Question

I have an assignment that requires me to first set up integer arrays to store arbitrarily large numbers. By using each element of an int array, I can hold one digit per element because int variable types are 32 bits and can only reach up to about 2 billion before failing. I know that there are other libraries using BigInt but I wanted to make my own arrays instead.

I tested my code out, but it doesn’t seem to be accepting values past 9 digits until it starts to fail.

int * toArray(int, int);
int main()
{
    int n, *res, counter, i;
    printf("Input: ");
    scanf("%d", &n);
    while(n>0){
         n/=10;
         counter++;
    }
    res = toArray(n, counter);
    for(i=1;i<=counter;i++){
         printf("%d", *(res)+i);
    }
    printf("\n");
}   

int * toArray(int n, int counter)
{
    int i;
    int *numberArray = malloc(sizeof(int) * counter);
    for(i=0; i< counter; i++)
    {   
         numberArray[i] = n % 10;
    }
    return numberArray;
}

I want to be able to accept close to twenty digits at the very least. I know this can be done with int arrays, although char arrays (or strings) are also a possibility. But I wanted to use int arrays instead. Any help in understanding why it fails around 9 digits and suggestions for going past that would be very much appreciated. Thank you.

Answer 1

The problem is that you are reading an int from keyboard

scanf("%d", &n);

so therefore no matter how many digits you enter, you still will only get 9 digits.

In order to be able to enter an arbitrary number you would have to read it as a string instead, then convert it to your int array.

EDIT:

this way (for instance) would allow for 20 digits

  char ch;
  int digits[20];
  int i = 0;
  do
  {
    ch = fgetc(stdin);
    if ( isdigit(ch) )
    {
      digits[i++] = ch - 48; // convert ASCII to int
    }
  }
  while (isdigit(ch) && i < sizeof(digits)/sizeof(digits[0]));