C Programming Language || Hands On Coding

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.

Check if a character is a vowel or consonant using if-else if-else

#include <stdio.h>
#include <ctype.h>

int main() {
    char input_char;

    scanf(" %c", &input_char);

    char lower_char = tolower(input_char);

    if (lower_char >= 'a' && lower_char <= 'z') {
        if (lower_char == 'a') {
            printf("Voweln");
        } else if (lower_char == 'e') {
            printf("Voweln");
        } else if (lower_char == 'i') {
            printf("Voweln");
        } else if (lower_char == 'o') {
            printf("Voweln");
        } else if (lower_char == 'u') {
            printf("Voweln");
        } else {
            printf("Consonantn");
        }
    } else {
        printf("Not an alphabetn");
    }

    return 0;
}

💡 Approach Step 1: Get character input: Read a character from the user using scanf. Step 2: Convert to lowercase: Convert the input character to lowercase using tolower() function. This simplifies the vowel check. Step 3: Check if it is an alphabet: Verify that the character is an alphabet (a-z). If not, display a message indicating it's not an alphabet and exit. Step 4: Check for vowels using if-else if-else: Use an if-else if-else statement to compare the lowercase character against the vowels ('a', 'e', 'i', 'o', 'u'). Step 5: Print the result: If the character matches a vowel, print that it's a vowel. Otherwise, print that it's a consonant. ───────────────────────────── Have you Understood? Drop a reaction: ❤️ Understood | 👎 Not Understood

📝 Check if a character is a vowel or consonant using if-else if-else Write a C program that takes a character as input and determines whether it is a vowel (a, e, i, o, u, case-insensitive). The program should use if-else if-else statements to output whether the input character is a vowel or a consonant.

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Find the largest among three numbers using nested if-else

#include <stdio.h>

int main() {
    int num1, num2, num3;

    printf("Enter three integers: ");
    scanf("%d %d %d", &num1, &num2, &num3);

    if (num1 > num2) {
        if (num1 > num3) {
            printf("Largest number: %dn", num1);
        } else {
            printf("Largest number: %dn", num3);
        }
    } else {
        if (num2 > num3) {
            printf("Largest number: %dn", num2);
        } else {
            printf("Largest number: %dn", num3);
        }
    }

    return 0;
}

💡 Approach Step 1: Declare three integer variables: Declare three integer variables (e.g., num1, num2, num3) to store the input numbers. Step 2: Read input from the user: Prompt the user to enter the values for the three integer variables and store them using scanf. Step 3: First if statement: Compare the first number (num1) with the second number (num2). If num1 is greater than num2, proceed to the nested if-else. Otherwise, proceed to the else part of the outer if. Step 4: Nested if-else (within outer if): If num1 was greater than num2, compare num1 with num3. If num1 is greater than num3, then num1 is the largest. Otherwise, num3 is the largest. Step 5: else block (of outer if): If num1 was not greater than num2, compare num2 with num3. If num2 is greater than num3, then num2 is the largest. Otherwise, num3 is the largest. Step 6: Print the largest number: After the nested if-else statements, print the variable that holds the largest number using printf. ───────────────────────────── Have you Understood? Drop a reaction: ❤️ Understood | 👎 Not Understood

📝 Find the largest among three numbers using nested if-else Write a C program that determines the largest of three integer numbers. Implement this using nested if-else statements to compare the numbers and identify the maximum value. The program should then print the largest number.

Check if a number is even or odd using if-else

#include <stdio.h>

int main() {
  int num;

  printf("Enter an integer: ");
  scanf("%d", &num);

  if (num % 2 == 0) {
    printf("Evenn");
  } else {
    printf("Oddn");
  }

  return 0;
}

💡 Approach Step 1: Get the integer input from the user. Store this number in a variable, say num. Step 2: Calculate the remainder when num is divided by 2 using the modulo operator (%). Step 3: Check if the remainder from Step 2 is equal to 0. Step 4: If the remainder is 0, print "Even". Otherwise (using else), print "Odd". ───────────────────────────── Have you Understood? Drop a reaction: ❤️ Understood | 👎 Not Understood

📝 Check if a number is even or odd using if-else Write a C program that takes an integer as input and determines whether it is even or odd. Use an if-else statement to check if the number is divisible by 2, and print "Even" if it is, or "Odd" if it is not.

📚 Control Flow Statements

Check operator precedence and associativity with an expression

#include <stdio.h>

int main() {
    int a = 10;
    int b = 5;
    int c = 2;

    int result = a + b * c;

    printf("Result of a + b * c: %dn", result);

    int a1 = 10;
    int b1 = 5;

    int result1 = a1 / b1 - 1;

    printf("Result of a / b - 1: %dn", result1);

    int x = 5;
    int y = 3;
    int z = 1;

    int result2 = x = y + z;

    printf("Result of x = y + z: %dn", result2);
    printf("Value of x after assignment: %dn", x);

    int i = 2;
    int j = 3;

    int result3 = i * j + i++;

    printf("Result of i * j + i++: %dn", result3);
    printf("Value of i after post-increment: %dn", i);

     int p = 5;
    int q = 2;

    int result4 = p % q * p + q;

    printf("Result of p %% q * p + q: %dn", result4);

    int num = 8;
    int shift_result = num << 2;

    printf("Result of num << 2: %dn", shift_result);

     int a2 = 1;
    int b2 = 2;
    int c2 = 3;

    int result5 = a2 < b2 ? b2 : c2;

    printf("Result of a2 < b2 ? b2 : c2: %dn", result5);

    return 0;
}

💡 Approach Here's a simple step-by-step approach to check operator precedence and associativity in C: Step 1: Understand the Expression: Carefully examine the C expression you want to evaluate. Identify all the operators involved. Step 2: Consult the Operator Precedence Table: Refer to a C operator precedence table. This table lists operators in order of priority (highest to lowest). Step 3: Group by Precedence (Highest to Lowest): Starting with the highest precedence operators, group the operands associated with those operators together. Think of these as mini-expressions to be evaluated first. Step 4: Handle Associativity within a Precedence Level: If multiple operators of the same precedence appear in the expression, use associativity (left-to-right or right-to-left) to determine the order of evaluation within that group. Step 5: Repeat Steps 3 & 4: Continue grouping and evaluating from highest precedence to lowest, resolving associativity at each level, until the entire expression is simplified to a single value. Use parentheses to explicitly show the order of evaluation if needed. Step 6: Write the C code: Express the evaluated expression in C, making sure that C compiler follows the same precedence. ───────────────────────────── Have you Understood? Drop a reaction: ❤️ Understood | 👎 Not Understood

📝 Check operator precedence and associativity with an expression Write a C program that evaluates a given arithmetic expression containing multiple operators (+, -, *, /, %) and parentheses. The program should correctly apply operator precedence and associativity rules to calculate and print the final result of the expression.

Calculate compound assignment operations

#include <stdio.h>

int main() {
    int initialValue = 10;
    int additionValue = 5;
    int subtractionValue = 3;
    int multiplicationValue = 2;
    float divisionValue = 2.0;
    int modulusValue = 3;

    int result = initialValue;
    printf("Initial value: %dn", result);

    result += additionValue;
    printf("After addition: %dn", result);

    result -= subtractionValue;
    printf("After subtraction: %dn", result);

    result *= multiplicationValue;
    printf("After multiplication: %dn", result);

    result /= (int)divisionValue;
    printf("After division: %dn", result);

    result %= modulusValue;
    printf("After modulus: %dn", result);

    return 0;
}