How to put an integer to an array of digits [closed]

Answer 1

"Can i define an array with unknown number of elements ?"

If the number is too large you can input it as string and then accordingly extract digits from it

Something like following :

char buf[128];
int *array;
//fscanf(stdin,"%s",buf);

array = malloc(strlen(buf) * sizeof(int)); //Allocate Memory
int i=0;
do{
 array[i] = buf[i]-'0'; //get the number from ASCII subtract 48
 }while(buf[++i]); // Loop till last but one 

Answer 2

int x[6];
int n=123456;
int i=0;
while(n>0){
   x[i]=n%10;
   n=n/10;
   i++;
}

Answer 3

Here are teh steps. First, get the size needed to store all the digits in the number -- do a malloc of an array. Next, take the mod of the number and then divide the number by 10. Keep doing this till you exhaust all digits in the number.

Answer 4

First calculate no of digits

int count = 0;
int n = number;

while (n != 0)
{
    n /= 10;
    cout++;
}

Now intialize the array and assign the size:

if(count!=0){
   int numberArray[count];

   count = 0;    
   n = number;

   while (n != 0){
       numberArray[count] = n % 10;
       n /= 10;
       count++;
   }
}

Answer 5

If you don't mind using char as the array element type, you can use snprintf():

char digits[32];
snprintf(digits, sizeof(digits), "%d", number);

Each digit will be represented as the character values '0' though '9'. To get the integer value, subtract the character value by '0'.

int digit_value = digits[x] - '0';