Example to find all Armstrong numbers between two integers (entered by the user) using loops and if...else statement.

To understand this example, you should have the knowledge of following C programming topics:

A positive integer is called an Armstrong number of order `n` if

abcd... = a^{n}+ b^{n}+ c^{n}+ d^{n}+ ...

In case of an Armstrong number of 3 digits, the sum of cubes of each digits is equal to the number itself. For example:

153 = 1*1*1 + 5*5*5 + 3*3*3 // 153 is an Armstrong number.

This program is built on the concept of how to check whether an integer is an Armstrong number or not.

```
#include <stdio.h>
#include <math.h>
int main()
{
int low, high, i, temp1, temp2, remainder, n = 0, result = 0;
printf("Enter two numbers(intervals): ");
scanf("%d %d", &low, &high);
printf("Armstrong numbers between %d an %d are: ", low, high);
for(i = low + 1; i < high; ++i)
{
temp2 = i;
temp1 = i;
// number of digits calculation
while (temp1 != 0)
{
temp1 /= 10;
++n;
}
// result contains sum of nth power of its digits
while (temp2 != 0)
{
remainder = temp2 % 10;
result += pow(remainder, n);
temp2 /= 10;
}
// checks if number i is equal to the sum of nth power of its digits
if (result == i) {
printf("%d ", i);
}
// resetting the values to check Armstrong number for next iteration
n = 0;
result = 0;
}
return 0;
}
```

**Output**

Enter two numbers(intervals): 999 99999 Armstrong numbers between 999 an 99999 are: 1634 8208 9474 54748 92727 93084