list of prime numbers between 1 and 100 - c++

Question

I want to find the list of prime numbers, I try this code but it doesn't show me anything:
#include <iostream>
using namespace std;
int main()
{
int n, i;
std::cout << "Liste des nombres premiers : " << std::endl;
for (n = 1; n < 100; n++) {
for (i = 2; i < n; i++) {
if (n % i == 0)
std::cout << n << " ";
}
}
return 0;
}

Answer 1

I don't like your code at all. Try to do this: create a function that tells you if a number is prime, and then go with a for loop through all numbers from 1 to 100, and check with the function if they are prime.
Here is an easy to remember function (for natural numbers):
bool isPrime(int n)
{
if (n <= 1)
return 0;
if (n == 2)
return 1;
if (n%2 == 0)
return 0;
for (int i=3; i*i <= n; i+=2)
if (n%i == 0)
return 0;
return 1;
}
This function returns 1 if the number is prime, and returns 0 if it isn't.
The idea is this:
If n is 0 or 1, it is not prime.
If n is 2, it is prime.
If n is a multiple of 2, but it isn't 2, it is not prime.
Then, you have to check if n has any divisors until you get to square root of n (i * i <= n)

Answer 2

To correct program errors:
Initialize the value of n as 2.
And print when i is equal to n not when n % i is 0. It should be rather n % i == 0 then break.
Change i < n to i <= n.
Try the first approach (used sqrt()):
#include <iostream>
#include <cmath>
int main(void)
{
for (int i = 2; i < 100; i++) {
if (i == 2 || i == 3) { // if number is 2 or 3, then prints it
std::cout << i << ' ';
}
for (int j = 2; j * j <= i; j++)
{
if (i % j == 0) break;
else if (j + 1 > sqrt(i)) std::cout << i << ' ';
}
}
return 0;
}
Alternative approach:
#include <iostream>
int main(void)
{
for (int n = 2; n < 100; n++)
for (int i = 2; i <= n; i++)
if (i == n) std::cout << i << ' ';
else if (n % i == 0) break;
return 0;
}