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.
Ko'proq ko'rsatish📈 Telegram kanali AI Coding & CSE Fundamentals analitikasi
AI Coding & CSE Fundamentals (@aicoding_csefundamentals) Ingliz til segmentidagi kanali faol ishtirokchi. Hozirda hamjamiyat 12 318 obunachidan iborat bo'lib, Texnologiyalar & Aralashmalar toifasida 10 235-o'rinni va Hindiston mintaqasida 31 257-o'rinni egallagan.
📊 Auditoriya ko‘rsatkichlari va dinamika
невідомо sanasidan buyon loyiha tez o‘sib, 12 318 obunachiga ega bo‘ldi.
13 Noyabr, 2026 dagi oxirgi ma’lumotlarga ko‘ra kanal barqaror faollikka ega. Oxirgi 30 kunda obunachilar soni -236 ga, so‘nggi 24 soatda esa -9 ga o‘zgardi va umumiy qamrov yuqori darajada qolmoqda.
- Tasdiqlash holati: Tasdiqlanmagan
- Jalb etish (ER): Auditoriya o‘rtacha 10.85% darajada jalb etiladi. Nashrdan keyingi dastlabki 24 soatda kontent odatda umumiy obunachilar sonining N/A% ini tashkil etuvchi reaksiyalarni to‘playdi.
- Post qamrovi: Har bir post o‘rtacha 1 337 marta ko‘riladi; birinchi sutkada odatda 0 ta ko‘rish yig‘iladi.
- Reaksiyalar va o‘zaro ta’sir: Auditoriya faol: har bir postga o‘rtacha 11 ta reaksiya keladi.
- Tematik yo‘nalishlar: Kontent input, string, scanf("%d, array, element kabi asosiy mavzularga jamlangan.
📝 Tavsif va kontent siyosati
Muallif resursni shaxsiy fikrni ifoda etish maydoni sifatida ta’riflaydi:
“🚀 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.”
Yuqori yangilanish chastotasi (oxirgi ma’lumot 14 Noyabr, 2026 da olingan) sababli kanal doimo dolzarb va katta qamrovli bo‘lib qoladi. Analitika auditoriya kontent bilan faol hamkorlik qilishini, uni Texnologiyalar & Aralashmalar toifasidagi muhim ta’sir nuqtasiga aylantirishini ko‘rsatadi.
#include <stdio.h>
int main() {
int choice, a, b, result;
printf("1. Add\n2. Subtract\n3. Multiply\n4. Divide\nEnter your choice: ");
scanf("%d", &choice);
printf("Enter two numbers: ");
scanf("%d %d", &a, &b);
switch (choice) {
case 1:
result = a + b;
printf("Result: %d\n", result);
break;
case 2:
result = a - b;
printf("Result: %d\n", result);
break;
case 3:
result = a * b;
printf("Result: %d\n", result);
break;
case 4:
if (b != 0) {
result = a / b;
printf("Result: %d\n", result);
} else {
printf("Cannot divide by zero!\n");
}
break;
default:
printf("Invalid choice!\n");
}
return 0;
}#include <stdio.h>
int main() {
int num1, num2;
scanf("%d %d", &num1, &num2);
if ((num1 ^ num2) == 0) {
printf("Numbers are equal!");
} else {
printf("Numbers are not equal!");
}
return 0;
}#include <stdio.h>
#include <math.h>
int main() {
float a, b, c, discriminant, root1, root2, realPart, imagPart;
printf("Enter coefficients a, b, and c: ");
scanf("%f %f %f", &a, &b, &c);
discriminant = b * b - 4 * a * c;
if (discriminant > 0) {
root1 = (-b + sqrt(discriminant)) / (2 * a);
root2 = (-b - sqrt(discriminant)) / (2 * a);
printf("Root 1 = %.2f and Root 2 = %.2f", root1, root2);
} else if (discriminant == 0) {
root1 = root2 = -b / (2 * a);
printf("Root 1 = Root 2 = %.2f;", root1);
} else {
realPart = -b / (2 * a);
imagPart = sqrt(-discriminant) / (2 * a);
printf("Root 1 = %.2f+%.2fi and Root 2 = %.2f-%.2fi", realPart, imagPart, realPart, imagPart);
}
return 0;
}#include <stdio.h>
int main() {
int year;
scanf("%d", &year);
if (year % 4 == 0) {
if (year % 100 == 0) {
if (year % 400 == 0) {
printf("Leap year\n");
} else {
printf("Not a leap year\n");
}
} else {
printf("Leap year\n");
}
} else {
printf("Not a leap year\n");
}
return 0;
}#include <stdio.h>
int main() {
int n, i, isPrime = 1;
scanf("%d", &n);
if (n <= 1) {
isPrime = 0;
} else {
if (n == 2) {
isPrime = 1;
} else {
for (i = 2; i * i <= n; i++) {
if (n % i == 0) {
isPrime = 0;
break;
}
}
}
}
if (isPrime)
printf("Prime");
else
printf("Not Prime");
return 0;
}#include <stdio.h>
int main() {
int num1, num2, num3, largest;
scanf("%d %d %d", &num1, &num2, &num3);
if (num1 >= num2 && num1 >= num3) {
largest = num1;
} else if (num2 >= num1 && num2 >= num3) {
largest = num2;
} else {
largest = num3;
}
printf("%d", largest);
return 0;
}
int age = 20;
if (age >= 18) {
printf("You are an adult! ✅n");
}
**2. The `else` Statement: Providing an Alternative**
The `else` statement is used in conjunction with `if` to provide an alternative code block to execute when the `if` condition is false.
`if (condition) {
// Code to execute if the condition is true
} else {
// Code to execute if the condition is false
}`
Example:
int age = 15;
if (age >= 18) {
printf("You are an adult! ✅n");
} else {
printf("You are not an adult yet. ⏳n");
}
**3. The `else if` Statement: Checking Multiple Conditions**
The `else if` statement allows you to check multiple conditions in a sequence.
`if (condition1) {
// Code to execute if condition1 is true
} else if (condition2) {
// Code to execute if condition1 is false AND condition2 is true
} else {
// Code to execute if all conditions are false
}`
Example:
int score = 75;
if (score >= 90) {
printf("Grade: A 🥇n");
} else if (score >= 80) {
printf("Grade: B 🥈n");
} else if (score >= 70) {
printf("Grade: C 🥉n");
} else {
printf("Grade: D or F 😥n");
}
**4. The `switch` Statement: Efficient Multi-Way Branching**
The `switch` statement provides a clean way to select one code block to execute from several options based on the value of an expression.
`switch (expression) {
case value1:
// Code to execute if expression == value1
break;
case value2:
// Code to execute if expression == value2
break;
default:
// Code to execute if expression doesn't match any of the cases
}`
- `expression`: An integer or character expression.
- `case`: Each `case` represents a specific value that `expression` might have.
- `break`: The `break` statement is crucial. It exits the `switch` statement after a match is found. Without `break`, the code will "fall through" to the next `case`. ⚠️
- `default`: The `default` case is optional and is executed if none of the other `case` values match the `expression`.
Example:
int day = 3;
switch (day) {
case 1:
printf("Mondayn");
break;
case 2:
printf("Tuesdayn");
break;
case 3:
printf("Wednesdayn");
break;
default:
printf("Invalid dayn");
}
💡 **Tips for Using Control Flow:**
- Keep your conditions clear and easy to understand.
- Use indentation to make your code readable. ✅
- Always include a `default` case in your `switch` statement to handle unexpected values.
- Be careful about "fall-through" in `switch` statements. Use `break` unless you specifically want this behavior. ⚠️
- When dealing with complex conditions, consider using logical operators (`&&` for AND, `||` for OR, `!` for NOT).
Control flow statements are essential for writing programs that can respond to different situations. Practice using `if`, `else`, and `switch` to master decision-making in your C programs! 💪#include <stdio.h>
int main() {
int x, y, max, min;
scanf("%d %d", &x, &y);
int diff = x - y;
int sign_bit = diff >> 31 & 1;
max = x - sign_bit * diff;
min = y + sign_bit * diff;
printf("Max: %d\n", max);
printf("Min: %d\n", min);
return 0;
}#include <stdio.h>
int main() {
int a = 10, b = 5;
a = a ^ b;
b = a ^ b;
a = a ^ b;
printf("a = %d, b = %d\n", a, b);
return 0;
}#include <stdio.h>
int main() {
unsigned int num = 10;
int bit_position = 1;
printf("Original number: %u\n", num);
unsigned int set_bit = num | (1 << bit_position);
printf("Number with bit set: %u\n", set_bit);
unsigned int clear_bit = num & ~(1 << bit_position);
printf("Number with bit cleared: %u\n", clear_bit);
unsigned int toggle_bit = num ^ (1 << bit_position);
printf("Number with bit toggled: %u\n", toggle_bit);
int bit_status = (num >> bit_position) & 1;
printf("Bit status (0 or 1): %d\n", bit_status);
return 0;
}