求素数为什么会有1问题在哪？

Question

#include <iostream>
using namespace std;
class Array
{
private:
    int* p, k;
    float s;
public:
    Array(int* ptr, int n);
    bool fun(int n);
    void sum();
    void show();
    ~Array();
};
Array::Array(int* ptr, int n)
{
    p = new int[n];
    k = n;
    for (int i = 0; i < k; i++)
    {
        p[i] = ptr[i];
    }
}
bool Array::fun(int n)
{
    bool flag = true;
    for (int i = 2; i < n; i++)
    {
        if (n % i == 0)
        {
            flag = false;
            break;
        }
    }
    return flag;

    
}
void Array::sum()
{
    int sum = 0;
    for (int i = 0; i < k; i++)
    {
        if (fun(p[i])==1)
        {
            sum = sum + p[i];
        }
    }
    s = sum / k;
}
void Array::show()
{
    int pn = 0;
    for (int i = 0; i < 9; i++)
    {
        if (fun(p[i]))
            {
                pn++;
                cout << p[i] << '\t';
                if (pn % 5 == 0)
                {
                    cout << endl;
                }
            }
    }
    if (pn % 5)cout << endl;
    
}
Array::~Array()
{
    delete[]p;

}
int main()
{
    int a[9] = { 5,2,7,4,8,23,65,1,40 };
    Array arr(a, sizeof(a)/sizeof(a[0]));
    arr.sum();
    arr.show();
    system("pause");
    return 0;

}

Answer 1

bool Array::fun(int n)
{
    bool flag = true;
    for (int i = 2; i < n; i++)
    {
        if (n % i == 0)
        {
            flag = false;
            break;
        }
    }
    return flag;
 
    
}

n值为1时，没进入循环，flag为true