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C Programming Codes

C Programming Codes

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C Programming Codes || Quizzes || DSA Learn along with the community Any queries admin - @Pradeep_saii

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πŸ“ˆ Analytical overview of Telegram channel C Programming Codes

Channel C Programming Codes (@c_programming_codes) in the English language segment is an active participant. Currently, the community unites 13 430 subscribers, ranking 9 534 in the Technologies & Applications category and 32 075 in the India region.

πŸ“Š Audience metrics and dynamics

Since its creation on Π½Π΅Π²Ρ–Π΄ΠΎΠΌΠΎ, the project has demonstrated rapid growth, gathering an audience of 13 430 subscribers.

According to the latest data from 11 June, 2026, the channel demonstrates stable activity. Although there has been a change in the number of participants by -239 over the last 30 days and by -9 over the last 24 hours, overall reach remains high.

  • Verification status: Not verified
  • Engagement rate (ER): The average audience engagement rate is 9.78%. Within the first 24 hours after publication, content typically collects N/A% reactions from the total number of subscribers.
  • Post reach: On average, each post receives 0 views. Within the first day, a publication typically gains 0 views.
  • Reactions and interaction: The audience actively supports content: the average number of reactions per post is 0.
  • Thematic interests: Content is focused on key topics such as input, string, scanf("%d, array, element.

πŸ“ Description and content policy

The author describes the resource as a platform for expressing subjective opinions:
β€œC Programming Codes || Quizzes || DSA Learn along with the community Any queries admin - @Pradeep_saii”

Thanks to the high frequency of updates (latest data received on 12 June, 2026), the channel maintains relevance and a high level of publication reach. Analytics show that the audience actively interacts with content, making it an important point of influence in the Technologies & Applications category.

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Posts Archive
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;
}

πŸ’‘ 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 | πŸ‘Ž Not Understood

πŸ“ 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.

πŸš€Everyone join below channel to prepare for interviewsπŸ‘‡ https://t.me/leetcode_problems_pool

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;
}

πŸ’‘ 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 | πŸ‘Ž Not Understood

πŸ“ 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;
}

πŸ’‘ 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 | πŸ‘Ž Not Understood

πŸ“ 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.

πŸ“š 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;
}

πŸ’‘ 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 | πŸ‘Ž Not Understood

πŸ“ 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;
}

πŸ’‘ Approach Step 1: Declare variables: Declare the variables you'll be using, including the initial values for the operands involved in the compound assignment operations. Make sure to select appropriate data types (e.g., `int`, `float`) based on the expected values. Step 2: Perform compound assignment operations: Use the compound assignment operators (+=, -=, =, /=, %=) on the variables. For example, `x += y` is equivalent to `x = x + y`. Perform the calculations sequentially as required by the problem. Step 3: Print the results:* After each compound assignment, or at the end of all operations, print the final values of the variables that were modified. Use `printf` with appropriate format specifiers (e.g., `%d` for integers, `%f` for floats) to display the values correctly. ───────────────────────────── Have you Understood\? Drop a reaction: ❀️ Understood | πŸ‘Ž Not Understood

πŸ“ Calculate compound assignment operations Write a C program that takes two integer inputs, x and y. Perform and print the result of the following compound assignment operations on x: addition assignment (+= y), subtraction assignment (-= y), multiplication assignment (*= y), division assignment (/= y), and modulo assignment (%= y).

Use ternary operator (conditional operator) #include <stdio.h> int main() { int number = 10; int abs_value; abs_value = (number >= 0) ? number : -number; printf("The absolute value of %d is %d\n", number, abs_value); int age = 20; const char *status; status = (age >= 18) ? "Adult" : "Minor"; printf("Age: %d, Status: %s\n", age, status); int a = 5, b = 10, max; max = (a > b) ? a : b; printf("The maximum of %d and %d is %d\n", a, b, max); return 0; }

πŸ’‘ Approach Here's a step-by-step approach to using the ternary operator in C: Step 1: Identify the Condition: Determine the condition you want to evaluate. This is the expression that will be tested for truthiness. Step 2: Define the 'True' Result: Decide what value or expression should be executed/returned if the condition is true (non-zero). Step 3: Define the 'False' Result: Decide what value or expression should be executed/returned if the condition is false (zero). Step 4: Construct the Ternary Operator: Use the following structure: (condition) ? (result_if_true) : (result_if_false); Step 5: Assign or Use the Result: Assign the result of the ternary operation to a variable, use it in a print statement, or in any other way you need to utilize the conditional outcome. ───────────────────────────── Have you Understood? Drop a reaction: ❀️ Understood | πŸ‘Ž Not Understood