java - 如何遍历数组哈希表中没有参数的方法？[爪哇]

Question

因此，我可以通过将索引设置为 c[h] 关闭 i 来查看具有参数的列表，但我不确定如何在没有参数的情况下执行此操作。如果没有这个索引，我将如何迭代？用于省略和打印。

  {

private int howmany;
private int[] c;
int counter = 0;

public Intcoll7()
  {
    c = new int[23];
    howmany = 0;
  }

public Intcoll7(int i)
  {
    double z = (i % 23);
    int y = (int) z;

    while (!isprime(y))
      {
        y--;
      }
    howmany = 0;
  }

public void insert(int i)
  {
    int h = i % 23;
    int H = h;

    if (c[h] == 0)
      {
        c[h] = i;
        howmany++;
      } else if (c[h] == i) ; else
      {
        if (h < c.length - 1)
          {
            h++;
          } else
          {
            h = 0; // why does h = 0?        
          }
        while ((h != H) && (c[h] != 0) && (c[h] != i))
          {
            if (h < c.length - 1)
              {
                h++;
              } else
              {
                h = 0;
              }
          }
        if (h != H)
          {
            if (c[h] == 0)
              {
                c[h] = i;
                howmany++;
              }
          } else
          {
            System.out.println("Table Full");
            System.exit(1);
          }
      }
  }

public boolean belongs(int i)
  {
    //address
    int h = i % 23;

    int H = h; // what

    if (c[h] == 0)
      {
        return false;
      } if (c[h] == H)
      {
        return true;
      } 
      else
        {
          return false;
        }


  }

public void omit(int i)
  {
    int h = i % 23;


    if (c[h] == i)
      {
        if (counter == 3)
          {
            //crate new array
          }
        c[h] = i * (-1);
        counter++;
      }
  }

private static boolean isprime(int n)
  {
    int i = 2;
    boolean result = true;

    while ((i <= n / 2) && result)
      {
        if (n % i == 0)
          {
            result = false;
          } else
          {
            i++;
          }
      }
    return (result);
  }

public void print()
  {
    System.out.println();
    //int x = c[z];
    //System.out.println(x);             
  }
  }