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pc-ip-02.c

Problem Statement

Question 2: Write a program to reverse a non-negative integer using a function.

Metadata

Property Detail
Author Amit Dutta (amitdutta4255@gmail.com)
License MIT
Difficulty Beginner (index: 2 / 10)

Concepts

Beta Feature

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  • Recursion
  • Pointers
  • Iteration

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Source Code

#include <stdio.h>

int reverse(int);

int main()
{
    int num;
    printf("Enter the number: ");
    scanf("%d", &num);
    if (num < 0)
    {
        printf("\nOnly poitive integers are allowed.");
        return 1;
    }
    printf("\nReverse of input %d is : %d", num, reverse(num));
    return 0;
}

int reverse(int num)
{
    int result = 0;
    while (num > 0)
    {
        result = (result * 10) + (num % 10);
        num /= 10;
    }
    return result;
}

Explanation

Explain with AI

Copy the prompt below and paste it into any AI assistant.

    You are explaining a C programming code to a beginner.

    STRICT RULES:

    - Only use the given code. Do NOT assume anything not present.

    - Do NOT add extra examples.

    - Keep explanation clear and short.

    - If something is unclear, say "Not clear from code".

    - Follow the exact format below. Do NOT change headings.

    FORMAT:

    [START]

    ## What it does

    (Explain the overall purpose in 1-2 sentences)

    ## Step-by-step

    (Explain how the code works in steps, simple language)

    ## Key Concepts

    (List concepts like loop, condition, function, etc.)

    ## Notes

    (Mention any limitations, errors, or assumptions)

    [END]

    CODE (pc-ip-02.c):

    #include <stdio.h>

    int reverse(int);

    int main()
    {
        int num;
        printf("Enter the number: ");
        scanf("%d", &num);
        if (num < 0)
        {
            printf("\nOnly poitive integers are allowed.");
            return 1;
        }
        printf("\nReverse of input %d is : %d", num, reverse(num));
        return 0;
    }

    int reverse(int num)
    {
        int result = 0;
        while (num > 0)
        {
            result = (result * 10) + (num % 10);
            num /= 10;
        }
        return result;
    }