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.
显示更多📈 Telegram 频道 AI Coding & CSE Fundamentals 的分析概览
频道 AI Coding & CSE Fundamentals (@aicoding_csefundamentals) 英语 语言赛道中的 是活跃参与者。目前社区聚集了 12 318 名订阅者,在 技术与应用 类别中位列第 10 235,并在 印度 地区排名第 31 257 位。
📊 受众指标与增长动态
自 невідомо 创建以来,项目保持高速增长,吸引了 12 318 名订阅者。
根据 13 十一月, 2026 的最新数据,频道保持稳定运转。过去 30 天订阅人数变化为 -236,过去 24 小时变化为 -9,整体触达仍然可观。
- 认证状态: 未认证
- 互动率 (ER): 平均受众互动率为 10.85%。内容发布后 24 小时内通常能获得 N/A% 的反应,占订阅者总量。
- 帖子覆盖: 每篇帖子平均可获得 1 337 次浏览,首日通常累积 0 次浏览。
- 互动与反馈: 受众积极参与,单帖平均反应数为 11。
- 主题关注点: 内容集中在 input, string, scanf("%d, array, element 等核心主题上。
📝 描述与内容策略
作者将该频道定位为表达主观观点的平台:
“🚀 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.”
凭借高频更新(最新数据采集于 14 十一月, 2026),频道始终保持新鲜度与高覆盖。分析显示受众积极互动,使其成为 技术与应用 类别中的关键影响点。
#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;
}