C Programming Codes
C Programming Codes || Quizzes || DSA Learn along with the community Any queries admin - @Pradeep_saii
Show moreπ Analytical overview of Telegram channel C Programming Codes
Channel C Programming Codes (@c_programming_codes) in the English language segment is an active participant. Currently, the community unites 13 417 subscribers, ranking 9 552 in the Technologies & Applications category and 32 040 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 13 417 subscribers.
According to the latest data from 13 June, 2026, the channel demonstrates stable activity. Although there has been a change in the number of participants by -228 over the last 30 days and by -2 over the last 24 hours, overall reach remains high.
- Verification status: Not verified
- Engagement rate (ER): The average audience engagement rate is 9.78%. 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 0 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 0.
- 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:
βC Programming Codes || Quizzes || DSA
Learn along with the community
Any queries
admin - @Pradeep_saiiβ
Thanks to the high frequency of updates (latest data received on 14 June, 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 num;
printf("Enter an integer: ");
scanf("%d", &num);
if (num % 2 == 0) {
printf("%d is even.\n", num);
} else {
printf("%d is odd.\n", num);
}
return 0;
}#include <stdio.h>
int main() {
int integer_variable = 10;
float float_variable = 3.14;
char char_variable = 'A';
double double_variable = 3.14159;
short short_variable = 5;
long long_variable = 1234567890;
unsigned int unsigned_variable = 4294967295;
printf("Integer: %d\n", integer_variable);
printf("Float: %f\n", float_variable);
printf("Character: %c\n", char_variable);
printf("Double: %lf\n", double_variable);
printf("Short: %hd\n", short_variable);
printf("Long: %ld\n", long_variable);
printf("Unsigned Integer: %u\n", unsigned_variable);
return 0;
}#include <stdio.h>
int main() {
char op;
double num1, num2;
printf("Enter an operator (+, -, *, /): ");
scanf(" %c", &op);
printf("Enter two operands: ");
scanf("%lf %lf", &num1, &num2);
switch (op) {
case '+':
printf("%.1lf + %.1lf = %.1lf", num1, num2, num1 + num2);
break;
case '-':
printf("%.1lf - %.1lf = %.1lf", num1, num2, num1 - num2);
break;
case '*':
printf("%.1lf * %.1lf = %.1lf", num1, num2, num1 * num2);
break;
case '/':
if (num2 != 0) {
printf("%.1lf / %.1lf = %.1lf", num1, num2, num1 / num2);
} else {
printf("Error: Division by zero");
}
break;
default:
printf("Error: Invalid operator");
}
return 0;
}#include <stdio.h>
int main() {
printf("Size of int: %lu bytes\n", sizeof(int));
printf("Size of float: %lu bytes\n", sizeof(float));
printf("Size of char: %lu byte\n", sizeof(char));
printf("Size of double: %lu bytes\n", sizeof(double));
return 0;
}#include <stdio.h>
int main() {
int a = 10;
int b = 20;
a = a + b;
b = a - b;
a = a - b;
printf("a = %d, b = %d\n", a, b);
return 0;
}#include <stdio.h>
int main() {
int a = 10;
int b = 20;
int temp;
printf("Before swap: a = %d, b = %d\n", a, b);
temp = a;
a = b;
b = temp;
printf("After swap: a = %d, b = %d\n", a, b);
return 0;
}#include <stdio.h>
int main() {
int num1, num2, sum;
printf("Enter two integers: ");
scanf("%d %d", &num1, &num2);
sum = num1 + num2;
printf("Sum = %d\n", sum);
return 0;
}#include <stdio.h>
int main() {
char name[50];
int age;
printf("Enter your name: ");
scanf("%s", name);
printf("Enter your age: ");
scanf("%d", &age);
printf("Hello, %s! You are %d years old.\n", name, age);
return 0;
}#include <stdio.h>
int main() {
printf("Hello, World!");
return 0;
}
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