" is not correct. Code was allocating 8 as in malloc(8), which was certainly insufficient memory for 8 int - hence the comment. IAC, the malloc(8) now is sufficient memory for 8 uint8_t. But now it has a new issue: returning a uint8_t * that is cast to a int *. With 4-byte int, the calling code could reference the first 2 of the int* array, but certainly OP was expecting 8.

takasoft · Accepted Answer · 2018-02-21 05:31:33Z

0

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);
}

answered Feb 21, 2018 at 5:31

takasoft

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Comments

Kaustav Ray · Accepted Answer · 2015-07-23 04:07:23Z

-1

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;
}

edited Jul 23, 2015 at 4:07

answered Jul 23, 2015 at 4:01

Kaustav Ray

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3 Comments

M. Shaw Over a year ago

Your array will still be reversed.

DanZimm Over a year ago

@KaustavRay possibly because recursive usually comes with being side effect free, but this is not side effect free

Kaustav Ray Over a year ago

Yes that is true ! it is always less efficient but it was nowhere wrong ! Anyways I learnt ! :)

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