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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Channel C Programming Language || Hands On Coding (@c_programming_language_coding) in the English language segment is an active participant. Currently, the community unites 12 818 subscribers, ranking 9 567 in the Technologies & Applications category and 30 989 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 12 818 subscribers.
According to the latest data from 28 August, 2026, the channel demonstrates stable activity. Although there has been a change in the number of participants by -213 over the last 30 days and by -1 over the last 24 hours, overall reach remains high.
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- Post reach: On average, each post receives 881 views. Within the first day, a publication typically gains 310 views.
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- 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:
βHands-on C programming language challenges for beginners. Learn building logic by solving programs.
Owner: @Pradeep_saiiβ
Thanks to the high frequency of updates (latest data received on 29 August, 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 n, i = 1;
printf("Enter a number: ");
scanf("%d", &n);
do {
printf("%d ", i);
i++;
} while (i <= n);
printf("\n");
return 0;
}#include <stdio.h>
int main() {
int n, i = 1;
printf("Enter a number: ");
scanf("%d", &n);
while (i <= n) {
printf("%d ", i);
i++;
}
printf("\n");
return 0;
}#include <stdio.h>
int main() {
int n, i;
printf("Enter a number: ");
scanf("%d", &n);
for (i = 1; i <= n; i++) {
printf("%d ", i);
}
printf("\n");
return 0;
}
#include <stdio.h>
int main() {
int size = 5;
for (int i = 0; i < size; i++) {
for (int j = 0; j < size; j++) {
printf("* ");
}
printf("n");
}
return 0;
}
2. **Inverted Right-Angled Triangle:**
#include <stdio.h>
int main() {
int rows = 5;
for (int i = rows; i >= 1; i--) {
for (int j = 1; j <= i; j++) {
printf("* ");
}
printf("n");
}
return 0;
}
Practice Tips:
- Start with simple patterns and gradually increase the complexity.
- Break down the pattern into rows and columns.
- Identify the relationship between the row and column numbers and the characters to be printed.
- Draw the pattern on paper first to visualize the output.
Practice Problems:
- Print a hollow square pattern.
- Print a diamond pattern.
- Print a pyramid pattern.
- Print patterns with numbers or characters instead of ''.
π Congratulations! You've taken the first steps into the world of loops and patterns in C. Keep practicing, and you'll be amazed at what you can create! Remember practice makes perfect. π¨βπ»
for (initialization; condition; increment/decrement) {
// Code to be executed repeatedly
}
Explanation:
1. `initialization`: This is executed only once at the beginning of the loop. Usually, it's used to declare and initialize a counter variable (e.g., `int i = 0;`).
2. `condition`: This is checked before each iteration of the loop. If the condition is true, the code inside the loop is executed. If it's false, the loop terminates. (e.g., `i < 10;`).
3. `increment/decrement`: This is executed after each iteration of the loop. It's usually used to update the counter variable (e.g., `i++`).
Example:
#include <stdio.h>
int main() {
for (int i = 0; i < 5; i++) {
printf("Iteration: %dn", i);
}
return 0;
}
Output:
```
Iteration: 0
Iteration: 1
Iteration: 2
Iteration: 3
Iteration: 4
```
β
Good Practice: Use `for` loops when you know the number of iterations beforehand.
βΎοΈ **2. `while` Loop**
The `while` loop executes a block of code as long as a condition is true. It's useful when you don't know the exact number of iterations.
Syntax:
while (condition) {
// Code to be executed repeatedly
}
Explanation:
The `condition` is checked before each iteration. If it's true, the code inside the loop is executed. If it's false, the loop terminates. β οΈ Make sure the condition eventually becomes false, or you'll end up with an infinite loop!
Example:
#include <stdio.h>
int main() {
int i = 0;
while (i < 5) {
printf("Iteration: %dn", i);
i++; // Increment i to avoid an infinite loop
}
return 0;
}
Output:
```
Iteration: 0
Iteration: 1
Iteration: 2
Iteration: 3
Iteration: 4
```
β
Good Practice: Use `while` loops when you need to repeat something until a certain condition is met.
π« **3. `do-while` Loop**
The `do-while` loop is similar to the `while` loop, but it guarantees that the code inside the loop is executed at least once.
Syntax:
do {
// Code to be executed repeatedly
} while (condition);
Explanation:
The code inside the loop is executed first, and then the `condition` is checked. If the condition is true, the loop continues. If it's false, the loop terminates. Notice the semicolon (`;`) at the end of the `while` condition.
Example:
#include <stdio.h>
int main() {
int i = 0;
do {
printf("Iteration: %dn", i);
i++;
} while (i < 5);
return 0;
}
Output:
```
Iteration: 0
Iteration: 1
Iteration: 2
Iteration: 3
Iteration: 4
```
β
Good Practice: Use `do-while` loops when you need to execute a block of code at least once, regardless of the initial condition.
π¨ **Patterns with Loops**
Now, let's combine loops to create patterns! This involves using nested loops (loops inside loops) to control the output.
Example: Printing a right-angled triangle
#include <stdio.h>
int main() {
int rows = 5;
for (int i = 1; i <= rows; i++) {
for (int j = 1; j <= i; j++) {
printf("* ");
}
printf("n");
}
return 0;
}
Output:
```
*
* *
* * *
* * * *
* * * * *
```
Explanation:
- The outer `for` loop controls the number of rows.
- The inner `for` loop controls the number of characters (in this case, ' ' )// Use switch for basic arithmetic operations
// Code will be generated in the interactive explainer
π Get AI Explanation#include <stdio.h>
int main() {
int month;
printf("Enter month number (1-12): ");
scanf("%d", &month);
switch (month) {
case 1: case 3: case 5: case 7: case 8: case 10: case 12:
printf("31 days\n");
break;
case 4: case 6: case 9: case 11:
printf("30 days\n");
break;
case 2:
printf("28 or 29 days\n");
break;
default:
printf("Invalid month\n");
}
return 0;
}#include <stdio.h>
#include <string.h>
int main() {
char username[50];
char password[50];
printf("Username: ");
scanf("%s", username);
printf("Password: ");
scanf("%s", password);
if (strcmp(username, "admin") == 0) {
if (strcmp(password, "secret123") == 0) {
printf("Login successful!
");
} else {
printf("Incorrect password.
");
}
} else {
printf("Incorrect username.
");
}
return 0;
}#include <stdio.h>
int main() {
char ch;
printf("Enter a character: ");
scanf(" %c", &ch);
switch (ch) {
case 'a':
case 'e':
case 'i':
case 'o':
case 'u':
case 'A':
case 'E':
case 'I':
case 'O':
case 'U':
printf("%c is a vowel.\n", ch);
break;
default:
printf("%c is a consonant.\n", ch);
}
return 0;
}#include <stdio.h>
int main() {
char grade;
printf("Enter your grade (A/B/C/D/F): ");
scanf(" %c", &grade);
switch (grade) {
case 'A':
printf("Excellent! π");
break;
case 'B':
printf("Good job! π");
break;
case 'C':
printf("Keep it up! π");
break;
case 'D':
printf("Needs improvement. π€");
break;
case 'F':
printf("Failed. π");
break;
default:
printf("Invalid grade. π");
}
printf("\n");
return 0;
}#include <stdio.h>
int main() {
int day = 4;
switch (day) {
case 1:
printf("Monday");
break;
case 2:
printf("Tuesday");
break;
case 3:
printf("Wednesday");
break;
case 4:
printf("Thursday");
break;
case 5:
printf("Friday");
break;
case 6:
printf("Saturday");
break;
case 7:
printf("Sunday");
break;
default:
printf("Invalid day");
}
return 0;
}#include <stdio.h>
int main() {
int choice, a, b, result;
printf("1. Add\n2. Subtract\n3. Multiply\n4. Divide\nEnter your choice: ");
scanf("%d", &choice);
printf("Enter two numbers: ");
scanf("%d %d", &a, &b);
switch (choice) {
case 1:
result = a + b;
printf("Result: %d\n", result);
break;
case 2:
result = a - b;
printf("Result: %d\n", result);
break;
case 3:
result = a * b;
printf("Result: %d\n", result);
break;
case 4:
if (b != 0) {
result = a / b;
printf("Result: %d\n", result);
} else {
printf("Cannot divide by zero!\n");
}
break;
default:
printf("Invalid choice!\n");
}
return 0;
}