C Programming Codes
C Programming Codes || Quizzes || DSA Learn along with the community Any queries admin - @Pradeep_saii
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El canal C Programming Codes (@c_programming_codes) en el segmento lingüístico de Inglés es un actor destacado. Actualmente la comunidad reúne a 13 422 suscriptores, ocupando la posición 9 537 en la categoría Tecnologías y Aplicaciones y el puesto 32 062 en la región India.
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Desde su creación el невідомо, el proyecto ha mostrado un crecimiento acelerado, reuniendo a 13 422 suscriptores.
Según los últimos datos del 12 junio, 2026, el canal mantiene una actividad estable. En los últimos 30 días la variación de miembros fue de -240, y en las últimas 24 horas de -9, conservando un alto alcance.
- Estado de verificación: No verificado
- Tasa de interacción (ER): El promedio de interacción de la audiencia es 9.78%. Durante las primeras 24 horas tras publicar, el contenido suele obtener N/A% de reacciones respecto al total de suscriptores.
- Alcance de las publicaciones: Cada publicación recibe en promedio 0 visualizaciones. En el primer día suele acumular 0 visualizaciones.
- Reacciones e interacción: La audiencia responde de forma activa: el promedio de reacciones por publicación es 0.
- Intereses temáticos: El contenido se centra en temas clave como input, string, scanf("%d, array, element.
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Gracias a la alta frecuencia de actualizaciones (últimos datos recibidos el 13 junio, 2026), el canal mantiene la vigencia y un amplio alcance. La analítica demuestra que la audiencia interactúa activamente con el contenido, lo que lo convierte en un punto de referencia dentro de la categoría Tecnologías y Aplicaciones.
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;
}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.
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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.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;
}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.
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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.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;
}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".
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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.#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;
}#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;
}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).(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.
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