Int array to store large numbers (over 20 digits) - c

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]));

Answer 2

#include<stdio.h>
#include<stdlib.h>
int *toArray(long long int n, int counter); int main() {
long long int n=0,copy_of_n=0;
int counter=0, i=0;
int *res=NULL;
printf("Input: ");
scanf("%lld", &n);//Refer note in the answer
copy_of_n=n;//see the while loop why use this.
while(n>0){
n/=10;//Good, but you are not storing it. copy_of_n does that.
counter++;//You didn't initialize this
}
res=toArray(copy_of_n, counter);
for(i=counter-1;i>=0;i--){//This ensures you get the numbers in a proper order and since index starts from 0 in the toArray function
printf("%d", res[i]);
}
printf("\n");
free(res);//You forgot this
return 0; }
int *toArray(long long int n, int counter) {
int i=0;
int *numberArray = malloc(sizeof(int) * counter);
memset(numberArray,0x00,counter*sizeof(int));//Good to do this
//Check for memory allocation
for(i=0; i< counter; i++)
{
numberArray[i] = n % 10;
n=n/10LL;//You have to remove the digits too
}
return numberArray; }
NOTE: Read in a while loop the integers in small parts because long long has limits of how many digits it can store. Another approach would be to store in a string and send split the string into exactly the max. no. of digits that can be stored in long long in the n variable and send it to the function, part by part.
long long size is implementation dependent, hence ensure you check the max. no. of digits it can accomodate.(It will ofcourse bypass your 9-digit limit as asked in question)