现已上线！2025 年 Telegram 研究 — 年度关键洞察 
C Programming Codes
前往频道在 Telegram
C Programming Codes || Quizzes || DSA Learn along with the community Any queries admin - @Pradeep_saii
显示更多📈 Telegram 频道 C Programming Codes 的分析概览
频道 C Programming Codes (@c_programming_codes) 英语 语言赛道中的 是活跃参与者。目前社区聚集了 13 370 名订阅者，在 技术与应用 类别中位列第 9 567，并在 印度 地区排名第 31 797 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 13 370 名订阅者。
根据 18 六月, 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 -226，过去 24 小时变化为 -3，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 9.82%。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 0 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 0。
- 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：
“C Programming Codes || Quizzes || DSA
Learn along with the community
Any queries
admin - @Pradeep_saii”
凭借高频更新（最新数据采集于 19 六月, 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。
13 370
订阅者
-324 小时
-567 天
-22630 天
帖子存档
13 369
Discussion Group for c programming 👇
https://t.me/C_programming_language_group
13 369
Discussion Group for c programming 👇
https://t.me/C_programming_language_group
13 369
/* Performing sum of two numbers using ,
Function with arguments and with return type.*/
#include<stdio.h>
int sum(int,int);
int main()
{ int a,b,res;
a=20;
b=5;
res=sum(a,b);
printf("Sum=%d",res);
return 0;
}
int sum(int x,int y)
{
int sum=0;
sum=x+y;
return sum;
}
13 369
/* Performing sum of two numbers using ,
Function with arguments and without return type.*/
#include<stdio.h>
void sum(int,int);
int main()
{ int a,b;
a=42;
b=32;
sum(a,b);
return 0;
}
void sum(int x,int y)
{
int sum=0;
sum=x+y;
printf("Sum=%d",sum);
}
13 369
/* Performing sum of two numbers using ,
Function with no arguments and with return type.*/
#include<stdio.h>
int sum(void);
int main()
{ int res;
res=sum();
printf("Sum=%d",res);
return 0;
}
int sum()
{
int a,b,sum=0;
a=10;
b=45;
sum=a+b;
return sum;
}
13 369
/* Performing sum of two numbers using ,
Function with no arguments and without return type.*/
#include<stdio.h>
void sum(void);
int main()
{
sum();
return 0;
}
void sum()
{
int a,b,sum=0;
a=4;
b=9;
sum=a+b;
printf("sum=%d",sum);
}
13 369
Wild Pointer-A wild pointer refers to a pointer that is uninitialized or has been assigned an arbitrary value that does not point to a valid memory address
13 369
// NULL pointer-NULL pointer is a special value that represents a pointer
// that does not point to any valid memory address.
// dereferencing of a NULL pointer can't be done.
#include<stdio.h>
void main()
{ int *ptr=NULL;
if(ptr==NULL){
printf("ptr is a NULL pointer");
}
else{
printf("ptr is not a NULL pointer ");
}
}
13 369
// void pointer- A void pointer is a special pointer type that can hold the
// address of any data type.
//dereferencing of void pointer can only be done after typecasting into other data type.
#include<stdio.h>
void main()
{ void *vp;
int a=10;
float b=5.5;
char ch='b';
vp=&a;
printf("%d\n",*(int*)vp);
vp=&b;
printf("%f\n",*(float*)vp);
vp=&ch;
printf("%c",*(char*)vp);
}
13 369
What will be the output of above program .(Do without running)
send output here👇
@c_programmerrr
13 369
#include<stdio.h>
void main()
{
int a[]={10,11,-1,56,67,5,4};
int *p,*q;
p=&a[0];
q=&a[0]+3;
printf("%d,%d,%d\n",(*p)++,(*p)++,*(++p));
printf("%d\n",*p);
printf("%d\n",(*p)++);
printf("%d\n",(*p)++);
q--;
printf("%d\n",(*(q+2))--);
printf("%d\n",*(p+2)-2);
printf("%d\n",*(p++-2)-1);
}
13 369
#include<stdio.h>
void main()
{
int a[7]={4,8,3,6,1,2,7};
int *p=&a[0],*q;
printf("%d\n",*p); //4
printf("%d,%d,%d\n",(*p)++,*p++,*++p); //3,8,8
printf("%d\n",*p); //4
q=p+3; //q is pointing to a[5]
printf("%d\n",*q-3); //-1
printf("%d\n",*--p+5); //13
printf("%d\n",*p+*q); //10
}
13 369
//Performing increment and decrement on pointer
#include<stdio.h>
void main()
{
int a[5]={5,4,6,8,3};
int *p=&a[0];
printf("Value:%d\n",*p); // Value:5
p++; // now,p is pointing to a[1]
printf("Value:%d\n",*p); // Value:4
++p; // now p is pointing to a[2]
printf("Value:%d\n",*p); // value:6
p--; // now p is pointing to a[1]
printf("Value:%d\n",*p); // value:4
--p; // now p is pointing to a[0]
printf("Value:%d\n",*p); // Value:5
}
