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# ft\_is\_prime

**Objective:**

Create a function that determines if a given number is a prime number. The function should return `1` if the number is prime, and `0` if it is not.

**Turn-in Directory:**

`ex06/`

**Files to Turn In:**

`ft_is_prime.c`

**Allowed Functions:**

None

**Prototype:**

```c
int ft_is_prime(int nb);
```

**Detailed Explanation:**

A prime number is a whole number greater than 1 that has only two divisors: 1 and itself. Your `ft_is_prime` function needs to check if the input number `nb` meets this condition. Consider these points:

* Numbers less than or equal to 1 are *not* prime.
* To determine if a number `nb` is prime, you only need to check for divisibility from 2 up to `nb / 2`. If `nb` is divisible by any number in this range, it is not prime.
* If the loop completes without finding any divisors, the number is prime.

**Code Implementation:**

```c
int	ft_is_prime(int nb)
{
	int	is_p;

	is_p = 2;
	if (nb <= 1)
		return (0);
	while (is_p <= nb / 2)
	{
		if (nb % is_p == 0)
			return (0);
		is_p++;
	}
	return (1);
}
```

**Example Usage:**

```c
#include <stdio.h>

int ft_is_prime(int nb);

int main(void)
{
    printf("Is 7 prime? %d\n", ft_is_prime(7));   // Output: 1
    printf("Is 10 prime? %d\n", ft_is_prime(10));  // Output: 0
    printf("Is 1 prime? %d\n", ft_is_prime(1));   // Output: 0
    printf("Is 2 prime? %d\n", ft_is_prime(2));   // Output: 1
    printf("Is -5 prime? %d\n", ft_is_prime(-5));  // Output: 0

    return 0;
}
```

**Edge Cases to Consider:**

* **Numbers Less Than or Equal to 1:** These are not prime.
  * Input: `1` → Returns `0`
  * Input: `0` → Returns `0`
  * Input: `-5` → Returns `0`
* **The Number 2:** 2 is the smallest prime number.
  * Input: `2` → Returns `1`
* **Prime Numbers:**
  * Input: `7` → Returns `1`
  * Input: `7919` → Returns `1`
* **Non-Prime Numbers:**
  * Input: `4` → Returns `0`
  * Input: `10` → Returns `0`
  * Input: `100` → Returns `0`

**Limitations of `ft_is_prime`:**

* **Integer Overflow:** For very large numbers, the `nb / 2` calculation could potentially lead to issues if `nb` is close to the maximum integer value, though this is unlikely to be a major concern for this exercise.

**How It Works:**

1. **Initialization:** The variable `is_p` is initialized to 2. This variable will be used to check for potential divisors of `nb`.

   ```c
   is_p = 2;
   ```
2. **Base Case:** If `nb` is less than or equal to 1, it's not prime, so the function immediately returns 0.

   ```c
   if (nb <= 1)
       return (0);
   ```
3. **Divisibility Check:** The `while` loop iterates from 2 up to `nb / 2`. In each iteration:
   * It checks if `nb` is divisible by `is_p` without any remainder (`nb % is_p == 0`).
   * If it finds a divisor, `nb` is not prime, and the function returns 0.

     ```c
     if (nb % is_p == 0)
         return (0);
     ```
   * If `nb` is not divisible by `is_p`, the loop continues to the next potential divisor by incrementing `is_p`.

     ```c
     is_p++;
     ```
4. **Prime Determination:** If the loop completes without finding any divisors, it means `nb` is only divisible by 1 and itself, so the function returns 1, indicating that `nb` is prime.

   ```c
   return (1);
   ```

**Key Points to Remember:**

* **Definition of Prime:** Understand the fundamental definition of a prime number.
* **Optimization:** The loop only iterates up to `nb / 2` because a number cannot have a divisor greater than half of itself (excluding itself).
* **Base Cases:** Handling numbers less than or equal to 1 is crucial.
* **Return Values:** Ensure your function returns `1` for prime numbers and `0` for non-prime numbers.
* **Iterative Approach:** The function utilizes a `while` loop for checking potential divisors.


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