C Programming Codes

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💻 Variable Swapping (Two Variables)

#include <stdio.h>

int main() {
  int a, b, temp;

  printf("Enter the value of a: ");
  scanf("%d", &a);

  printf("Enter the value of b: ");
  scanf("%d", &b);

  printf("Before swapping: a = %d, b = %dn", a, b);

  temp = a;
  a = b;
  b = temp;

  printf("After swapping: a = %d, b = %dn", a, b);

  return 0;
}

📤 Output:

Input: 5
Input: 10
Output: Enter the value of a: Enter the value of b: Before swapping: a = 5, b = 10
After swapping: a = 10, b = 5

💻 Basic Calculator (Add, Subtract, Multiply, Divide)

#include <stdio.h>

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

    printf("Enter first number: ");
    scanf("%f", &num1);

    printf("Enter an operator (+, -, *, /): ");
    scanf(" %c", &operator);

    printf("Enter second number: ");
    scanf("%f", &num2);

    switch (operator) {
        case '+':
            result = num1 + num2;
            printf("%.2f + %.2f = %.2fn", num1, num2, result);
            break;
        case '-':
            result = num1 - num2;
            printf("%.2f - %.2f = %.2fn", num1, num2, result);
            break;
        case '*':
            result = num1 * num2;
            printf("%.2f * %.2f = %.2fn", num1, num2, result);
            break;
        case '/':
            if (num2 == 0) {
                printf("Error! Division by zero.n");
            } else {
                result = num1 / num2;
                printf("%.2f / %.2f = %.2fn", num1, num2, result);
            }
            break;
        default:
            printf("Error! Invalid operator.n");
    }

    return 0;
}

📤 Output:

Input: 10
Input: +
Input: 5
Output: 10.00 + 5.00 = 15.00

Input: 20
Input: -
Input: 7
Output: 20.00 - 7.00 = 13.00

Input: 4
Input: *
Input: 6
Output: 4.00 * 6.00 = 24.00

Input: 15
Input: /
Input: 3
Output: 15.00 / 3.00 = 5.00

Input: 8
Input: /
Input: 0
Output: Error! Division by zero.

Input: 9
Input: %
Input: 2
Output: Error! Invalid operator.

💻 Hello World Program

#include <stdio.h>

int main() {
    printf("Hello, World!n");
    return 0;
}

📤 Output:

Hello, World!

🔧 C Basics & Syntax

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🚀100 Days of LeetCode 2025 Challenge Join below channel to be part of this challenge 🔗https://t.me/+utZsK17rSXRmZWQ1

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Print days of the week using switch statement

#include <stdio.h>

int main() {
    int dayNumber;

    printf("Enter a number (1-7): ");
    scanf("%d", &dayNumber);

    switch (dayNumber) {
        case 1:
            printf("Sundayn");
            break;
        case 2:
            printf("Mondayn");
            break;
        case 3:
            printf("Tuesdayn");
            break;
        case 4:
            printf("Wednesdayn");
            break;
        case 5:
            printf("Thursdayn");
            break;
        case 6:
            printf("Fridayn");
            break;
        case 7:
            printf("Saturdayn");
            break;
        default:
            printf("Invalid inputn");
    }

    return 0;
}

💡 Approach Step 1: Get integer input from the user representing the day of the week (1 for Sunday, 2 for Monday, ..., 7 for Saturday). Step 2: Create a switch statement. The switch expression will be the integer input from the user. Step 3: Inside the switch statement, create case labels for each day of the week (1 to 7). Step 4: Within each case label, use printf to print the corresponding day of the week (e.g., case 1: printf("Sunday"); break;). Include a break statement after each printf to exit the switch statement after a match. Step 5: Add a default case to handle invalid input (numbers outside the range 1-7). Print an error message like "Invalid input". ───────────────────────────── Have you Understood? Drop a reaction: ❤️ Understood | 👎 Not Understood

📝 Print days of the week using switch statement Write a C program that takes an integer input representing a day of the week (1 for Sunday, 2 for Monday, etc.). Using a switch statement, print the corresponding day of the week's name based on the input number.

Implement a simple calculator using `switch` statement for operations (+, -, , /) #include <stdio.h> int main() { float num1, num2, result; char operator; printf("Enter first number: "); scanf("%f", &num1); printf("Enter operator (+, -, , /): "); scanf(" %c", &operator); printf("Enter second number: "); scanf("%f", &num2); switch (operator) { case '+': result = num1 + num2; break; case '-': result = num1 - num2; break; case '': result = num1 num2; break; case '/': if (num2 == 0) { printf("Error: Division by zero is not allowed.\n"); return 1; } result = num1 / num2; break; default: printf("Error: Invalid operator.\n"); return 1; } printf("Result: %.2f\n", result); return 0; }

💡 Approach Step 1: Include Header and Declare Variables: Include the standard input/output library (`stdio.h`). Declare variables to store the two numbers (e.g., `num1`, `num2` as `float` for decimal support), the operator (e.g., `operator` as `char`), and the result (e.g., `result` as `float`). Step 2: Get User Input: Prompt the user to enter the two numbers and the operator (+, -, , /). Use `scanf` to read these values from the console. Step 3: Implement the `switch` Statement: Use a `switch` statement to perform the calculation based on the operator entered by the user. Each `case` will correspond to a specific operator. Step 4: Handle Each Case: Inside each `case`, perform the corresponding arithmetic operation. For example, in the `case '+'`, calculate `result = num1 + num2`. Step 5: Handle Division by Zero: In the `case '/'`, add a check to prevent division by zero. If `num2` is zero, print an error message and possibly exit the `switch` statement or program. Step 6: Provide a `default` Case: Include a `default` case in the `switch` statement to handle invalid operator inputs. Print an error message if the operator is not recognized. Step 7: Print the Result: After the `switch` statement (assuming no errors occurred), print the calculated `result` to the console with appropriate formatting. Step 8: Return 0:* At the end of the `main` function, return 0 to indicate successful execution of the program. ───────────────────────────── Have you Understood\? Drop a reaction: ❤️ Understood | 👎 Not Understood

📝 Implement a simple calculator using switch statement for operations (+, -, , /) Write a C program that takes two numbers and an arithmetic operator (+, -, , /) as input. Using a switch statement, perform the corresponding operation and print the result. Handle the case of division by zero by printing an appropriate error message.

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Determine if a year is a leap year

#include <stdio.h>

int main() {
  int year;

  printf("Enter a year: ");
  scanf("%d", &year);

  if (year % 4 != 0) {
    printf("%d is not a leap year.n", year);
  } else {
    if (year % 100 == 0) {
      if (year % 400 == 0) {
        printf("%d is a leap year.n", year);
      } else {
        printf("%d is not a leap year.n", year);
      }
    } else {
      printf("%d is a leap year.n", year);
    }
  }

  return 0;
}

💡 Approach Step 1: Get the year as input from the user. This is the year we will check. Step 2: Check if the year is divisible by 4. If it is NOT, then it's NOT a leap year, and the process ends. Step 3: If the year IS divisible by 4, then check if it's divisible by 100. Step 4: If the year IS divisible by 100, then check if it's also divisible by 400. If it IS, then it's a leap year. If it's NOT, then it's NOT a leap year. Step 5: If the year is divisible by 4 but NOT divisible by 100, then it IS a leap year. ───────────────────────────── Have you Understood? Drop a reaction: ❤️ Understood | 👎 Not Understood

📝 Determine if a year is a leap year Write a C program that takes a year as input and determines whether it is a leap year. The program should use control flow statements (if/else) to implement the leap year rules: divisible by 4, but not divisible by 100 unless also divisible by 400.