java - Eratosthenes 故障筛

Question

我正在尝试用 Java 创建埃拉托色尼筛法，但我编写的代码似乎有缺陷。我想知道是否有其他人能发现我的错误，因为我不能。我得到的输出只是 [2]，这意味着我的主循环不起作用。我才刚刚接触 Java，所以如果您能给出详细的答案，我将不胜感激。

我的代码:

public static int[] primes(int n)
{
    //Variable assignement
    double sqrt;
    List<Integer> primes= new ArrayList<Integer>();
    //Adds 2 to primes so i don't need to include the even numbers in my for loop.
    if(n>1)
    {
        primes.add(2);
    }
    //For loop that goes through the oneven numbers up to n
    for(int counter1=3;counter1<=n;counter1+=2)
    {
        sqrt=Math.floor(Math.sqrt(0.0+counter1));
        //for loop that tests if the first for loops number is prime
        for(int counter2=0;sqrt<=primes.get(counter2);counter2++)
        {
            if(counter1 % primes.get(counter2) != 0 && counter2 ==sqrt)
            {
                primes.add(counter1);
            }

            if(counter1 % primes.get(counter2)==0)
            {
                break;
            }
        }
    }
    return convertIntegers(primes);
}
//Converts the list to an array
public static int[] convertIntegers(List<Integer> integers)
{
    int[] ret = new int[integers.size()];
    for (int i=0; i < ret.length; i++)
    {
        ret[i] = integers.get(i).intValue();
    }
    return ret;
}

Answer 1

if(counter1 % primes.get(counter2)==0);{
    break;}
}

不太可能帮助解决问题。您在左大括号之前有一个分号 - 您的 break 将始终执行。