8051 microcontroller tutorial

Write a program that displays the position of a character ch in the string S or – 1 if S doesn’t contain ch. || C Lab Program

noreply@blogger.com (belgaumboy) — Fri, 3 Apr 2026 17:26:00 +0530

WAP_E04: Write a C program that displays the position of a character ch in the string S or – 1 if S doesn’t contain ch. || Strings

WAP_E04: C Lab Program

//position of a character ch in the string S or - 1 if S doesn't contain ch.
#include <stdio.h>
#include <string.h>

int main() {
char str[100], ch;
char *position;

// Input string and character
printf("Enter the string: ");
fgets(str, sizeof(str), stdin);

printf("Enter the character to find: ");
scanf("%c", &ch);

// Use strchr to find the first occurrence of ch in string S
position = strchr(str, ch);

// Check if character is found
if (position) {
// Output the index of the character
printf("Position of '%c' in string: %d\n", ch, position - str);
} else {
printf("-1\n"); // If the character is not found
}

return 0;
}

OUTPUT

Enter the string: HelloWorld!
Enter the character to find: W
Position of 'W' in string: 5

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Credit Card Validator || 16 Intermediate Level C++ Projects

noreply@blogger.com (belgaumboy) — Fri, 3 Apr 2026 12:37:00 +0530

Summary:

A high-precision C++ utility designed to verify the structural validity of credit card numbers using the Luhn checksum. This project demonstrates string manipulation, algorithmic logic, and robust input sanitization.

Learning Objectives:

Algorithmic Mastery: Implement the Luhn Algorithm logic (doubling, digit-summing, and modulo-10).
String Processing: Learn to handle numeric data stored as std::string to bypass long long overflow limits.
Input Sanitization: Differentiate between "Invalid Format" (non-digits) and "Invalid Checksum" (bad math).
OOP Encapsulation: Organize logic into a cohesive CCValidator class.

The Source Code (main.cpp)

This code is optimized for the MinGW compiler in Code::Blocks. It includes <limits> to prevent the "console closing immediately" bug and <iomanip> for clean output formatting.

#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
#include <limits> // Required for clearing input buffer in Code::Blocks

using namespace std;

class CCValidator {
private:
// Helper: Sums digits of a number (e.g., 14 becomes 1+4=5)
int getDigitSum(int n) {
return (n < 10) ? n : (n / 10 + n % 10);
}

public:
bool isValid(string cardNo) {
int nSum = 0;
bool isSecond = false;

// Iterate from right to left
for (int i = cardNo.length() - 1; i >= 0; i--) {
// Basic sanitization: check if character is a digit
if (!isdigit(cardNo[i])) return false;

int d = cardNo[i] - '0';

if (isSecond == true) {
d = d * 2;
d = getDigitSum(d);
}

nSum += d;
isSecond = !isSecond;
}

return (nSum % 10 == 0);
}
};

int main() {
CCValidator validator;
string input;

cout << "========================================" << endl;
cout << " ACADEMIC CREDIT CARD VALIDATOR " << endl;
cout << "========================================" << endl;

while (true) {
cout << "\nEnter card number (or 'exit' to quit): ";
cin >> input;

if (input == "exit") break;

if (validator.isValid(input)) {
cout << ">> Result: [VALID] The checksum is correct." << endl;
} else {
cout << ">> Result: [INVALID] Checksum mismatch or bad format." << endl;
}

// Professor's Note: Standard Code::Blocks/MinGW buffer clear
cin.clear();
cin.ignore(numeric_limits<streamsize>::max(), '\n');
}

cout << "\nProgram terminated. Press Enter to close...";
cin.get();
return 0;
}

Execution Trace (Sample Session)

========================================
ACADEMIC CREDIT CARD VALIDATOR
========================================

-- Case 1: Happy Path (Valid Input) --
Enter card number (or 'exit' to quit): 79927398713
>> Result: [VALID] The checksum is correct.

-- Case 2: Checksum Mismatch (Invalid Card) --
Enter card number (or 'exit' to quit): 79927398710
>> Result: [INVALID] Checksum mismatch or bad format.

-- Case 3: Error Handling (Non-numeric Input) --
Enter card number (or 'exit' to quit): 1234-5678-ABCD
>> Result: [INVALID] Checksum mismatch or bad format.

Enter card number (or 'exit' to quit): exit

Academic "Learning Corner"

The String Over Long Long Debate

In many introductory courses, students try to use long long cardNum;. Warning: A standard credit card is 16 digits. While long long can technically hold up to $\approx 1.8 \times 10^{19}$, reading it via cin >> longLongVar often fails if the user enters spaces or dashes. Using std::string is the "Architect's Choice" because it allows us to iterate through digits easily and handle cards of any length (like 15-digit Amex or 19-digit specialty cards).

The MinGW cin.ignore() Trick

If you run this in Code::Blocks and the window vanishes instantly, it's because the trailing newline character from your last input is still in the buffer. The line cin.ignore(numeric_limits<streamsize>::max(), '\n'); effectively "flushes" the toilet, ensuring the final cin.get() actually waits for your keypress.

Logic Tip: The "Subtract 9" Shortcut

In the Luhn algorithm, if a doubled digit is $> 9$ (like $8 \times 2 = 16$), we add the digits ($1+6=7$). Mathematically, for any doubled single digit, $d \times 2 - 9$ yields the exact same result as adding the digits of the product. It’s a cleaner way to write the logic!

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Write a program to determine if the given string is a palindrome or not || C Lab Program

noreply@blogger.com (belgaumboy) — Thu, 2 Apr 2026 17:23:00 +0530

WAP_E03: Write a C program to determine if the given string is a palindrome or not (Spelled same in both directions with or without a meaning like madam, civic, noon, abcba, etc.) || Strings

WAP_E03: C Lab Program

//to find the given string is a palindrome or not.
#include <stdio.h>
#include <string.h>

int main() {
char str[20];
int i,j;
int flag = 0;

printf("Enter a string: ");
scanf("%s", str);
// Calculate the string length, -1 for '\0'
j=strlen(str)-1;
// Compare characters from the start and end of the string
// and stop if a mismatch is found or the middle of the string is reached.
for(i=0;i<=j;i++,j--)
{
if(str[i]!=str[j])
{
flag=1;
break;
}
}
if (flag) {
printf("%s is not a palindrome\n", str);
} else {
printf("%s is a palindrome\n", str);
}
}

OUTPUT

Enter a string: madam
madam is a palindrome

Enter a string: abcdba
abcdba is not a palindrome

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Student Management System || 16 Intermediate Level C++ Projects

noreply@blogger.com (belgaumboy) — Thu, 2 Apr 2026 12:37:00 +0530

Summary.

This project is a foundational implementation of a CRUD (Create, Read, Update, Delete) application designed for the Code::Blocks environment. It focuses on managing student records using Object-Oriented principles, ensuring data persistence during the session via dynamic memory or container management.

Learning Objectives:

Encapsulation: Implementing private data members with public getter/setter interfaces.
Object Composition: Managing a collection of objects within a controller class.
Input Validation: Utilizing <limits> to handle stream errors and prevent infinite loops.
Formatted Output: Mastering <iomanip> for professional, tabular console reports.
Memory Safety: Understanding object lifecycles within the scope of a management system.

The Source Code

This code is optimized for the MinGW compiler. Ensure your Code::Blocks project is set to the ISO C++11 standard or higher (Project > Build Options > Compiler Flags).

#include <iostream>
#include <vector>
#include <string>
#include <iomanip>
#include <limits>

using namespace std;

// --- Entity Class ---
class Student {
private:
int id;
string name;
float gpa;

public:
Student(int id, string name, float gpa)
: id(id), name(name), gpa(gpa) {}

int getId() const { return id; }
string getName() const { return name; }
float getGpa() const { return gpa; }

void displayRow() const {
cout << left << setw(10) << id
<< setw(25) << name
<< fixed << setprecision(2) << setw(10) << gpa << endl;
}
};

// --- Controller Class ---
class ManagementSystem {
private:
vector<Student> database;

void clearBuffer() {
cin.clear();
cin.ignore(numeric_limits<streamsize>::max(), '\n');
}

public:
void addStudent() {
int id;
string name;
float gpa;

cout << "\n--- Add New Student ---" << endl;
cout << "Enter ID (Integer): ";
while (!(cin >> id)) {
cout << "Invalid input. Enter a numeric ID: ";
clearBuffer();
}

clearBuffer(); // Clean buffer before getline
cout << "Enter Name: ";
getline(cin, name);

cout << "Enter GPA (0.0 - 4.0): ";
while (!(cin >> gpa) || gpa < 0 || gpa > 4.0) {
cout << "Invalid GPA. Enter a value between 0.0 and 4.0: ";
clearBuffer();
}

database.emplace_back(id, name, gpa);
cout << "Student record added successfully!\n";
}

void viewAll() const {
if (database.empty()) {
cout << "\n[!] No records found.\n";
return;
}

cout << "\n" << string(45, '-') << endl;
cout << left << setw(10) << "ID" << setw(25) << "Name" << setw(10) << "GPA" << endl;
cout << string(45, '-') << endl;

for (const auto& s : database) {
s.displayRow();
}
cout << string(45, '-') << endl;
}

void searchById() const {
int searchId;
cout << "Enter Student ID to search: ";
cin >> searchId;

for (const auto& s : database) {
if (s.getId() == searchId) {
cout << "\nRecord Found:\nName: " << s.getName() << "\nGPA: " << s.getGpa() << endl;
return;
}
}
cout << "\n[!] Student ID " << searchId << " not found.\n";
}

void run() {
int choice;
do {
cout << "\n=== STUDENT MANAGEMENT SYSTEM ===";
cout << "\n1. Add Student\n2. View All\n3. Search by ID\n4. Exit";
cout << "\nSelection: ";

if (!(cin >> choice)) {
cout << "Invalid selection. Please use numbers 1-4.\n";
clearBuffer();
continue;
}

switch (choice) {
case 1: addStudent(); break;
case 2: viewAll(); break;
case 3: searchById(); break;
case 4: cout << "Exiting system...\n"; break;
default: cout << "Choice out of range.\n";
}
} while (choice != 4);
}
};

int main() {
ManagementSystem sms;
sms.run();
return 0;
}

Execution Trace (Sample Session)

=== STUDENT MANAGEMENT SYSTEM ===
1. Add Student
2. View All
3. Search by ID
4. Exit
Selection: 1

--- Add New Student ---
Enter ID (Integer): 101
Enter Name: Alice Hamilton
Enter GPA (0.0 - 4.0): 3.85
Student record added successfully!

=== STUDENT MANAGEMENT SYSTEM ===
Selection: 2

---------------------------------------------
ID Name GPA
---------------------------------------------
101 Alice Hamilton 3.85
---------------------------------------------

=== STUDENT MANAGEMENT SYSTEM ===
Selection: 1

--- Add New Student ---
Enter ID (Integer): ABC
Invalid input. Enter a numeric ID: 102
Enter Name: Bob Marley
Enter GPA (0.0 - 4.0): 9.5
Invalid GPA. Enter a value between 0.0 and 4.0: 3.20
Student record added successfully!

Academic Learning Corner

1. The cin.clear() and numeric_limits Fix

In Code::Blocks (MinGW), if a user enters a character like 'A' into an int variable, the input stream enters a "fail state." Without cin.clear(), the program will skip all future inputs, creating an infinite loop. Using numeric_limits<streamsize>::max() ensures we purge the entire keyboard buffer so the next getline() doesn't capture leftover newline characters.

2. Formatting with <iomanip>

We use left and setw(n) to create columns. Note that setw only applies to the very next output operation, which is why it must be repeated for every column. fixed and setprecision(2) ensure that even a GPA of 4 is displayed as 4.00, maintaining the professional "Report" look.

3. Encapsulation Logic

The Student class keeps data private. This prevents the ManagementSystem from accidentally modifying a student's ID, which should ideally be immutable once set. We provide "Getters" to access the data for display purposes only.

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Quiz Game || 14 Beginner Level C++ Projects

noreply@blogger.com (belgaumboy) — Thu, 2 Apr 2026 12:11:00 +0530

Summary:

A robust, console-based educational tool designed to manage and execute multiple-choice examinations. The system emphasizes modular design, separating the question logic from the engine's execution flow.

Learning Objectives:

Encapsulation: Managing question data and validation through private members and public interfaces.
Object Composition: Utilizing a std::vector of Class Objects to build a dynamic database.
Input Validation: Implementing "fail-safe" console handling to prevent infinite loops during data entry.

The Source Code

This main.cpp is optimized for Code::Blocks using the MinGW toolchain. It includes <limits> for buffer clearing and <iomanip> for score formatting.

#include <iostream>
#include <vector>
#include <string>
#include <limits>
#include <iomanip>

using namespace std;

// Class representing a single Multiple Choice Question
class Question {
private:
string text;
string options[4];
int correctAnswer; // Index 1-4

public:
Question(string q, string opts[4], int ans) {
text = q;
for(int i = 0; i < 4; i++) options[i] = opts[i];
correctAnswer = ans;
}

void display(int qNum) const {
cout << "\nQuestion " << qNum << ": " << text << endl;
for(int i = 0; i < 4; i++) {
cout << " " << (i + 1) << ". " << options[i] << endl;
}
}

bool checkAnswer(int userChoice) const {
return userChoice == correctAnswer;
}

int getCorrectAnswer() const { return correctAnswer; }
};

class QuizEngine {
private:
vector<Question> bank;
int score;

public:
QuizEngine() : score(0) {}

void addQuestion(const Question& q) {
bank.push_back(q);
}

void run() {
cout << "========================================" << endl;
cout << " WELCOME TO THE ACADEMIC QUIZ GAME " << endl;
cout << "========================================" << endl;

for(size_t i = 0; i < bank.size(); i++) {
bank[i].display(i + 1);
int choice = getValidInput();

if(bank[i].checkAnswer(choice)) {
cout << ">> Correct!" << endl;
score++;
} else {
cout << ">> Wrong. The correct answer was " << bank[i].getCorrectAnswer() << "." << endl;
}
}

displayResults();
}

private:
int getValidInput() {
int input;
while (true) {
cout << "Your Answer (1-4): ";
if (cin >> input && input >= 1 && input <= 4) {
return input;
} else {
cout << "[Error] Invalid input. Please enter a number between 1 and 4." << endl;
cin.clear(); // Clear error flags
cin.ignore(numeric_limits<streamsize>::max(), '\n'); // Discard buffer
}
}
}

void displayResults() {
double percentage = (static_cast<double>(score) / bank.size()) * 100;
cout << "\n----------------------------------------" << endl;
cout << "QUIZ FINISHED!" << endl;
cout << "Final Score: " << score << "/" << bank.size() << endl;
cout << "Grade: " << fixed << setprecision(2) << percentage << "%" << endl;
cout << "----------------------------------------" << endl;
}
};

int main() {
QuizEngine myQuiz;

string o1[] = {"C++", "Java", "Python", "Assembly"};
myQuiz.addQuestion(Question("Which language is known for its manual memory management?", o1, 1));

string o2[] = {"Encapsulation", "Polymorphism", "Recursion", "Inheritance"};
myQuiz.addQuestion(Question("Which of these is NOT a primary pillar of OOP?", o2, 3));

string o3[] = {"MinGW", "Visual Studio", "Code::Blocks", "GCC"};
myQuiz.addQuestion(Question("Which of the following is an Integrated Development Environment (IDE)?", o3, 3));

myQuiz.run();

cout << "\nPress Enter to exit...";
cin.ignore();
cin.get();

return 0;
}

Execution Trace (Sample User Session)

========================================
WELCOME TO THE ACADEMIC QUIZ GAME
========================================

Question 1: Which language is known for its manual memory management?
1. C++
2. Java
3. Python
4. Assembly
Your Answer (1-4): 1
>> Correct!

Question 2: Which of these is NOT a primary pillar of OOP?
1. Encapsulation
2. Polymorphism
3. Recursion
4. Inheritance
Your Answer (1-4): x
[Error] Invalid input. Please enter a number between 1 and 4.
Your Answer (1-4): 3
>> Correct!

Question 3: Which of the following is an Integrated Development Environment (IDE)?
1. MinGW
2. Visual Studio
3. Code::Blocks
4. GCC
Your Answer (1-4): 1
>> Wrong. The correct answer was 3.

----------------------------------------
QUIZ FINISHED!
Final Score: 2/3
Grade: 66.67%
----------------------------------------

Academic "Learning Corner"

1. The MinGW "Input Hang" Fix

In standard C++ console applications, if a user enters a character (like 'x') when the code expects an int, std::cin enters a fail state. If you don't clear it, the program will spin in an infinite loop.

cin.clear() resets the internal error state flags.
cin.ignore(numeric_limits<streamsize>::max(), '\n') wipes the "junk" character out of the keyboard buffer so the next cin >> actually waits for new input.

2. Header Justification

<limits>: Essential for numeric_limits, providing a platform-independent way to clear the input buffer.
<iomanip>: Used for setprecision(2). In academic reporting, providing raw doubles (like 66.666667%) is considered unprofessional; fixed-point notation ensures clean UI.

3. Object Composition

The QuizEngine does not inherit from Question. Instead, it "has-a" collection of questions. This is a fundamental OOP design pattern that makes the code more flexible—you can add 10 or 100 questions without changing the Engine's logic.

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Write a program that uses functions to delete n Characters from a given position in a given string || C Lab Program

noreply@blogger.com (belgaumboy) — Wed, 1 Apr 2026 17:20:00 +0530

WAP_E02: Write a C program that uses functions to delete n Characters from a given position in a given string || Strings

WAP_E02: C Lab Program

//delete n Characters from a given position in a given string.
#include <stdio.h>
#include <string.h>

int main() {
char str[20],finalStr[20];
int position,n;

// Input a string
printf("Enter a string: ");
fgets(str, sizeof(str), stdin);

// Input the position
printf("Enter the position from where you want to delete: ");
scanf("%d", &position);

// Input the no.of characters to delete
printf("Enter the number of characters to delete: ");
scanf("%d", &n);

// Check if the position is valid
int strLen = strlen(str);
if (position < 0 || position > strLen) {
printf("Invalid position.\n");
return 1;
}
strncpy(finalStr,str,position); // Copy string up to position
finalStr[position] = '\0'; // Important null terminate
strcat(finalStr,str+position+n); // Concatenate remaining characters

// Print the modified string
printf("Modified string: %s\n", finalStr);

return 0;
}

OUTPUT

Enter a string: Welcome to MyBlog
Enter the position from where you want to delete: 7
Enter the number of characters to delete: 4
Modified string: WelcomeMyBlog

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To-Do List Manager || 14 Beginner Level C++ Projects

noreply@blogger.com (belgaumboy) — Wed, 1 Apr 2026 12:09:00 +0530

Summary:

A command-line productivity tool that allows users to create, track, and manage tasks with priority levels. This project emphasizes the separation of concerns between data storage and user interface logic.

Learning Objectives:

Object-Oriented Design: Implementing a class with private attributes and public interfaces.
STL Containers: Utilizing std::vector for dynamic array management.
Input Validation: Handling stream errors (e.g., non-numeric input) to prevent infinite loops.
Formatting: Using <iomanip> to create clean, tabular console outputs.

The Source Code (main.cpp)

#include <iostream>
#include <vector>
#include <string>
#include <limits>
#include <iomanip>

using namespace std;

// --- Task Entity Class ---
class Task {
private:
string title;
int priority;

public:
Task(string t, int p) : title(t), priority(p) {}

string getTitle() const { return title; }
int getPriority() const { return priority; }
};

// --- Manager Logic Class ---
class ToDoManager {
private:
vector<Task> tasks;

void clearInput() {
cin.clear();
cin.ignore(numeric_limits<streamsize>::max(), '\n');
}

public:
void addTask() {
string title;
int priority;

cout << "\nEnter Task Description: ";
getline(cin >> ws, title);

while (true) {
cout << "Enter Priority (1-5): ";
if (cin >> priority && priority >= 1 && priority <= 5) break;
cout << "Invalid input. Please enter a number between 1 and 5.\n";
clearInput();
}
tasks.emplace_back(title, priority);
cout << "Task added successfully!\n";
}

void viewTasks() const {
if (tasks.empty()) {
cout << "\n[!] Your list is currently empty.\n";
return;
}

cout << "\n" << setfill('=') << setw(40) << "" << endl;
cout << setfill(' ') << left << setw(5) << "ID" << setw(25) << "Task" << "Priority" << endl;
cout << setfill('-') << setw(40) << "" << setfill(' ') << endl;

for (size_t i = 0; i < tasks.size(); ++i) {
cout << left << setw(5) << i + 1
<< setw(25) << tasks[i].getTitle()
<< tasks[i].getPriority() << endl;
}
cout << setfill('=') << setw(40) << "" << setfill(' ') << endl;
}

void removeTask() {
viewTasks();
if (tasks.empty()) return;

int index;
cout << "Enter Task ID to remove: ";
if (!(cin >> index) || index < 1 || index > (int)tasks.size()) {
cout << "Invalid ID. Operation cancelled.\n";
clearInput();
} else {
tasks.erase(tasks.begin() + (index - 1));
cout << "Task removed.\n";
}
}

void runMenu() {
int choice;
do {
cout << "\n--- TO-DO ARCHITECT V1.0 ---" << endl;
cout << "1. Add Task\n2. View Tasks\n3. Remove Task\n4. Exit\nChoice: ";

if (!(cin >> choice)) {
cout << "Numeric input required.\n";
clearInput();
continue;
}

switch (choice) {
case 1: addTask(); break;
case 2: viewTasks(); break;
case 3: removeTask(); break;
case 4: cout << "Exiting program...\n"; break;
default: cout << "Invalid choice.\n";
}
} while (choice != 4);
}
};

int main() {
ToDoManager myApp;
myApp.runMenu();
return 0;
}

Execution Trace

--- TO-DO ARCHITECT V1.0 ---
1. Add Task
2. View Tasks
3. Remove Task
4. Exit
Choice: 1

Enter Task Description: Finish C++ Lab 4
Enter Priority (1-5): 5
Task added successfully!

--- TO-DO ARCHITECT V1.0 ---
1. Add Task
2. View Tasks
3. Remove Task
4. Exit
Choice: 2

========================================
ID Task Priority
----------------------------------------
1 Finish C++ Lab 4 5
========================================

--- ERROR HANDLING CASE: Non-numeric Input ---
Choice: abc
Numeric input required.

--- ERROR HANDLING CASE: Out of Bounds ---
Choice: 3
Enter Task ID to remove: 99
Invalid ID. Operation cancelled.

Academic Learning Corner

The cin.clear() and cin.ignore() Combo: In MinGW, when a user types a character where an int is expected, the stream enters a "fail state." Without clearing this state and flushing the buffer, your program would enter an infinite loop. We use numeric_limits to ensure we wipe the entire buffer.
The ws Manipulator: Notice getline(cin >> ws, title). The ws (whitespace) extractor is vital here; it consumes any leftover newline characters from previous cin >> calls so your program doesn't "skip" the input prompt.
Vector Erasing: tasks.erase(tasks.begin() + (index - 1)) is the standard way to remove elements by index. Remember that vectors are 0-indexed, while users see 1-based IDs.

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Write a program that uses functions to insert a sub-string into a given main string from a given position || C Lab Program

noreply@blogger.com (belgaumboy) — Tue, 31 Mar 2026 17:16:00 +0530

WAP_E01: Write a C program that uses functions to insert a sub-string into a given main string from a given position|| Strings

WAP_E01: C Lab Program

//insert a sub-string into a given main string from a given position.
#include <stdio.h>
#include <string.h>

int main() {
char mainStr[50], subStr[50],finalStr[100];
int position;

// Input the main string
printf("Enter the main string: ");
fgets(mainStr, sizeof(mainStr), stdin);

// Input the substring
printf("Enter the substring: ");
fgets(subStr, sizeof(subStr), stdin);
subStr[strcspn(subStr, "\n")] = '\0'; // Remove newline

// Input the position
printf("Enter the position to insert the substring: ");
scanf("%d", &position);

// Check if the position is valid
int mainLen = strlen(mainStr);
if (position < 0 || position > mainLen) {
printf("Invalid position.\n");
return 1;
}

strncpy(finalStr,mainStr,position); // Copy string up to position
strcat(finalStr,subStr); // Insert substring
strcat(finalStr,mainStr+position); // Append remaining part

// Print the result string
printf("Modified string: %s\n", finalStr);

return 0;
}

OUTPUT

Enter the main string: welcome to Blog
Enter the substring: My
Enter the position to insert the substring: 11
Modified string: welcome to MyBlog

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Simple Calculator || 14 Beginner Level C++ Projects

noreply@blogger.com (belgaumboy) — Tue, 31 Mar 2026 12:19:00 +0530

Summary:

A robust, console-based arithmetic engine designed to perform fundamental mathematical operations while maintaining high data integrity. This project serves as a foundational exercise in implementing the Model-View-Controller (MVC) pattern within a procedural-object-oriented hybrid structure, specifically optimized for the MinGW/GCC toolchain.

Learning Objectives:

Encapsulation: Implementing a logic class to separate mathematical operations from I/O.
Input Validation: Mastering the use of std::cin.clear() and std::cin.ignore() to prevent infinite loops on invalid data.
Stream Formatting: Using <iomanip> for professional decimal precision in scientific outputs.

The Source Code

This code is optimized for Code::Blocks. It includes necessary headers for buffer management and precision control.

#include <iostream>
#include <limits> // Required for clearing input buffers
#include <iomanip> // Required for setprecision

using namespace std;

// Logic Layer: The Calculator Class
class Calculator {
public:
double add(double a, double b) { return a + b; }
double subtract(double a, double b) { return a - b; }
double multiply(double a, double b) { return a * b; }
double divide(double a, double b) {
if (b == 0) throw "Division by zero error!";
return a / b;
}
};

// Interface Layer: Utility functions
void clearBuffer() {
cin.clear();
cin.ignore(numeric_limits<streamsize>::max(), '\n');
}

int main() {
Calculator calc;
int choice;
double num1, num2;

cout << fixed << setprecision(2); // Set global precision to 2 decimal places

while (true) {
cout << "\n===============================" << endl;
cout << " PRECISION CALC-ARCHITECT " << endl;
cout << "===============================" << endl;
cout << "1. Addition (+)\n2. Subtraction (-)\n3. Multiplication (*)\n4. Division (/)\n5. Exit" << endl;
cout << "Enter choice: ";

if (!(cin >> choice)) {
cout << "Invalid input! Please enter a number (1-5)." << endl;
clearBuffer();
continue;
}

if (choice == 5) break;
if (choice < 1 || choice > 5) {
cout << "Out of range! Select 1-5." << endl;
continue;
}

cout << "Enter first number: ";
while (!(cin >> num1)) {
cout << "Error: Enter a valid number: ";
clearBuffer();
}

cout << "Enter second number: ";
while (!(cin >> num2)) {
cout << "Error: Enter a valid number: ";
clearBuffer();
}

try {
switch (choice) {
case 1: cout << "Result: " << calc.add(num1, num2) << endl; break;
case 2: cout << "Result: " << calc.subtract(num1, num2) << endl; break;
case 3: cout << "Result: " << calc.multiply(num1, num2) << endl; break;
case 4: cout << "Result: " << calc.divide(num1, num2) << endl; break;
}
} catch (const char* msg) {
cerr << "MATH ERROR: " << msg << endl;
}
}

cout << "Program terminated safely." << endl;
return 0;
}

Execution Trace (Sample Session)

===============================
PRECISION CALC-ARCHITECT
===============================
1. Addition (+)
2. Subtraction (-)
3. Multiplication (*)
4. Division (/)
5. Exit
Enter choice: 1

[HAPPY PATH]
Enter first number: 12.5
Enter second number: 7.25
Result: 19.75

===============================

Enter choice: 4

[ERROR HANDLING - MATH]
Enter first number: 10
Enter second number: 0
MATH ERROR: Division by zero error!

===============================

Enter choice: ABC

[ERROR HANDLING - INPUT]
Invalid input! Please enter a number (1-5).

Academic "Learning Corner"

1. The MinGW "Buffer Problem"

In many IDEs like Code::Blocks, if a user enters a character (like 'A') into an integer variable, the std::cin stream enters a fail state. Subsequent calls to cin will be ignored, often resulting in an infinite loop. We solve this using:

cin.clear(): Resets the error flags.
cin.ignore(...): Discards the "junk" characters still sitting in the keyboard buffer.

2. Precision Formatting

By using std::fixed and std::setprecision(2), we ensure the output looks like a professional financial or scientific tool (e.g., 5.00 instead of 5), which is critical for user experience in technical software.

eBook ‘14 Beginner Level C++ Projects’ purchase Link: Google Play Store || Google Books

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Level Up Your Coding Skills: Master C++ with 14 Hands-On Academic Projects!

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Write a program to merge two files into a third file || C Lab Program

noreply@blogger.com (belgaumboy) — Mon, 30 Mar 2026 16:02:00 +0530

WAP_D02: Write a C program to merge two files into a third file (i.e., the contents of the first file followed by those of the second are put in the third file).|| Files

file1.txt

Welcome Belgaum, Boy.

file2.txt

How are you?

WAP_D02: C Lab Program

//to merge two files into a third file.
#include <stdio.h>
#include <ctype.h>

int main() {
FILE *file1, *file2, *file3;
char ch;

// Open first file
file1 = fopen("file1.txt", "r");
if (file1 == NULL) {
printf("Error opening first file.\n");
return 1;
}

// Open second file
file2 = fopen("file2.txt", "r");
if (file2 == NULL) {
printf("Error opening second file.\n");
fclose(file1);
return 1;
}

// Open third file (output)
file3 = fopen("file3.txt", "w");
if (file3 == NULL) {
printf("Error creating destination file.\n");
fclose(file1);
fclose(file2);
return 1;
}

// Copy contents of first file
while ((ch = fgetc(file1)) != EOF) {
fputc(ch, file3);
}

// Copy contents of second file
while ((ch = fgetc(file2)) != EOF) {
fputc(ch, file3);
}

printf("Files merged successfully.");

// Close all files
fclose(file1);
fclose(file2);
fclose(file3);

return 0;
}

OUTPUT

Files merged successfully

file3.txt

Welcome Belgaum, Boy. How are you?

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Write a program which copies one file to another, replacing all lowercase characters with their uppercase equivalents || C Lab Program

noreply@blogger.com (belgaumboy) — Sun, 29 Mar 2026 16:00:00 +0530

WAP_D01: Write a C program which copies one file to another, replacing all lowercase characters with their uppercase equivalents.|| Files

file1.txt

Welcome Belgaum, Boy

WAP_D01: C Lab Program

//copies one file to another, replacing all lowercase characters with their uppercase equivalents.
#include <stdio.h>
#include <ctype.h>

int main() {
FILE *src, *dest;
char ch;

// Open source file
src = fopen("file1.txt", "r");
if (src == NULL) {
printf("Error: Cannot open source file.\n");
return 1;
}

// Open destination file
dest = fopen("file2.txt", "w");
if (dest == NULL) {
printf("Error: Cannot create destination file.\n");
fclose(src);
return 1;
}

// Copy characters with conversion
while ((ch = fgetc(src)) != EOF) {
fputc(toupper(ch), dest);
}

printf("File copied successfully with uppercase conversion.\n");

// Close files
fclose(src);
fclose(dest);

return 0;
}

OUTPUT

File copied successfully with uppercase conversion.

file2.txt

WELCOME BELGAUM, BOY

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Write a program for display values reverse order from an array using a pointer || C Lab Program

noreply@blogger.com (belgaumboy) — Sat, 28 Mar 2026 15:38:00 +0530

WAP_C05: Write a program for display values reverse order from an array using a pointer.|| Arrays, Pointers and Functions

WAP_C05: C Lab Program

//display values reverse order from an array using a pointer.
#include <stdio.h>

int main() {
int n, sum = 0, arr[10];

// Input the number of elements
printf("Enter the number of elements: ");
scanf("%d", &n);

// Input the array elements
printf("Enter the elements:\n");
for (int i = 0; i < n; i++) {
scanf("%d", &arr[i]);
}

int *ptr = arr;

// Display the array elements in reverse order
printf("The elements in reverse order are:\n");
for (int i = n-1; i >= 0; i--) {
printf("%2d", *(ptr+i));
}

return 0;
}

OUTPUT

Enter the number of elements: 5
Enter the elements:
2 4 6 8 10
The elements in reverse order are:
10 8 6 4 2

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Write a program for reading elements using a pointer into an array and display the values using the array || C Lab Program

noreply@blogger.com (belgaumboy) — Fri, 27 Mar 2026 15:35:00 +0530

WAP_C04 Write a program for reading elements using a pointer into an array and display the values using the array.|| Arrays, Pointers and Functions

WAP_C04: C Lab Program

//reading elements using a pointer into an array and display the values using the array.
#include <stdio.h>

int main() {
int n, sum = 0, arr[10];

// Input the number of elements
printf("Enter the number of elements: ");
scanf("%d", &n);

int *ptr = arr;

printf("Enter the elements:\n");
for (int i = 0; i < n; i++) {
// Read element using the pointer
scanf("%d", ptr + i); // Pointer arithmetic to read the value
}

// Display the values using the array
printf("The elements are:\n");
for (int i = 0; i < n; i++) {
printf("%d ", arr[i]); // Display elements using the array
}

return 0;
}

OUTPUT

Enter the number of elements: 5
Enter the elements:
2 4 6 8 10
The elements are:
2 4 6 8 10

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Write a program that uses functions to perform Multiplication of two matrices || C Lab Program

noreply@blogger.com (belgaumboy) — Thu, 26 Mar 2026 15:32:00 +0530

WAP_C03: Write a C program that uses functions to perform Multiplication of two matrices.|| Arrays, Pointers and Functions

WAP_C03: C Lab Program

// Multiplication of Two Matrices

#include <stdio.h>

// Function to perform matrix multiplication
void multiplyMatrices(int A[10][10], int B[10][10], int result[10][10], int rowA, int colA, int rowB, int colB) {
int i,j,k;
for (i = 0; i < rowA; i++) {
for (j = 0; j < colB; j++) {
result[i][j] = 0; // Initializing the result matrix with 0
for (k = 0; k < colA; k++) {
result[i][j] += A[i][k] * B[k][j];
}
}
}
}

// Function to input matrix elements
void inputMatrix(int matrix[10][10], int rows, int cols) {
int i,j;
printf("Enter elements of the matrix (%dx%d):\n", rows, cols);
for (i = 0; i < rows; i++) {
for (j = 0; j < cols; j++) {
scanf("%d", &matrix[i][j]);
}
}
}

// Function to display a matrix
void displayMatrix(int matrix[10][10], int rows, int cols) {
int i,j;
for (i = 0; i < rows; i++) {
for (j = 0; j < cols; j++) {
printf("%d ", matrix[i][j]);
}
printf("\n");
}
}

int main() {
int A[10][10], B[10][10], result[10][10];
int rowA, colA, rowB, colB;

// Input matrix dimensions
printf("Enter the number of rows and columns for Matrix A: ");
scanf("%d %d", &rowA, &colA);

printf("Enter the number of rows and columns for Matrix B: ");
scanf("%d %d", &rowB, &colB);

// Check if multiplication is possible (columns of A must be equal to rows of B)
if (colA != rowB) {
printf("Matrix multiplication is not possible. The number of columns of A must be equal to the number of rows of B.\n");
return 1; // Exit the program if multiplication is not possible
}

// Input matrices A and B
printf("Input matrix A:\n");
inputMatrix(A, rowA, colA);

printf("Input matrix B:\n");
inputMatrix(B, rowB, colB);

// Perform matrix multiplication
multiplyMatrices(A, B, result, rowA, colA, rowB, colB);

// Display the result
printf("\nMatrix A:\n");
displayMatrix(A, rowA, colA);

printf("\nMatrix B:\n");
displayMatrix(B, rowB, colB);

printf("\nResultant Matrix (A * B):\n");
displayMatrix(result, rowA, colB);

return 0;
}

OUTPUT

Enter the number of rows and columns for Matrix A: 3 3
Enter the number of rows and columns for Matrix B: 3 3
Input matrix A:
Enter elements of the matrix (3x3):
1 2 3
4 5 6
7 8 9
Input matrix B:
Enter elements of the matrix (3x3):
1 0 0
0 1 0
0 0 1

Matrix A:
1 2 3
4 5 6
7 8 9

Matrix B:
1 0 0
0 1 0
0 0 1

Resultant Matrix (A * B):
1 2 3
4 5 6
7 8 9

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Write a program that uses functions to perform Addition of two matrices. || C Lab Program

noreply@blogger.com (belgaumboy) — Wed, 25 Mar 2026 15:29:00 +0530

WAP_C02: Write a C program that uses functions to perform Addition of two matrices.|| Arrays, Pointers and Functions

WAP_C02: C Lab Program

// Addition of Two Matrices

#include <stdio.h>
// Function to perform matrix addition
void addMatrices(int A[10][10], int B[10][10], int result[10][10], int rows, int cols) {
int i,j;
for (i = 0; i < rows; i++) {
for (j = 0; j < cols; j++) {
result[i][j] = A[i][j] + B[i][j];
}
}
}

// Function to input matrix elements
void inputMatrix(int matrix[10][10], int rows, int cols) {
printf("Enter elements of the matrix (%dx%d):\n", rows, cols);
int i,j;
for (i = 0; i < rows; i++) {
for (j = 0; j < cols; j++) {
scanf("%d", &matrix[i][j]);
}
}
}

// Function to display a matrix
void displayMatrix(int matrix[10][10], int rows, int cols) {
int i,j;
for (i = 0; i < rows; i++) {
for (j = 0; j < cols; j++) {
printf("%2d ", matrix[i][j]);
}
printf("\n");
}
}

int main() {
int A[10][10], B[10][10], result[10][10];
int rows, cols;

// Input matrix dimensions
printf("Enter the number of rows and columns: ");
scanf("%d %d", &rows, &cols);

// Input matrices A and B
printf("Input matrix A:\n");
inputMatrix(A, rows, cols);

printf("Input matrix B:\n");
inputMatrix(B, rows, cols);

// Add matrices A and B
addMatrices(A, B, result, rows, cols);

// Display the results
printf("\nMatrix A:\n");
displayMatrix(A, rows, cols);

printf("\nMatrix B:\n");
displayMatrix(B, rows, cols);

printf("\nResultant Matrix (A + B):\n");
displayMatrix(result, rows, cols);

return 0;
}

OUTPUT

Enter the number of rows and columns: 3 3
Input matrix A:
Enter elements of the matrix (3x3):
1 2 3
4 5 6
7 8 9
Input matrix B:
Enter elements of the matrix (3x3):
1 0 0
0 1 0
0 0 1

Matrix A:
1 2 3
4 5 6
7 8 9

Matrix B:
1 0 0
0 1 0
0 0 1

Resultant Matrix (A + B):
2 2 3
4 6 6
7 8 10

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Write a program to find the minimum, maximum and average in an array of integers || C Lab Program

noreply@blogger.com (belgaumboy) — Tue, 24 Mar 2026 15:24:00 +0530

WAP_C01: Write a C program to find the minimum, maximum and average in an array of integers..|| Arrays, Pointers and Functions

WAP_C01: C Lab Program

//to find the minimum, maximum and average in an array of integers
#include <stdio.h>
#include <math.h>

int main() {
int n,arr[10],min,max,sum;
float average;

// Read the size of the array
printf("Enter the number of elements in the array: ");
scanf("%d", &n);

// Read the elements of the array
printf("Enter %d integers:\n", n);
for (int i = 0; i < n; i++) {
scanf("%d", &arr[i]);
}

// Initialize min, max and sum
min = arr[0];
max = arr[0];
sum = arr[0]; // Using long long to avoid overflow

// Iterate through the array to find min, max, and sum
for (int i = 1; i < n; i++) {
if (arr[i] < min) {
min = arr[i];
}
if (arr[i] > max) {
max = arr[i];
}
sum += arr[i];
}

// Calculate the average
average = (float)sum / n;

// Print the results
printf("Minimum: %d\n", min);
printf("Maximum: %d\n", max);
printf("Average: %.2f\n", average);

return 0;
}

OUTPUT

Enter the number of elements in the array: 5
Enter 5 integers:
5
15
25
10
20
Minimum: 5
Maximum: 25
Average: 15.00

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Write a C program to generate the first n terms of the Fibonacci sequence..|| C Lab Program

noreply@blogger.com (belgaumboy) — Mon, 23 Mar 2026 19:00:00 +0530

WAP_B04: A Fibonacci sequence is defined as follows: the first and second terms in the sequence are 0 and 1. Subsequent terms are found by adding the preceding two terms in the sequence. Write a C program to generate the first n terms of the sequence....|| Expression Evaluation

Algorithm

start
step1: a := 0, b:= 1
step2: Read n
step3: Print "Fibonacci sequence" a,b
step4: for i := 3 to n step 1 do
step5: next := a + b
step6: Print next
step7: a := b
step8: b := next
stop

WAP_B04: C Lab Program

//to generate the first n terms of the Fibonacci sequence
#include <stdio.h>

int main() {
int n, a = 0, b = 1, next,i;

printf("Enter the number of terms you want in the Fibonacci sequence: ");
scanf("%d", &n);
printf("Fibonacci Sequence:\n");
printf("%d %d ",a,b);
for (i = 3; i <= n; i++) {
next = a + b;
printf("%d ", next);
a = b;
b = next;
}
return 0;
}

OUTPUT

Enter the number of terms you want in the Fibonacci sequence: 9
Fibonacci Sequence:
0 1 1 2 3 5 8 13 21

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Write a C program to find the sum of individual digits of a positive integer and test given number is palindrome.|| C Lab Program

noreply@blogger.com (belgaumboy) — Sun, 22 Mar 2026 18:57:00 +0530

WAP_B03: Write a C program to find the sum of individual digits of a positive integer and test given number is palindrome....|| Expression Evaluation

Algorithm

start
step1: sum := 0
step2: Read num
step3: temp := num
step4: while temp > 0 then
step5: digit := temp % 10
step6: sum := sum + digit
step7: reversed := reversed * 10 + digit
step8: temp := temp/10
step9: Print sum
step10:if num = teversed then
step11: Print "Num is a palindrome"
step12:else Print "Num is not a palindrome"
stop

WAP_B03: C Lab Program

//sum of individual digits of a positive integer and test given numberis palindrome.
#include <stdio.h>

int main() {
int num,sum = 0,reversed,temp,digit;

printf("Enter a number: ");
scanf("%d", &num);
temp = num;
while (temp > 0) {
digit=temp%10;
sum += digit; // Add the last digit to sum
reversed = reversed * 10 + digit; // Build the reversed number
temp /= 10; // Remove the last digit
}
printf("Sum of individual digits: %d\n", sum);
// Check if given number and reversed are the same
if (num == reversed) {
printf("%d is a palindrome.\n", num);
} else {
printf("%d is not a palindrome.\n", num);
}

return 0;
}

OUTPUT

Enter a number: 12321
Sum of individual digits: 9
12321 is a palindrome

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Write a program that finds if a given number is a prime number.|| C Lab Program

noreply@blogger.com (belgaumboy) — Sat, 21 Mar 2026 18:54:00 +0530

WAP_B02: Write a program that finds if a given number is a prime number...|| Expression Evaluation

Algorithm

start
step1: Read number
step2: for i := 2 to number/2 step 1 do
step3: if number % i = 0 then
step4: count := count +1
step5: break
step6: if count = 0 then
step7: Print "number is prime"
step8: else Print "number is not a prime"
stop

WAP_B02: C Lab Program

//a given number is a prime number or not
#include <stdio.h>

int main() {
int number,count=0;

printf("Enter a number: ");
scanf("%d", &number);
for (int i = 2; i<= number/2; i++) {
if (number % i == 0) {
count++; // Found a divisor, not prime
break;
}
}
if (count == 0) {
printf("%d is a prime number.\n", number);
} else {
printf("%d is not a prime number.\n", number);
}

return 0;
}

OUTPUT

Enter a number: 31
31 is a prime number.

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Write a program to prints the result (user input: two integer operands and one operator) || C Lab Program

noreply@blogger.com (belgaumboy) — Fri, 20 Mar 2026 18:52:00 +0530

WAP_B01: Write a C program, which takes two integer operands and one operator from the user, performs the operation and then prints the result. (Consider the operators +,-,*, /, % and use Switch Statement).|| Expression Evaluation

Algorithm

start
step1: Read num1,num2 and operator
step2: switch (operator)
step3: case '+': result = num1 + num2
step4: Print result
step5: case '-': result = num1 - num2
step6: Print result
step7: case '*': result = num1 * num2
step8: Print result
step9: case '/': if num2 ≠ 0 then
step10: result = num1 / num2;
step11: Print result
step12: else Print "Error: Division by zero is not allowed"
step13: case '%': if num2 ≠ 0 then
step14: result = num1 % num2;
step15: Print result
step16: else Print "Error: Division by zero is not allowed"
step17:default: Print "invalid operator"
step18:end switch
stop

WAP_B01: C Lab Program

//takes two integer operands and one operator from the user, performs the operation and then prints the result. (Consider the operators +,-,*, /, % and use Switch Statement).
#include <stdio.h>

int main() {
int num1, num2, result;
char operator;

// Take input from the user
printf("Enter two numbers: ");
scanf("%d%d", &num1,&num2);
printf("choose an operator (+,-,*,/,%): ");
scanf(" %c", &operator); // Note the space before %c to consume any whitespace

// Perform the operation based on the operator
switch (operator) {
case '+':
result = num1 + num2;
printf("%d + %d = %d\n", num1, num2, result);
break;
case '-':
result = num1 - num2;
printf("%d - %d = %d\n", num1, num2, result);
break;
case '*':
result = num1 * num2;
printf("%d * %d = %d\n", num1, num2, result);
break;
case '/':
if (num2 != 0) {
result = num1 / num2;
printf("%d / %d = %d\n", num1, num2, result);
} else {
printf("Error: Division by zero is not allowed.\n");
}
break;
case '%':
if (num2 != 0) {
result = num1 % num2;
printf("%d %% %d = %d\n", num1, num2, result);
} else {
printf("Error: Division by zero is not allowed for modulus.\n");
}
break;
default:
printf("Invalid operator\n");
break;
}

return 0;
}

OUTPUT

Enter two numbers: 55 3
choose an operator (+,-,*,/,%): %
55 % 3 = 1

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Write a program that shows the binary equivalent of a given positive number between 0 to 255 || C Lab Program

noreply@blogger.com (belgaumboy) — Thu, 19 Mar 2026 16:30:00 +0530

WAP_A04: Write a program that shows the binary equivalent of a given positive number between 0 to 255 || Simple Numeric Problems

Algorithm

start
step1: Read number
step2: if (number < 0 OR number > 255) then
step3: Print "Number is out of range." stop
step4: for i:=0 to number ≠ 0 step 1 do
step5: bit[i] := number % 2
step6: number := number / 2
step7: Print "Binary equivalent"
step8: for j := i-1 to j ≥ 0 step -1 do
step9: print bit[j]
stop

WAP_A04: C Lab Program

//Binary equivalent of a given positive number between 0 to 255.

#include <stdio.h>

int main() {
static int number,bit[8],i,j;

// Prompt the user for input
printf("Enter a number: ");
scanf("%d", &number);

// Validate the input
if (number < 0 || number > 255) {
printf("Number is out of range.\n");
return 1; // Exit with error code
}

for(i=0;number!=0;i++)
{
bit[i]=number%2;
number=number/2;
}
printf("Binary equivalent: ");
for(j=i-1;j>=0;j--)
printf("%d",bit[j]);
printf("\n");
return 0; // Successful execution
}

OUTPUT

Enter a number: 56
Binary equivalent: 111000

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Write a program that prints a multiplication table for a given number and the number of rows in the table || C Lab Program

noreply@blogger.com (belgaumboy) — Wed, 18 Mar 2026 16:27:00 +0530

WAP_A03: . Write a program that prints a multiplication table for a given number and the number of rows in the table.|| Simple Numeric Problems

Algorithm

start
step1: Read a,r
step2: for i := 1 to r step 1 do
step3: Print a * i
stop

WAP_A03: C Lab Program

// multiplication table for a given number and the number of rows in the table

#include <stdio.h>
int main()
{
int a,r,i;
printf("Enter an integer number : ");
scanf("%d",&a);
printf("Enter number of rows : ");
scanf("%d",&r);
for(i=1;i<=r;i++)
printf("%d * %2d = %2d\n",a,i,a*i);
return(0);
}

OUTPUT

Enter an integer number : 6
Enter number of rows : 10
6 * 1 = 6
6 * 2 = 12
6 * 3 = 18
6 * 4 = 24
6 * 5 = 30
6 * 6 = 36
6 * 7 = 42
6 * 8 = 48
6 * 9 = 54
6 * 10 = 60

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Write the program for the simple, compound interest || C Lab Program

noreply@blogger.com (belgaumboy) — Tue, 17 Mar 2026 16:13:00 +0530

WAP_A02: Write the program for the simple, compound interest. || Simple Numeric Problems

Algorithm

start
step1: Read p, rate, time
step2: SI := (p * rate * time)/100
step3: CI := p * (1+(rate/100))time - p
step4: Print SI, CI
stop

WAP_A02: C Lab Program

/* calculate simple interest & compound interest by reading principal amount, rate of interest & time

The formula to find simple interest is simpleInterest = (principal * rate * time) / 100.

The formula to find compound interest is compoundInterest = principal * pow(1 + (rate / 100), time) - principal. */

#include <stdio.h>
#include <math.h>

int main() {
float P, rate, time, SI, CI;

// Input P amount, rate of interest and time
printf("Enter P,R,T: ");
scanf("%f%f%f", &P,&rate,&time);

// Calculate simple interest
SI = (P * rate * time) / 100;

// Calculate compound interest
CI = P * pow(1 + (rate / 100), time) - P;

// Display results
printf("SI= %f\n", SI);
printf("CI= %f\n", CI);

return 0;
}

OUTPUT

Enter P,R,T: 1000 5 10
SI= 500.000000
CI= 628.893860

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Write a program for finding the max and min from the three numbers || C Lab Program

noreply@blogger.com (belgaumboy) — Mon, 16 Mar 2026 16:00:00 +0530

WAP_A01: Write a program for finding the max and min from the three numbers || Simple Numeric Problems

Algorithm

start
step1: Read a,b,c
step2: max, min := a
step3: if b > max then
step4: max := b
step5: if b < min then
step6: min := b
step7: if c > max then
step8: max := c
step9: if c < max then
step10: min := c
step11: Print max,min
Stop

WAP_A01: C Lab Program

//finding the max and min from the three numbers

#include <stdio.h>
int main()
{
int a,b,c;
int max, min;
printf("Enter 3 integers: ");
scanf("%d%d%d",&a,&b,&c);
max = min = a;
if (b > max) {
max = b;
}
if (b < min) {
min = b;
}

if (c > max) {
max = c;
}
if (c < min) {
min = c;
}

// Output the results
printf("%d is max\n", max);
printf("%d is min\n", min);

return(0);
}

OUTPUT

Enter 3 integers: 60 20 30
60 is max
20 is min

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The Ultimate ‘Chicken or the Egg’ Paradox: How C Was Actually Written in C

noreply@blogger.com (belgaumboy) — Sun, 15 Mar 2026 19:15:00 +0530

Every junior developer eventually hits a wall when they first encounter the statement: "C is written in C." It sounds like a glitch in the matrix—a recursive loop that defies the laws of causality. If you need a C compiler to turn C code into a functional program, but the compiler itself is written in C, how did the first one ever come to be?

It feels like a technical impossibility, a "black box" mystery that suggests the tool must exist before it can be created. However, this isn't magic; it is one of the most elegant demonstrations of logic in computer science. Let’s dismantle this paradox once and for all and uncover the process of "bootstrapping" that allows a language to pull itself up by its own bootstraps.

The Language Barrier: Why Compilers Exist

To understand the solution, we must first accept a fundamental truth about hardware: your CPU is a monoglot. It does not understand C, C++, or any other high-level language. It speaks only one language—Machine Code, the raw sequence of 0s and 1s known as binary.

When we write C code, we use human-readable keywords like int main or if. Because the CPU is physically incapable of executing these English-like strings, we require a "middleman" to bridge the gap. This middleman is the compiler. Crucially, a compiler is not a physical device or a special piece of hardware; it is simply a piece of code—a software tool designed to perform a complex translation.

Compilers Aren't Magic; They’re Logic

The key to unlocking the paradox is realizing that a compiler is just another program. It functions exactly like any other software you might write, such as a program that checks if a number is prime or reverses a string. It takes an input, processes it according to specific rules, and produces an output.

In the case of a compiler, the "input" is your C source code, and the "output" is a machine code executable. As the source context explains:

"Compiler is also just your some piece of code... it is also a code which runs, which takes [C code] in input and generates [machine code] in output."

Once you stop viewing the compiler as a mysterious entity and start seeing it as a logical process, you realize that the instructions for "how to translate text into binary" can be written in any language capable of handling complex logic—including C itself.

Text is Text: How C Processes Itself

If you can write a C program to perform a linear search, you can also write a C program that contains the logic for translation. This leads to the "Aha!" moment: To a compiler written in C, a C-source file is just data.

To the compiler, there is no functional difference between the string "Uday" and the keyword int. Both are merely sequences of characters to be processed. The compiler doesn't "know" C in a sentient way; it simply follows a logical map. When the "Special Code" inside the compiler encounters the characters i, n, and t, it triggers a specific logic branch that tells it to generate a corresponding sequence of binary. Because C is a powerful, low-level language, it is perfectly suited to describe the intricate rules of its own translation.

The Great Translation: The One-Time Bridge

When people say "C is written in C," they are referring to the language in which the compiler's logic was authored. But how did the first C compiler ever run if there wasn't a C compiler to translate it? This is where we witness "The Great Handoff."

The process follows a specific evolutionary path:

The Logic: An engineer writes the logic for a C compiler using the C language.
The Bridge: To turn that logic into something the CPU can run, they use an existing compiler written in a different, older language (such as Assembly or B).
The Result: That older compiler generates a machine-code version of the new C compiler.

This is the "One-Time Bridge." Once you have that first executable version of the C compiler in machine code, you no longer need the older language. The C compiler can now compile its own source code to create newer, better versions of itself. This is the moment the language becomes "self-hosting." You have kicked away the ladder used to climb to the roof.

The Recursive World of Programming

This concept of bootstrapping is a cornerstone of modern systems engineering. It isn’t an quirk unique to C; languages like C++, Rust, and Java have all reached this state of self-sufficiency, where the language is used to maintain and evolve its own tools.

This logic invites us to look deeper into the foundations of our software. While systems languages often pull themselves up by their own bootstraps, other environments operate differently. For example, is JavaScript written in JavaScript, or is it built upon a foundation of C++?

I challenge you to research the architecture of your favorite language. Does it rely on a "middleman" from another language, or has it achieved the ultimate elegance of self-hosting? Understanding this "chicken or the egg" logic is the first step toward mastering the true power of software engineering.

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