Program to reverse a number without using an array

Question

This is the code that I used. It works perfectly, but I don't understand why it works. I just kept changing my original logic until I started using -1 in the counter loop.

#include<stdio.h>
#include<math.h>

int main(){
    int number, reverse, sum =0;
    scanf("%d", &number);
    int temp = number;
    int ctr;

    for(ctr = -1; temp!=0; ctr++)
        temp = temp/10;

    while(number)
    {
        sum = sum + (number%10 * (pow(10, ctr--)));
        number = number/10; 
    }
    printf("%d", sum);

    return 0;
}

Answer 1

same basic math but so much easier to understand:

unsigned int number = 123456789;
unsigned int reversed = 0;
do {
    reversed *= 10;
    reversed += number % 10;
    number /= 10; 
} while (number > 0);

Answer 2

Although this is kind of awkward to explain the logic of the code that you wrote it yourself, I understood the logic behind it.

Indeed, I was looking for a solution to solve this problem: How to reverse any positive integer without using arrays or indexed variables to which your code was the solution I was looking for.

The logic of your programme is like this:

you first count the digits of the given number in the FOR-loop in the shortest possible way. you need this because you have to 10-power each remainder up to digits of the given number in the next WHILE-loop and you store/add the result of this each time in the variable named SUM. The final result of this would be the whole integer to be reversed.

PS1:

You do not need the variable named REVERSE. Just drop it or replace SUM with it.

PS2: You really do not need to go that long way to do this. Here's another shorter version:

#include <stdio.h>

void main (void){
    unsigned short int uNum, nReversed;

    puts ("Enter a positive integer number: ");
    scanf ("%hu", &uNum);

    while (uNum != 0){
        nReversed = uNum%10 + nReversed*10 ;
        uNum /= 10;
    }
    printf ("\n\n\nThat is %d", nReversed);
}