结构数组还是数组结构？

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问题描述

嗯。我有一个表，该表是我需要存储在Java中的结构数组。天真的不用担心内存方法就是这样做的：

Hmmm. I have a table which is an array of structures I need to store in Java. The naive don't-worry-about-memory approach says do this:

public class Record {
  final private int field1;
  final private int field2;
  final private long field3;
  /* constructor & accessors here */
}

List<Record> records = new ArrayList<Record>();

如果我最终使用了大量（> 10 ）记录，一次只偶尔访问一条记录，我如何才能将前面的方法（ArrayList）与优化的存储成本方法进行比较：

If I end up using a large number (> 10 ) of records, where individual records are accessed occasionally, one at a time, how would I figure out how the preceding approach (an ArrayList) would compare with an optimized approach for storage costs:

public class OptimizedRecordStore {
  final private int[] field1;
  final private int[] field2;
  final private long[] field3;

  Record getRecord(int i) { return new Record(field1[i],field2[i],field3[i]); }
  /* constructor and other accessors & methods */
}

编辑：

假设记录的数量很少或永远不会更改

我可能不会使用
很明显，如果我在上面的OptimizedRecordStore方法中添加/更改了记录数，我会很自信地做出决定。

要么必须用一个新对象替换整个对象，要么删除 final关键字。

kd304提出了一个在我脑海中浮现的优点。在与此类似的其他情况下，我需要对记录进行列访问，例如如果field1和field2是时间和位置，对我来说重要的是将这些值作为数组与MATLAB一起使用，这样我就可以有效地对其进行图形化/分析。

assume the # of records is something that is changed infrequently or never
I'm probably not going to use the OptimizedRecordStore approach, but I want to understand the storage cost issue so I can make that decision with confidence.
obviously if I add/change the # of records in the OptimizedRecordStore approach above, I either have to replace the whole object with a new one, or remove the "final" keyword.
kd304 brings up a good point that was in the back of my mind. In other situations similar to this, I need column access on the records, e.g. if field1 and field2 are "time" and "position", and it's important for me to get those values as an array for use with MATLAB, so I can graph/analyze them efficiently.

OptimizedRecordStore

问题描述

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