convert integer to 32-bit binary in c

Question

I'm writing a program to convert an integer to 32-bit binary. The problem is with the output - it comes backwards.

#include <stdio.h>

int main() {
    long number, binary, num2;

    printf("Enter an integer: ");
    scanf("%ld", &number);

    for (num2 = (number * 2) / 2; num2 > 0; num2 /= 2) {
        binary = num2 % 2;
        printf("%ld", binary);
    }
    putchar('\n');
    return 0;
}

So if I put '6' it shows as 011 and it has to be 110

Also, how do I output the rest of '0's? So that the whole output in this case would be:

00000000 00000000 00000000 00000110 

Answer 1

You compute digits starting from the right, which is why your output shows the right-most digit first. Here is a way that starts from the left, using a bitmask, and does not convert your value to unsigned which may change the bits:

#include <stdio.h>
#include <limits.h>

int main()
{
    long number;
    if ( 1 != scanf("%ld", &number) )
        return 1;

    // sign bit  (cannot use 1L left-shift as that causes UB)
    putchar( '0' + (number < 0) );

    // value bits
    for (long bit = 1L << (CHAR_BIT * sizeof number - 2); bit; bit >>= 1)
        putchar( '0' + !!(number & bit) );

    putchar('\n');
}

Answer 2

It's so much easier to use a recursive function for this:

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

void printInBinary(long num, int bit)
{
   if ( bit >= 8*sizeof(num) )
   {
      return;
   }

   printInBinary(num/2, bit+1);
   printf("%ld", num%2);

   if ( bit%8 == 0 )
   {
      printf(" ");
   }
   if ( bit == 0 )
   {
      printf("\n");
   }
}

int main()
{
   int y = 31;
   uint32_t x1 = (1 << y );
   uint32_t x2 = (1u << y );
   printf("x1: %u\n", x1);
   printInBinary(x1, 0);

   printf("x2: %u\n", x2);
   printInBinary(x2, 0);
}

Output:

x1: 2147483648
00000000 00000000 00000000 00000000 10000000 00000000 00000000 00000000 
x2: 2147483648
00000000 00000000 00000000 00000000 10000000 00000000 00000000 00000000 

PS If you used uint32_t for num instead of long, you will get the 32 bits of output.

Answer 3

If you want to print the binary number, you should print the bits backwards. Watch this:

6 (10):

6 / 2 = 3, rem => 0

3 / 2 = 1, rem => 1

1 / 2 = 0, rem => 1

Hence 6 (2) = 110

You are printing it the forward order which will give you 011. So you should keep the binary bits by putting them in a variable and finally print them back.

Try This

#include <stdio.h>
  int main(){
  long number, binary, num2;
  int i = 0, j;
  char num[100];

  printf("Enter an integer: ");
  scanf("%ld", &number);

  while (number != 0){
    num[i] = num2 % 2;
    number /= 2;
    i++;
  }
  for (j = 0; j < 32; j++) {
    if (j > i) {
      printf("0");
    }
    else {
      printf("%c", num[i]);
      i--
    }
  }
  printf("\n");
  return 0;
}

Answer 4

Here is a simple portable implementation for 32-bit numbers:

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

int main() {
    long input;  // at least 32 bits
    unsigned long number;
    int i;

    printf("Enter an integer: ");
    if (scanf("%ld", &input) != 1) 
        return 1;
    /* copy the bit pattern to an unsigned long */
    memcpy(&number, &input, sizeof number);
    for (i = 32; i-- > 0;) {
        if (i > 0 && (i & 7) == 0)
            putchar(' ');
        putchar('0' + (int)((number >> i) & 1));
    }
    putchar('\n');
    return 0;
}

Answer 5

Simply examine the bits in the order in which you want them output. I have used a unsigned long cast for the shifting, because the result of shifting bits of a signed value into the sign bit is undefined.

    #include <stdio.h>
    #include <limits.h>

    int main() {
        long number = 0;
        int i;

        printf("Enter an integer: ");
        scanf("%ld", &number);
        for(i=0; i<sizeof(number)*CHAR_BIT; i++) {
            if (number < 0)
                printf ("1");
            else
                printf ("0");
            if ((i % CHAR_BIT) == CHAR_BIT - 1)
                printf (" ");
            number = (long)((unsigned long)(number) << 1);
        }
        printf("\n");
    return 0;
    }

Program output:

Enter an integer: 6
00000000 00000000 00000000 00000110

Answer 6

#include <stdio.h>

int main(){
    // Assuming 32 bit architecture.
    unsigned long number;

    // Initialize a null-terminated char-array
    // of zeros (ASCII value 48).
    char binary[33] = {[0 ... 31] = 48, [32] = 0}; 

    printf("Enter an integer: ");
    scanf("%lu", &number);

    for(int i = 0; number > 0; ++i) {
        binary[31 - i] = number % 2;
        number /= 2;
    }
    printf("%s\n", binary);
    return 0;
}