偶数和奇数的数组

Question

我有一个问题，请帮助我创建2 array s 1）偶数数组
2）奇数数组

在标签中我看到一切都好。数组很好。但它们并没有像我预期的那样被创造出来。

和数组2

我的数组代码是

for (indexi = 2; indexi < masivi1.Length; indexi = indexi + 2)
{
    masivi1[indexi] = indexi;

}
for (k = 2; k < masivi1.Length; k = k + 2)
{


    label2.Text += masivi1[k].ToString;
}

第二个数组

for (indexi1 = 0; indexi1 < masivi2.Length; indexi1++)
{
    if (indexi1 % 2 != 0)
    {
        masivi2[indexi1] = indexi1;
    }
}
for (k1 = 0; k1 < masivi2.Length; k1++)
{
    if (k1 % 2 != 0)
    {
        label3.Text += masivi2[k1].ToString() + " | ";
    }
}

有人可以帮助显示正确创建数组的解决方案

  

2,4,6 ...
  1,3,5 ...

Answer 1

你可以用这样的奇数，偶数来填写两个数组

var evenArray = new int[10];
var oddArray = new int[10];

for (int i = 0, even = 0; i < evenArray.Length; i++, even += 2)
{
    evenArray[i] = even;
}

for (int i = 0, odd = 1; i < oddArray.Length; i++, odd += 2)
{
    oddArray[i] = odd;
}

可生产

evenArray = [ 0, 2, 4, 6, 8, 10, 12, 14, 16, 18 ]
oddArray = [ 1, 3, 5, 7, 9, 11, 13, 15, 17, 19 ]

Answer 2

不确定你想要实现的目标，但这看起来不对：

for (indexi = 2; indexi < masivi1.Length; indexi = indexi + 2)
{
     masivi1[indexi] = indexi;
}

因为你只是在每个第二个数组元素中加入一些东西，即2,4,6，......

也许这就是你想要的：

for (indexi = 0; indexi < masivi1.Length; ++indexi)
{
       masivi1[indexi] = 2*indexi + 2;

}

第二个阵列有同样的问题：

for (indexi1 = 0; indexi1 < masivi2.Length; indexi1++)
{
      if (indexi1 % 2 != 0)   // This line means that you only put
                              // elements in the array when index1 = 1, 3, 5, ...
      {
            masivi2[indexi1] = indexi1;
      }
}

Answer 3

for (int indexi = 0; indexi < masivi1.Length && indexi < masivi2.Length; ++indexi)
{
    if (indexi % 2 == 0)
    {
        masivi1[indexi / 2] = indexi;
    }
    else
    {
        masivi2[indexi / 2] = indexi;
    }
}