Digit sum, reverse and palindrome

LAB A04 · 01

Understand the task.

Aim: Digit sum, reverse and palindrome.

Review the related theory in Unit 2.

LAB A04 · 02

Know what goes in and what comes out.

One integer from 0 to 999999999. This bound keeps the reverse within the range of standard long. Leading zeros are not retained as part of an integer value.

121

Digit sum = 4
Reverse = 121
Palindrome = YES

Source programs read the documented input directly, without prompt text, so the same sample works in the embedded compiler’s STDIN field.

LAB A04 · 03

Plan the steps before writing code.

1. Store the input in original and copy it into remaining. Initialize sum and reverse to zero.
2. Extract the last digit with remaining % 10.
3. Add the digit to sum and append it to reverse using reverse * 10 + digit.
4. Remove the last digit with remaining / 10 and repeat while remaining is positive.
5. Compare reverse with original and display all three results.

LAB A04 · 04

One complete program. Ready to run.

#include <stdio.h>
int main(void)
{
    long original, remaining, digit, sum = 0, reverse = 0;
    if (scanf("%ld", &original) != 1 || original < 0 || original > 999999999L) {
        puts("Enter an integer from 0 to 999999999.");
        return 1;
    }
    remaining = original;
    do {
        digit = remaining % 10;
        sum += digit;
        reverse = reverse * 10 + digit;
        remaining /= 10;
    } while (remaining > 0);
    printf("Digit sum = %ld\nReverse = %ld\n", sum, reverse);
    printf("Palindrome = %s\n", original == reverse ? "YES" : "NO");
    return 0;
}

gcc A04-digits-reverse-palindrome.c -o program

LAB A04 · 05

Connect each statement to its purpose.

• The do-while loop processes zero once, correctly giving sum 0, reverse 0 and palindrome YES.
• original stays unchanged because the loop repeatedly shortens remaining.
• The reverse accumulator shifts previously collected digits one decimal position before appending the new digit.
• A number ending in zeros loses those zeros at the front of its reversed numeric value: 1200 reverses to 21.

LAB A04 · 06

Follow the sample calculation.

LAB A04 · 07

Check more than one input.

Time and storage

O(d) time for d decimal digits and O(1) extra space.

Start with the sample, then test a boundary and a case that changes the result. A correct sample alone does not establish that every input is handled correctly.

LAB A04 · 08

Use the real C compiler.

Choose an example and load it into the embedded C compiler. Select I/O inside the compiler, check the sample STDIN, and press Run. Change the input or edit the source to explore a different case.

121

Digit sum = 4
Reverse = 121
Palindrome = YES

#include <stdio.h>
int main(void)
{
    long original, remaining, digit, sum = 0, reverse = 0;
    if (scanf("%ld", &original) != 1 || original < 0 || original > 999999999L) {
        puts("Enter an integer from 0 to 999999999.");
        return 1;
    }
    remaining = original;
    do {
        digit = remaining % 10;
        sum += digit;
        reverse = reverse * 10 + digit;
        remaining /= 10;
    } while (remaining > 0);
    printf("Digit sum = %ld\nReverse = %ld\n", sum, reverse);
    printf("Palindrome = %s\n", original == reverse ? "YES" : "NO");
    return 0;
}

LAB A04 · 09

Explain why the program works.

LAB A04 · 10

Build the next skill.

Read the viva answers, try the extra cases, and return to the lab index to mark this exercise as practised.