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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 13 412 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 13 412 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 = 7;
if (num & 1) {
printf("%d is odd.\n", num);
} else {
printf("%d is even.\n", num);
}
return 0;
}#include <stdio.h>
#include <stdint.h>
unsigned char setBits[256] = {
0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 4,
1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5,
1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5,
2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5,
2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7,
1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5,
2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7,
2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7,
3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7,
4, 5, 5, 6, 5, 6, 6, 7, 5, 6, 6, 7, 6, 7, 7, 8
};
int countSetBits(int n) {
int count = 0;
count += setBits[n & 0xFF];
n >>= 8;
count += setBits[n & 0xFF];
n >>= 8;
count += setBits[n & 0xFF];
n >>= 8;
count += setBits[n & 0xFF];
return count;
}
int main() {
int num = 0b10110011101011001101010111100010;
printf("Number of set bits in %d is %d\n", num, countSetBits(num));
return 0;
}#include <stdio.h>
int isolateRightmostSetBit(int n) {
return n & -n;
}
int main() {
int num = 20; // Example number (binary 10100)
int isolatedBit = isolateRightmostSetBit(num);
printf("Original number: %d\n", num);
printf("Rightmost set bit: %d\n", isolatedBit);
return 0;
}#include <stdio.h>
int turnOffRightmostSetBit(int n) {
return n & (n - 1);
}
int main() {
int num = 12;
int result = turnOffRightmostSetBit(num);
printf("Original number: %d\n", num);
printf("Number with rightmost set bit turned off: %d\n", result);
return 0;
}#include <stdio.h>
int findRightmostSetBit(unsigned int n) {
if (n == 0) {
return 0;
}
int pos = 1;
while ((n & 1) == 0) {
n = n >> 1;
pos++;
}
return pos;
}
int main() {
unsigned int num = 12; // Example number (1100 in binary)
int position = findRightmostSetBit(num);
printf("Rightmost set bit is at position: %d\n", position);
return 0;
}#include <stdio.h>
int findRightmostSetBit(unsigned int n) {
if (n == 0)
return 0;
int pos = 1;
while ((n & 1) == 0) {
n = n >> 1;
pos++;
}
return pos;
}
int main() {
unsigned int num = 12; // Example: 1100 in binary
int position = findRightmostSetBit(num);
printf("Rightmost set bit is at position: %d\n", position);
return 0;
}#include <stdio.h>
#include <stdbool.h>
bool isOnlyOneBitSet(int n) {
if (n <= 0) {
return false;
}
return (n & (n - 1)) == 0;
}
int main() {
int num = 8;
if (isOnlyOneBitSet(num)) {
printf("%d has only one bit set.\n", num);
} else {
printf("%d does not have only one bit set.\n", num);
}
return 0;
}#include <stdio.h>
int main() {
int num = 16;
int multiplyBy2 = num << 1;
int divideBy2 = num >> 1;
printf("Original number: %d\n", num);
printf("Multiplied by 2: %d\n", multiplyBy2);
printf("Divided by 2: %d\n", divideBy2);
return 0;
}#include <stdio.h>
#include <stdbool.h>
bool isPowerOfTwo(int n) {
if (n <= 0) {
return false;
}
return (n & (n - 1)) == 0;
}
int main() {
int num = 16;
if (isPowerOfTwo(num)) {
printf("%d is a power of 2\n", num);
} else {
printf("%d is not a power of 2\n", num);
}
return 0;
}#include <stdio.h>#include <stdbool.h>bool isPowerOfTwo(int n) { if (n <= 0) { return false; } return (n & (n - 1)) == 0;}int main() { int num = 16; if (isPowerOfTwo(num)) { printf("%d is a power of 2\n", num); } else { printf("%d is not a power of 2\n", num); } return 0;}