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¶
Beta Feature
This concept detection system is still in beta and may occasionally show incorrect or incomplete results.
- Array
- Sorting (possible)
- Iteration
Actions¶
You can print or save this file by opening Raw and using your browser.
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¶
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 (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;
}