回到第一个索引到达阵列中的最后一个之后

回到第一个索引到达阵列中的最后一个之后

本文介绍了回到第一个索引到达阵列中的最后一个之后的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

在我的数组循环到达最后一个索引,我得到一个异常说,该指数超出范围。我想要的是让它回到第一个索引,直到以Z 等于点击率。我该怎么做?

我的code:

 字符资源;
INT点击率= 10
的char [] =烈焰{'F','L','A','M','E','S'};对于(INT Z = 0; z,其中,CTR-1; Z ++){
    RES =(火焰[Z]);
    jLabel1.setText(将String.valueOf(RES));
}


解决方案

您需要使用一个索引仅限于数组的大小。更多precisely和esoterically来说,你需要在for循环迭代{} 0..9映射为火焰阵{0 .. flames.length()的有效指标 - 1 },这是相同的,在这种情况下,向{0..5}

在从0到5的循环迭代,映射是微不足道的。当循环迭代一个第6次,那么你需要回到它映射到数组索引0,当迭代到第7的时候,你把它映射到数组索引1,依此类推。

==原始的方法==

 的for(int Z = 0,J = 0; z,其中,CTR-1; Z ++,J ++)
{
      如果(J> = flames.length())
      {
         J = 0; //重新回到起点
      }
      RES =(火焰[J]);
      jLabel1.setText(将String.valueOf(RES));
}

==一个更合适的方式==

然后,你可以通过实现完善这一 flames.length()是不变的,你搬出一个for循环的。

 最终诠释N = flames.length();
对于(INT Z = 0,J = 0; z,其中,CTR-1; Z ++,J ++)
{
      如果(J> = N)
      {
         J = 0; //重新回到起点
      }
      RES =(火焰[J]);
      jLabel1.setText(将String.valueOf(RES));
}

==如何做它==

现在,如果你注意,你可以看到我们只做模运算上的索引。所以,如果我们采用模块化(%)经营者,我们可以简化您的code:

 最终诠释N = flames.length();
对于(INT Z = 0; z,其中,CTR-1; Z ++)
{
      RES =(火焰[Z%N]);
      jLabel1.setText(将String.valueOf(RES));
}

在像这样的问题的工作,想想函数映射,从一个域(在这种情况下,循环迭代),以一个范围(有效的数组索引)。

更重要的是,解决它在纸上之前,你甚至开始code。这将带你很长的路要走朝着解决这些类型的元素的问题。

After my array in the for loop reaches the last index, I get an exception saying that the index is out of bounds. What I wanted is for it to go back to the first index until z is equal to ctr. How can I do that?

My code:

char res;
int ctr = 10
char[] flames = {'F','L','A','M','E','S'};

for(int z = 0; z < ctr-1; z++){
    res = (flames[z]);
    jLabel1.setText(String.valueOf(res));
}
解决方案

You need to use an index that is limited to the size of the array. More precisely, and esoterically speaking, you need to map the for-loop iterations {0..9} to the valid indexes for the flame array {0..flames.length()-1}, which are the same, in this case, to {0..5}.

When the loop iterates from 0 to 5, the mapping is trivial. When the loop iterates a 6th time, then you need to map it back to array index 0, when it iterates to the 7th time, you map it to array index 1, and so on.

== Naïve Way ==

for(int z = 0, j = 0; z < ctr-1; z++, j++)
{
      if ( j >= flames.length() )
      {
         j = 0; // reset back to the beginning
      }
      res = (flames[j]);
      jLabel1.setText(String.valueOf(res));
}

== A More Appropriate Way ==

Then you can refine this by realizing flames.length() is an invariant, which you move out of a for-loop.

final int n = flames.length();
for(int z = 0, j = 0; z < ctr-1; z++, j++)
{
      if ( j >= n )
      {
         j = 0; // reset back to the beginning
      }
      res = (flames[j]);
      jLabel1.setText(String.valueOf(res));
}

== How To Do It ==

Now, if you are paying attention, you can see we are simply doing modular arithmetic on the index. So, if we use the modular (%) operator, we can simplify your code:

final int n = flames.length();
for(int z = 0; z < ctr-1; z++)
{
      res = (flames[z % n]);
      jLabel1.setText(String.valueOf(res));
}

When working with problems like this, think about function mappings, from a Domain (in this case, for loop iterations) to a Range (valid array indices).

More importantly, work it out on paper before you even begin to code. That will take you a long way towards solving these type of elemental problems.

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09-02 02:15