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luc100.c

Problem Statement

Determine if an animal is Carnivore/Herbivore and its type (Canine, Feline, Cetacean, Marsupial) based on bits in an integer.

Metadata

Property Detail
Author Amit Dutta amitdutta4255@gmail.com
Date 08 Feb 2026
License MIT License (See the LICENSE file for details)
Difficulty Beginner (index: 2 / 10)

Concepts

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  • Array
  • Sorting (possible)
  • Iteration

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Source Code

#include <stdio.h>
#include <stdlib.h>

struct animal
{
    char name[30];
    int type;
};

int main()
{
    struct animal a = {"OCELOT", 18};
    int type = a.type;

    printf("Animal: %s\n", a.name);

    // Check Bit 4 for Diet (Assuming 1=Carnivore, 0=Herbivore based on context)
    if (type & (1 << 4))
        printf("Diet: Carnivore\n");
    else
        printf("Diet: Herbivore\n");

    // Check Bits 0-3 for Family
    printf("Family: ");
    if (type & (1 << 0)) printf("Canine ");
    if (type & (1 << 1)) printf("Feline ");
    if (type & (1 << 2)) printf("Cetacean ");
    if (type & (1 << 3)) printf("Marsupial ");
    printf("\n");

    return 0;
}

Explanation

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    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 (luc100.c):

    #include <stdio.h>
    #include <stdlib.h>

    struct animal
    {
        char name[30];
        int type;
    };

    int main()
    {
        struct animal a = {"OCELOT", 18};
        int type = a.type;

        printf("Animal: %s\n", a.name);

        // Check Bit 4 for Diet (Assuming 1=Carnivore, 0=Herbivore based on context)
        if (type & (1 << 4))
            printf("Diet: Carnivore\n");
        else
            printf("Diet: Herbivore\n");

        // Check Bits 0-3 for Family
        printf("Family: ");
        if (type & (1 << 0)) printf("Canine ");
        if (type & (1 << 1)) printf("Feline ");
        if (type & (1 << 2)) printf("Cetacean ");
        if (type & (1 << 3)) printf("Marsupial ");
        printf("\n");

        return 0;
    }