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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C Programming Language || Hands On Coding (@c_programming_language_coding) Ingliz til segmentidagi kanali faol ishtirokchi. Hozirda hamjamiyat 12 818 obunachidan iborat bo'lib, Texnologiyalar & Aralashmalar toifasida 9 567-o'rinni va Hindiston mintaqasida 30 989-o'rinni egallagan.
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невідомо sanasidan buyon loyiha tez o‘sib, 12 818 obunachiga ega bo‘ldi.
28 Avgust, 2026 dagi oxirgi ma’lumotlarga ko‘ra kanal barqaror faollikka ega. Oxirgi 30 kunda obunachilar soni -213 ga, so‘nggi 24 soatda esa -1 ga o‘zgardi va umumiy qamrov yuqori darajada qolmoqda.
- Tasdiqlash holati: Tasdiqlanmagan
- Jalb etish (ER): Auditoriya o‘rtacha 6.87% darajada jalb etiladi. Nashrdan keyingi dastlabki 24 soatda kontent odatda umumiy obunachilar sonining 2.42% ini tashkil etuvchi reaksiyalarni to‘playdi.
- Post qamrovi: Har bir post o‘rtacha 881 marta ko‘riladi; birinchi sutkada odatda 310 ta ko‘rish yig‘iladi.
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- Tematik yo‘nalishlar: Kontent input, string, scanf("%d, array, element kabi asosiy mavzularga jamlangan.
📝 Tavsif va kontent siyosati
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 29 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() {
int num1, num2;
scanf("%d %d", &num1, &num2);
if ((num1 ^ num2) == 0) {
printf("Numbers are equal!");
} else {
printf("Numbers are not equal!");
}
return 0;
}#include <stdio.h>
#include <math.h>
int main() {
float a, b, c, discriminant, root1, root2, realPart, imagPart;
printf("Enter coefficients a, b, and c: ");
scanf("%f %f %f", &a, &b, &c);
discriminant = b * b - 4 * a * c;
if (discriminant > 0) {
root1 = (-b + sqrt(discriminant)) / (2 * a);
root2 = (-b - sqrt(discriminant)) / (2 * a);
printf("Root 1 = %.2f and Root 2 = %.2f", root1, root2);
} else if (discriminant == 0) {
root1 = root2 = -b / (2 * a);
printf("Root 1 = Root 2 = %.2f;", root1);
} else {
realPart = -b / (2 * a);
imagPart = sqrt(-discriminant) / (2 * a);
printf("Root 1 = %.2f+%.2fi and Root 2 = %.2f-%.2fi", realPart, imagPart, realPart, imagPart);
}
return 0;
}#include <stdio.h>
int main() {
int year;
scanf("%d", &year);
if (year % 4 == 0) {
if (year % 100 == 0) {
if (year % 400 == 0) {
printf("Leap year\n");
} else {
printf("Not a leap year\n");
}
} else {
printf("Leap year\n");
}
} else {
printf("Not a leap year\n");
}
return 0;
}#include <stdio.h>
int main() {
int n, i, isPrime = 1;
scanf("%d", &n);
if (n <= 1) {
isPrime = 0;
} else {
if (n == 2) {
isPrime = 1;
} else {
for (i = 2; i * i <= n; i++) {
if (n % i == 0) {
isPrime = 0;
break;
}
}
}
}
if (isPrime)
printf("Prime");
else
printf("Not Prime");
return 0;
}#include <stdio.h>
int main() {
int num1, num2, num3, largest;
scanf("%d %d %d", &num1, &num2, &num3);
if (num1 >= num2 && num1 >= num3) {
largest = num1;
} else if (num2 >= num1 && num2 >= num3) {
largest = num2;
} else {
largest = num3;
}
printf("%d", largest);
return 0;
}
int age = 20;
if (age >= 18) {
printf("You are an adult! ✅n");
}
**2. The `else` Statement: Providing an Alternative**
The `else` statement is used in conjunction with `if` to provide an alternative code block to execute when the `if` condition is false.
`if (condition) {
// Code to execute if the condition is true
} else {
// Code to execute if the condition is false
}`
Example:
int age = 15;
if (age >= 18) {
printf("You are an adult! ✅n");
} else {
printf("You are not an adult yet. ⏳n");
}
**3. The `else if` Statement: Checking Multiple Conditions**
The `else if` statement allows you to check multiple conditions in a sequence.
`if (condition1) {
// Code to execute if condition1 is true
} else if (condition2) {
// Code to execute if condition1 is false AND condition2 is true
} else {
// Code to execute if all conditions are false
}`
Example:
int score = 75;
if (score >= 90) {
printf("Grade: A 🥇n");
} else if (score >= 80) {
printf("Grade: B 🥈n");
} else if (score >= 70) {
printf("Grade: C 🥉n");
} else {
printf("Grade: D or F 😥n");
}
**4. The `switch` Statement: Efficient Multi-Way Branching**
The `switch` statement provides a clean way to select one code block to execute from several options based on the value of an expression.
`switch (expression) {
case value1:
// Code to execute if expression == value1
break;
case value2:
// Code to execute if expression == value2
break;
default:
// Code to execute if expression doesn't match any of the cases
}`
- `expression`: An integer or character expression.
- `case`: Each `case` represents a specific value that `expression` might have.
- `break`: The `break` statement is crucial. It exits the `switch` statement after a match is found. Without `break`, the code will "fall through" to the next `case`. ⚠️
- `default`: The `default` case is optional and is executed if none of the other `case` values match the `expression`.
Example:
int day = 3;
switch (day) {
case 1:
printf("Mondayn");
break;
case 2:
printf("Tuesdayn");
break;
case 3:
printf("Wednesdayn");
break;
default:
printf("Invalid dayn");
}
💡 **Tips for Using Control Flow:**
- Keep your conditions clear and easy to understand.
- Use indentation to make your code readable. ✅
- Always include a `default` case in your `switch` statement to handle unexpected values.
- Be careful about "fall-through" in `switch` statements. Use `break` unless you specifically want this behavior. ⚠️
- When dealing with complex conditions, consider using logical operators (`&&` for AND, `||` for OR, `!` for NOT).
Control flow statements are essential for writing programs that can respond to different situations. Practice using `if`, `else`, and `switch` to master decision-making in your C programs! 💪#include <stdio.h>
int main() {
int x, y, max, min;
scanf("%d %d", &x, &y);
int diff = x - y;
int sign_bit = diff >> 31 & 1;
max = x - sign_bit * diff;
min = y + sign_bit * diff;
printf("Max: %d\n", max);
printf("Min: %d\n", min);
return 0;
}#include <stdio.h>
int main() {
int a = 10, b = 5;
a = a ^ b;
b = a ^ b;
a = a ^ b;
printf("a = %d, b = %d\n", a, b);
return 0;
}#include <stdio.h>
int main() {
unsigned int num = 10;
int bit_position = 1;
printf("Original number: %u\n", num);
unsigned int set_bit = num | (1 << bit_position);
printf("Number with bit set: %u\n", set_bit);
unsigned int clear_bit = num & ~(1 << bit_position);
printf("Number with bit cleared: %u\n", clear_bit);
unsigned int toggle_bit = num ^ (1 << bit_position);
printf("Number with bit toggled: %u\n", toggle_bit);
int bit_status = (num >> bit_position) & 1;
printf("Bit status (0 or 1): %d\n", bit_status);
return 0;
}#include <stdio.h>
int main() {
int a = 60;
int b = 13;
int result = 0;
result = a & b;
printf("a & b = %d\n", result);
result = a | b;
printf("a | b = %d\n", result);
result = a ^ b;
printf("a ^ b = %d\n", result);
result = ~a;
printf("~a = %d\n", result);
result = a << 2;
printf("a << 2 = %d\n", result);
result = a >> 2;
printf("a >> 2 = %d\n", result);
return 0;
}