基本的嵌套循环计算1 - 239之间的素数(basic nestled loop calculate prime numbers between 1 - 239, inclusive)

 我正在制定一个计划，在这个计划中，我必须从1 - 239中打印出包括1和239在内的素数（我知道其中一个或两个可能不是素数，但我们会在此计划中考虑它们）它必须是一个非常简单的程序，因为我们只是讨论了一些基础知识。 到目前为止，我的代码是这样的，这对我来说似乎是体面的逻辑流程，但不产生输出。 
 
#include <iostream>

using namespace std;
int main()
{
int x;
int n = 1;
int y = 1;
int i = 0;

while (n<=239)
    {x = n % y;
    if (x = 0)
        i++;
    if (y < n)
    y++;
    n++;
    while (i == 2)
        cout << n;
    }
return 0;
}
 
 我希望这个工作的方式是取n，只要n等于或小于239，并且预制型模数除以从1到n的每个数字。 每当数字y均匀地进入n时，计数器将增加1.如果计数器等于2，则数字为素数，我们将其打印到屏幕。 任何帮助将非常感激。 谢谢 

I am working on a program in which I must print out the number of primes, including 1 and 239, from 1 - 239 ( I know one and or two may not be prime numbers, but we will consider them as such for this program) It must be a pretty simple program because we have only gone over some basics. So far my code is as such, which seems like decent logical flow to me, but doesnt produce output.
 
#include <iostream>

using namespace std;
int main()
{
int x;
int n = 1;
int y = 1;
int i = 0;

while (n<=239)
    {x = n % y;
    if (x = 0)
        i++;
    if (y < n)
    y++;
    n++;
    while (i == 2)
        cout << n;
    }
return 0;
}
 
The way I want this to work is to take n, as long as n is 239 or less, and preform modulus division with every number from 1 leading up to n. Every time a number y goes evenly into n, a counter will be increased by 1. if the counter is equal to 2, then the number is prime and we print it to the screen. Any help would be so greatly appreciated. Thanks 

原文：https://stackoverflow.com/questions/33176155