AI Coding & CSE Fundamentals
π Explore AI, Coding & Computer Science Fundamentals. π» Programming β’ DSA β’ DBMS β’ OS β’ Networks π€ AI & Emerging Technologies π οΈ Projects β’ Tutorials β’ Learning Resources π Your freedom to gain knowledge for free.
Show moreπ Analytical overview of Telegram channel AI Coding & CSE Fundamentals
Channel AI Coding & CSE Fundamentals (@aicoding_csefundamentals) in the English language segment is an active participant. Currently, the community unites 12 318 subscribers, ranking 10 235 in the Technologies & Applications category and 31 257 in the India region.
π Audience metrics and dynamics
Since its creation on Π½Π΅Π²ΡΠ΄ΠΎΠΌΠΎ, the project has demonstrated rapid growth, gathering an audience of 12 318 subscribers.
According to the latest data from 13 November, 2026, the channel demonstrates stable activity. Although there has been a change in the number of participants by -236 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 10.85%. 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 1 337 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 11.
- Thematic interests: Content is focused on key topics such as input, string, scanf("%d, array, element.
π Description and content policy
The author describes the resource as a platform for expressing subjective opinions:
βπ Explore AI, Coding & Computer Science Fundamentals.
π» Programming β’ DSA β’ DBMS β’ OS β’ Networks
π€ AI & Emerging Technologies
π οΈ Projects β’ Tutorials β’ Learning Resources
π Your freedom to gain knowledge for free.β
Thanks to the high frequency of updates (latest data received on 14 November, 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.
#include <stdio.h>
int main() {
int a = 60;
int b = 13;
int result = 0;
result = a & b;
printf("a & b = %d\n", result);
result = a | b;
printf("a | b = %d\n", result);
result = a ^ b;
printf("a ^ b = %d\n", result);
result = ~a;
printf("~a = %d\n", result);
result = a << 2;
printf("a << 2 = %d\n", result);
result = a >> 2;
printf("a >> 2 = %d\n", result);
return 0;
}#include <stdio.h>
int main() {
int integer_value = 10;
float float_value;
float_value = (float)integer_value;
printf("Integer: %d\n", integer_value);
printf("Float: %.1f\n", float_value);
float another_float = 3.14;
int another_integer;
another_integer = (int)another_float;
printf("Float: %.2f\n", another_float);
printf("Integer: %d\n", another_integer);
return 0;
}#include <stdio.h>
int main() {
int num = 10;
int *ptr;
ptr = #
printf("Value of num: %d\n", num);
printf("Address of num: %p\n", &num);
printf("Value of ptr: %p\n", ptr);
printf("Value pointed to by ptr: %d\n", *ptr);
*ptr = 20;
printf("New value of num: %d\n", num);
return 0;
}
#include <stdio.h>
int main() {
int age = 20;
int isAdult = (age >= 18) ? 1 : 0;
printf("Is adult: %d\n", isAdult);
return 0;
}#include <stdio.h>
int main() {
int a, b, c;
a = (b = 5, c = 10, b + c);
printf("a = %d, b = %d, c = %d\n", a, b, c);
return 0;
}#include <stdio.h>
int main() {
printf("Size of int: %lu bytes\n", sizeof(int));
printf("Size of char: %lu byte\n", sizeof(char));
printf("Size of float: %lu bytes\n", sizeof(float));
printf("Size of double: %lu bytes\n", sizeof(double));
return 0;
}#include <stdio.h>
int main() {
int x = 10;
int y = 5;
x += y;
printf("x += y: %d\n", x);
x -= y;
printf("x -= y: %d\n", x);
x *= y;
printf("x *= y: %d\n", x);
x /= y;
printf("x /= y: %d\n", x);
x %= y;
printf("x %%= y: %d\n", x);
return 0;
}#include <stdio.h>
int main() {
int a = 5, b = 3, c = 2;
int result1 = a + b * c;
int result2 = a << 1 + c;
printf("a + b * c = %d\n", result1);
printf("a << 1 + c = %d\n", result2);
return 0;
}#include <stdio.h>
int main() {
int num1 = 10;
int num2 = 3;
int sum = num1 + num2;
int difference = num1 - num2;
int product = num1 * num2;
int quotient = num1 / num2;
int remainder = num1 % num2;
printf("Sum: %d\n", sum);
printf("Difference: %d\n", difference);
printf("Product: %d\n", product);
printf("Quotient: %d\n", quotient);
printf("Remainder: %d\n", remainder);
return 0;
}