C: Best way to index the digits in an integer - c

Question

If I have an
int i = 11110001
How would I be able to convert this int into an int array where
int array[8] = {1, 1, 1, 1, 0, 0, 0, 1}

Answer 1

Using a little different approach and snprintf:
#include <stdio.h>
int main (void) {
int i = 11110001;
char arr[9]; //8 digits + \0
int array[8];
if ((snprintf(arr,9,"%d", i) == 8) { //return the 8 characters that were printed
int c;
for(c = 0; c < 8; c++)
array[c] = arr[c] - '0';
}
return 0;
}
P.S: I'm assuming positive values only

Answer 2

You may try like this:
#include <math.h>
char * convertNumber(unsigned int i) {
/* unsigned int length = (int)(log10((float)i)) + 1; */
/* char * arr = (char *) malloc(length * sizeof(char)), * x = arr; */
char * arr = malloc(8);
char * x = arr;
do
{
*x++ = i% 10;
i/= 10;
} while (i != 0);
return arr;
}

Answer 3

Try this :
#include<stdio.h>
void convert_int_to_array(unsigned int);
int main()
{
unsigned int a = 12345678;
convert_int_to_array(a);
return 0;
}
void convert_int_to_array(unsigned int a)
{
int array[25]; // array large enough for an integer
int i = 0, count = 0;
unsigned int num = a;
memset(array, '\0', 20); // I've not included the header file for this.
// gives a warning on compilation.
while(num > 0)
{
array[i] = num % 10;
num = num / 10;
++i;
++count;
}
for(i = count; i>=0;--i)
{
printf("array[%d] = %d\n",i, array[i]);
// or
printf("%d", array[i]);
// dont use both the printf statements, else you will see a
// messed up output.
}
}
BINARY REPRESENTATION :
#include<stdio.h>
struct bit
{
int a : 1;
};
int main()
{
struct bit b;
int d ,f,i;
d=f=256; // take any number of your choice
printf("binary representation of 256:\n");
for(i = 15; i>=0 ; i--) //assuming that the number wont have more than
// 15 DIGITS
{
f=f>>i;
b.a = f;
//d= d>>1;
f=d;
printf("%d",b.a);
}
return 0;
}