AI Coding & CSE Fundamentals
🚀 Explore AI, Coding & Computer Science Fundamentals. 💻 Programming • DSA • DBMS • OS • Networks 🤖 AI & Emerging Technologies 🛠️ Projects • Tutorials • Learning Resources 📚 Your freedom to gain knowledge for free.
Mostrar más📈 Análisis del canal de Telegram AI Coding & CSE Fundamentals
El canal AI Coding & CSE Fundamentals (@aicoding_csefundamentals) en el segmento lingüístico de Inglés es un actor destacado. Actualmente la comunidad reúne a 12 324 suscriptores, ocupando la posición 10 232 en la categoría Tecnologías y Aplicaciones y el puesto 31 271 en la región India.
📊 Métricas de audiencia y dinámica
Desde su creación el невідомо, el proyecto ha mostrado un crecimiento acelerado, reuniendo a 12 324 suscriptores.
Según los últimos datos del 12 noviembre, 2026, el canal mantiene una actividad estable. En los últimos 30 días la variación de miembros fue de -238, y en las últimas 24 horas de -8, conservando un alto alcance.
- Estado de verificación: No verificado
- Tasa de interacción (ER): El promedio de interacción de la audiencia es 10.45%. 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 1 288 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 11.
- Intereses temáticos: El contenido se centra en temas clave como input, string, scanf("%d, array, element.
📝 Descripción y política de contenido
El autor describe el recurso como un espacio para expresar opiniones subjetivas:
“🚀 Explore AI, Coding & Computer Science Fundamentals.
💻 Programming • DSA • DBMS • OS • Networks
🤖 AI & Emerging Technologies
🛠️ Projects • Tutorials • Learning Resources
📚 Your freedom to gain knowledge for free.”
Gracias a la alta frecuencia de actualizaciones (últimos datos recibidos el 13 noviembre, 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.
#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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❤️ Understood | 👎 Not Understoodx and y, and initialize x with a starting value (e.g., 10).
Step 2: Demonstrate post-increment: Print the current value of x. Then, use the post-increment operator (x++) and print the value of x again. Explain the observed change in value.
Step 3: Demonstrate pre-increment: Reset x to its original value (e.g., 10). Then, use the pre-increment operator (++x) and print the value of x. Explain the observed change in value.
Step 4: Demonstrate post-decrement: Initialize another integer variable y (e.g., with 20). Print the current value of y. Then, use the post-decrement operator (y--) and print the value of y again. Explain the observed change in value.
Step 5: Demonstrate pre-decrement: Reset y to its original value (e.g., 20). Then, use the pre-decrement operator (--y) and print the value of y. Explain the observed change in value.
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❤️ Understood | 👎 Not Understooda, b, c, d, e, f and initialize them with a value)
Step 2: Demonstrate the assignment operator (=). (Assign a value to variable a using the = operator)
Step 3: Demonstrate the addition assignment operator (+=). (Add a value to variable b using the += operator)
Step 4: Demonstrate the subtraction assignment operator (-=). (Subtract a value from variable c using the -= operator)
Step 5: Demonstrate the multiplication assignment operator (=). (Multiply variable d by a value using the = operator)
Step 6: Demonstrate the division assignment operator (/=). (Divide variable e by a value using the /= operator)
Step 7: Demonstrate the modulus assignment operator (%=). (Find the remainder of dividing variable f by a value using the %= operator)
Step 8: Print the values of all variables. (Print the values of a, b, c, d, e, and f to display the results of the assignment operations.)
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❤️ Understood | 👎 Not Understoodnum1, num2) and assign them some initial values. These will be the operands for our bitwise operations.
Step 2: Demonstrate the AND Operator (&): Print the result of num1 & num2. Explain in a comment that this performs a bitwise AND operation.
Step 3: Demonstrate the OR Operator (|): Print the result of num1 | num2. Explain in a comment that this performs a bitwise OR operation.
Step 4: Demonstrate the XOR Operator (^): Print the result of num1 ^ num2. Explain in a comment that this performs a bitwise XOR operation.
Step 5: Demonstrate the NOT Operator (~): Print the result of ~num1. Explain in a comment that this performs a bitwise NOT operation (one's complement).
Step 6: Demonstrate the Left Shift Operator (<<): Print the result of num1 << 2. Explain in a comment that this shifts the bits of num1 two positions to the left.
Step 7: Demonstrate the Right Shift Operator (>>): Print the result of num1 >> 2. Explain in a comment that this shifts the bits of num1 two positions to the right.
Step 8: Add Clear Output with Explanations: For each operation, use printf to display both the operation being performed (e.g., "num1 & num2 = ") and the result, including comments explaining what each operator does.
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❤️ Understood | 👎 Not Understooda, b, and result. Assign a and b some initial integer values (e.g., a = 5, b = 10).
Step 2: Demonstrate the AND (&&) Operator: Use an if statement with the && operator to check if a is greater than 0 AND b is less than 20. Print a message to the console based on whether the condition is true or false. Store the result of the conditional expression (a > 0 && b < 20) in the result variable and print its value (1 for true, 0 for false).
Step 3: Demonstrate the OR (||) Operator: Use another if statement with the || operator to check if a is less than 0 OR b is greater than 5. Print a message to the console based on whether the condition is true or false. Store the result of the conditional expression (a < 0 || b > 5) in the result variable and print its value (1 for true, 0 for false).
Step 4: Demonstrate the NOT (!) Operator: Use the ! operator to negate the value of a > b. Print the result to the console. Store the result of the expression !(a > b) in the result variable and print its value (1 for true, 0 for false).
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