Sieve of Eratosthenes - to find the prime numbers till n

Given a number N, calculate the prime numbers up to N using Sieve of Eratosthenes.  

Example 1:

Input:
N = 10

Output:
2 3 5 7

Explanation:
Prime numbers less than equal to N 
are 2 3 5 and 7.

Example 2:

Input:
N = 35

Output:
2 3 5 7 11 13 17 19 23 29 31

Explanation:
Prime numbers less than equal to 35 are
2 3 5 7 11 13 17 19 23 29 and 31.

code: c++ implementation

void SieveOfEratosthenes(int n)
{
    bool prime[n + 1];
    memset(prime, true, sizeof(prime));
 
    for (int p = 2; p * p <= n; p++)
    {
        if (prime[p] == true)
        {
            for (int i = p * p; i <= n; i += p)
                prime[i] = false;
        }
    }
    for (int p = 2; p <= n; p++)
        if (prime[p])
            cout << p << " ";
}