C Programming Language || Hands On Coding
Hands-on C programming language challenges for beginners. Learn building logic by solving programs. Owner: @Pradeep_saii
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Muallif resursni shaxsiy fikrni ifoda etish maydoni sifatida ta’riflaydi:
“Hands-on C programming language challenges for beginners. Learn building logic by solving programs.
Owner: @Pradeep_saii”
Yuqori yangilanish chastotasi (oxirgi ma’lumot 30 Avgust, 2026 da olingan) sababli kanal doimo dolzarb va katta qamrovli bo‘lib qoladi. Analitika auditoriya kontent bilan faol hamkorlik qilishini, uni Texnologiyalar & Aralashmalar toifasidagi muhim ta’sir nuqtasiga aylantirishini ko‘rsatadi.
#include <stdio.h>
int main() {
printf("Hello, World!");
return 0;
}#include <stdio.h>int main() { int n = 10; int sum = (n * (n + 1)) / 2; printf("Sum of numbers from 1 to %d is: %d\n", n, sum); return 0;}#include <stdio.h>#include <stdbool.h>bool isPowerOfTwo(int n) { return (n > 0) && ((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>
void increment(int *n) {
(*n)++;
}
int main() {
int num = 10;
printf("Before: %d\n", num);
increment(&num);
printf("After: %d\n", num);
return 0;
}#include <stdio.h>
#include <stdlib.h>
int* create_dynamic_array(int size) {
if (size <= 0) {
return NULL;
}
int* arr = (int*)malloc(size * sizeof(int));
if (arr == NULL) {
return NULL;
}
for (int i = 0; i < size; i++) {
arr[i] = i + 1;
}
return arr;
}
int main() {
int size = 5;
int* my_array = create_dynamic_array(size);
if (my_array != NULL) {
for (int i = 0; i < size; i++) {
printf("%d ", my_array[i]);
}
printf("\n");
free(my_array);
my_array = NULL;
}
return 0;
}#include <stdio.h>
#include <stdlib.h>
int main() {
int x = 10;
int *ptr = &x;
int **ptr_to_ptr = &ptr;
printf("Value of x: %d\n", x);
printf("Address of x: %p\n", &x);
printf("Value of ptr: %p\n", ptr);
printf("Value pointed to by ptr: %d\n", *ptr);
printf("Address of ptr: %p\n", &ptr);
printf("Value of ptr_to_ptr: %p\n", ptr_to_ptr);
printf("Value pointed to by ptr_to_ptr: %p\n", *ptr_to_ptr);
printf("Value pointed to by *ptr_to_ptr: %d\n", **ptr_to_ptr);
printf("Address of ptr_to_ptr: %p\n", &ptr_to_ptr);
// Example: Dynamically allocate an array of strings
int num_strings = 3;
char **string_array = (char **)malloc(num_strings * sizeof(char *));
if (string_array == NULL) {
perror("malloc failed");
return 1;
}
string_array[0] = (char *)malloc(10 * sizeof(char));
string_array[1] = (char *)malloc(15 * sizeof(char));
string_array[2] = (char *)malloc(20 * sizeof(char));
if (string_array[0] == NULL || string_array[1] == NULL || string_array[2] == NULL) {
perror("malloc failed");
// Free previously allocated memory to avoid memory leaks
for(int i = 0; i < num_strings; ++i){
if(string_array[i] != NULL){
free(string_array[i]);
}
}
free(string_array);
return 1;
}
sprintf(string_array[0], "Hello");
sprintf(string_array[1], "World!");
sprintf(string_array[2], "Double Pointers");
for (int i = 0; i < num_strings; i++) {
printf("string_array[%d] = %s\n", i, string_array[i]);
}
// Free dynamically allocated memory
for (int i = 0; i < num_strings; i++) {
free(string_array[i]);
}
free(string_array);
return 0;
}#include <stdio.h>
int main() {
int arr[] = {10, 20, 30, 40, 50};
int *ptr = arr;
int size = sizeof(arr) / sizeof(arr[0]);
printf("Array elements:\n");
for (int i = 0; i < size; i++) {
printf("Element %d: %d\n", i, *ptr);
ptr++;
}
ptr = arr;
printf("Array elements (using pointer arithmetic):\n");
for (int i = 0; i < size; i++) {
printf("Element %d: %d\n", i, *(arr + i));
}
return 0;
}#include <stdio.h>
void swap(int *x, int *y) {
int temp = *x;
*x = *y;
*y = temp;
}
int main() {
int a = 10;
int b = 20;
printf("Before swap: a = %d, b = %d\n", a, b);
swap(&a, &b);
printf("After swap: a = %d, b = %d\n", a, b);
return 0;
}#include <stdio.h>
int stringLength(const char *str) {
const char *p = str;
while (*p != '\0') {
p++;
}
return p - str;
}
int main() {
char str[] = "Hello, World!";
int len = stringLength(str);
printf("Length of the string: %d\n", len);
return 0;
}