8 replies to this topic

[sobersaber]

Hi, this is a program to detect prime numbers or not. I understand what the program does except for the i++ part. Can someone tell me why does the program have i++ included?

#include <stdio.h>

int main()

{

    int num,i;

    printf("Enter a number");

    scanf("%d",&num);

    i=2;

    while(i<=num-1)

    {

        if(num%i==0)

        {

        printf("Not a prime number\n");

        break;

    }

[B]    i++;

[/B]}



if(i==num)

    printf("Prime number\n");

}   

[Vladimir]

What is about this? :)

#include <stdio.h>

int main()

{

    int num,i;

    printf("Enter a number");

    scanf("%d",&num);

    i=2;

    while(i<=num-1)

    {

        if(num%i==0)

        {

            printf("Not a prime number\n");

            break;

        }

        i++;

    }



    if(i==num)

        printf("Prime number\n");

}

[Flying Dutchman]

i++ is just fancy way of saying i = i + 1

Also, when checking for prime's, the fastest way I know of is checking up to square of that number.w

#include <iostream>
#include <cmath>

bool is_prime(int n) {
    if (n < 2)  // 0 and 1 and not primes
        return false;

    int sn = sqrt(abs(n));  // abs if negative values is passed, and sqrt for speed

    for (int i = 2; i < sn + 1; ++i)
        if ((n % i) * n == 0)
            return false;
    return true;
}

int main() {

    for (int i = 0; i < 20; ++i)
        if (is_prime(i))
            std::cout << i << " is prime\n";

    return 0;
}

[nerio]

Flying Dutchman said:

Also, when checking for prime's, the fastest way I know of is checking up to square of that number.

Also, when checking for prime's, the fastest way I know of is checking up to square root of that number.

[xle_camry]

i++ here is increment of that number. i++ also same with i=i+1. For example, it starts from 2. And increments by one. It checks until num-1. Firstly it checkes 2, then 3,4,5, etc. If you dont put increment, then your program wont work. It ll only work with 2.

[mnirahd]

Hi,

The program is running in a loop where it determines whether given number is a prime or not.
The logic is that it loop goes on until one of the following happens. i++ provides a next number each turn the loop runs

1. number is dividable by i
2. i reaches equal to number.

if you remove i++: the loop would go on forever:

Munir

[Flying Dutchman]

nerio said:

Also, when checking for prime's, the fastest way I know of is checking up to square root of that number.

Yes, I meant that and didn't notice I forgot. Thank you for correcting :)

[zoranh]

Just to add a reference to Sieve of Eratosthenes algorithm for finding all prime numbers in range 2-N for any given N. It operates very fast, and can be memory optimized by spending 1 bit per number, meaning that N/8 bytes are required to find all prime numbers up to N.

[sobersaber]

Thanks all! I now understand the entire program. It took me a while to understand the logic, it wasn't easy!