Prime Number assignment basic c++ - c++

Question

Here is my assignment:
A prime number is a number greater than 1 which is only evenly divisible by 1 and itself. For this assignment you will find which numbers from 2 to n (where n is a user-specified number) are prime.
Ask the user for a number, n, greater than 2. Keep asking for a number until a number greater than 2 is provided. Assume that the user will only enter numbers (that is, you do not need to check if a user enters text).
Use a loop to iterate on a variable, i, from 2 through n. For each iteration, check all numbers from 2 through i to determine whether the number is prime. If it is prime, print out i and the word "Prime".
Use the modulus operator, %, to determine if a number is prime
Here is what I have so far. It doesnt work. And I dont know why. please help, im a business student taking basic programming as an elective.
#include <iostream>
using namespace std;
int main()
{
int n;
int i;
int x;
while (n<=2)
{
cout << "Enter a number greater then 2: \n";
cin >> n;
for (x=n; x>=2; x--)
{
bool prime = false;
for (i=2; i<x; i++)
{
if (x%i==0)
{
prime = true;
}
}
if (prime==false)
{
cout << x << " Prime.\n";
}
}
}
return 0;
}

Answer 1

I didn't actually used your code because of indentication it was little bit of hard to read. But I wrote a new method for you. I suggest always divide your code into methods to make it more managable. You can call this in your main method
bool checkPrime(int number)
{ // input: num an integer > 1
// Returns: true if num is prime
// false otherwise.
int i;
for (i=2; i<number; i++)
{
if (number % i == 0)
{
return false;
}
}
return true;
}
And here is how can you call this method in the main:
int main()
{
int number;
cout << "Enter an integer (>1): ";
cin >> number;
if (checkPrime(number))
{
cout << number << " is prime." << endl;
}
else
{
cout << number << " is not prime." << endl;
}
// I think this is more convention than anything.
return 0;
}

Answer 2

It may not be the optimal program out there, but this should work:
#include <iostream>
using namespace std;
int main()
{
int n;
int i;
int x;
cout << "Enter a number greater then 2: \n";
cin >> n;
while (n<=2)
{
cout << "Enter a number greater then 2: \n";
cin >> n;
}
for (x=n; x>=2; --x)
{
for (i=2; i<x; ++i)
{
bool prime = true;
for (j=2; j<i/2; ++j)
{
if (i%j==0)
{
prime = false;
break;
}
}
if (prime)
{
cout << j << " Prime.\n";
}
}
}
return 0;
}

Answer 3

There are two easy means to go faster: first there is no need to test potential divisors that are too big (as pointed out by arne), and second, there is no need to test even numbers except 2.
Something like this:
#include <cassert>
bool is_prime(unsigned n)
{
if (n == 2)
return true;
if (n <= 1
|| n % 2 == 0)
return false;
for (int d = 3; d * d < n; ++d)
if (n % d == 0)
return false;
return true;
}
int main()
{
assert(!is_prime(0));
assert(!is_prime(1));
assert(is_prime(2));
assert(is_prime(3));
assert(!is_prime(4));
assert(is_prime(5));
assert(!is_prime(6));
assert(!is_prime(256));
assert(is_prime(257));
}
Of course, even faster is building a table of primes, and using this table as potential divisors, instead of every odd number. Makes sense if you have several numbers to check.

Answer 4

Please, say, why it let say you it does not work? Among others I get this output.
Enter a number greater then 2:
100
97 Prime.
89 Prime.
83 Prime.
79 Prime.
73 Prime.
71 Prime.
67 Prime.
61 Prime.
59 Prime.
53 Prime.
47 Prime.
43 Prime.
41 Prime.
37 Prime.
31 Prime.
29 Prime.
23 Prime.
19 Prime.
17 Prime.
13 Prime.
11 Prime.
7 Prime.
5 Prime.
3 Prime.
2 Prime.
But as int n; leaves n uninitialized, the while loop might not be entered.

Answer 5

I think the Answer 1 function checkprime(int number) can improved but purely on preformance basis, consider the fact that prime numbers cannot be even.So if add an extra check to see if (number % 2 == 0) will reduce a lot of iteration of the for loop, and for the remaining i think iterating the loop from 2 to 9 is enough rather than 2 to n. Too many iterations will slow you down on larger numbers.