Reverse Number In C

[ReaseySo]

#include <stdio.h>

Reverse Number In C Program
int main()
{
int n, reverse = 0;

   printf("Enter a number to reverse\n");
   scanf("%d",&n);

   while (n != 0)
   {
      reverse = reverse * 10;
      reverse = reverse + n%10;
      n = n/10;
   }

   printf("Reverse of entered number is = %d\n", reverse);

   return 0;
}

[Banfa]

What is your question, your code seems to do what is intended?

[shashi rai]

please explain me about step 6 and 7.

[BobS0327]

I hope the following explanation is helpful. Unfortunately, I can't find any preview option, so I'm not sure how the post will appear

I've iterated line by line thru the while loop in your posted coded as indicated below.

Assume n (input) is 1234

Since reverse is initially zero

 reverse = reverse * 10;

will result in reverse still being equal to zero

Since n is equal to 1234, using the modulus operator will produce 123.4 and the modulus operator will return 4 which is entered into the reverse variable
Remember the reverse is initially set to zero. So, zero plus 4 equals 4

reverse = reverse + n%10;

The moudulus operator (percent sign) computes the remainder that results from performing integer division.

At this point reverse now contains 4

We now divide n by 10. Integer division will only return the whole excluding the remainder. So, 1234/10 equals 123. n now contains 123.

n = n/10;

We now loop back up to...

reverse = reverse * 10;

Since reverse contains 4, multiplying it by 10 will result in reverse now containing 40

Now we again use the modulus operator on n. Remember that n now contains 123. Using the modulus operator will return a remainder of 3 (123/10 gives 12 with a remainder of 3). To reiterate, the modulus operator will ONLY return the remainder value of 3. Thus, the following statement is adding 3 to the value of reverse which is 40, making reverse now equal to 43

reverse = reverse + n%10;

The next statement divides n which contains 123 by 10 resulting in n containing the whole number 12. The remainder 3 is discarded,

 n = n/10;

We again loop back up to..

reverse = reverse * 10;

reverse contains 43 prior to executing the above statement. Thus, multiplying reverse by 10 will make reverse equal 430

Keep in mind prior to executing the next statement that n is equal to 12 and reverse is equal to 430. So, using the modulus operator on n will result in it generating a remainder of 2 which is added to reverse, making reverse equal to 432

reverse = reverse + n%10;

The current value of n before executing the next statement is 12. After executing the next statement, n will contain 1 size integer division will return whole numbers exclusive of remainders (12/10 = 1)

n = n/10;

Finally, we loop back one last time to

reverse = reverse * 10;

reverse contains 432 prior to executing the previous statement. So, when we multiple reverse by 10, reverse will now contain 4320

At this point n contains 1. using the modulus operator will generate a remainder of 1 which is added to reverse, making reverse equal to 4321

 reverse = reverse + n%10;

Executing the next statement returns an integer value of ZERO which will cause the while loop to break since n is equal to ZERO

n = n/10;

The end result is having reverse contain 4321.

[jnawrocki]

Cool but probably slow to execute due to all the math. Seems like handling it as an array of characters and using xor swap would be faster.