Convert integer input to digit using array in C - arrays

Question

#include <stdio.h>
void seperate(int intNum);
int main()
{
int x;
printf("Please enter an integer: \n");
scanf("%d",&x);
seperate(x);
}
void seperate(int intNum)
{
int i,count=0;
while (intNum!=0)
{
intNum/=10;
++count; //to calculate the number of digits for user input = size of array
}
int array[count];
printf("The number on seperate line as follows:\n");
for(i=count-1; i>=0; i--)
{
array[i]= intNum%10;
intNum /= 10;
}
for(i=0; i<=count-1; i++)
{
printf("%d\n", array[i]);
}
}
The expected output(it works when the array size is fixed):
Please enter an integer:
42568
The number on separate line as follows:
4
2
5
6
8
The output:
Please enter an integer:
42568
The number on separate line as follows:
0
0
0
0
0
The code works only if the array sized is fixed, how to solve this problem? Does it not work if I create an array without declaring the size?
EDIT: the value for variable intNum changed in while loop, nothing to do with the declaration of the array

Answer 1

Your while loop, in which you count the number of digits, is changing the given intNum variable, which will be zero at the end of that loop.
You should make a copy of intNum and use/modify that in the first (while) loop, like this:
void seperate(int intNum)
{
int i, count = 0, temp = intNum; // Make a copy to use in the "count" loop
while (temp != 0) {
temp /= 10;
++count; //to calculate the number of digits for user input = size of array
}
if (count == 0) count = 1; // To take care of the case when intNum == 0
int array[count];
printf("The number on seperate line as follows:\n");
for (i = count - 1; i >= 0; i--) {
array[i] = intNum % 10;
intNum /= 10;
}
for (i = 0; i <= count - 1; i++) {
printf("%d\n", array[i]);
}
}
Alternatively, you could save the value of intNum and restore that after the while loop has completed:
void seperate(int intNum)
{
int i, count = 0, save = intNum; // Save the original value of intNum ...
while (intNum != 0) {
intNum /= 10;
++count; //to calculate the number of digits for user input = size of array
}
intNum = save; // ... and restore it after counting the digits
if (!count) count = 1;
int array[count];
//...
EDIT: There is a much simpler and shorter solution, with only one loop, using the sprintf function to get the digits (in the right order) for you:
void seperate(int intNum)
{
char output[100];// Should be long enough for the largest integer.
sprintf(output, "%d", intNum);
printf("The number on seperate line as follows:\n");
for (char* cp = output; *cp; ++cp) {
printf("%c\n", *cp);
}
}

Answer 2

You could simplify this problem with a little math. First,
#include <math.h>
Then, the count of digits can be determined by calculating log10intNum. And you only need a single loop. Something like,
void seperate(int intNum)
{
printf("The number on seperate line as follows:\n");
double log10 = log(10);
while (intNum > 0) {
int count = (int) (log(intNum) / log10);
int pow10 = pow(10, count);
int digit = (int) (intNum / pow10);
printf("%d\n", digit);
intNum -= digit * pow10;
}
}