为什么数组大小声明中使用＆QUOT; 1 QUOT;作为第一索引？ [英] Why does array-size declaration use "1" as the first index?

问题描述

东西我注意到了有关C＃/ Java是这个看似（我此刻的）问题与数组大小的声明和数组大小的默认第一个指数不一致。

Something that I noticed about C#/Java is this seemingly (to me at the moment) inconsistent issue with array size declaration and the default first-index of array sizes.

在使用数组时，说你要创建一个新的整数数组大小 3 ，它应该是这样的：

When working with arrays, say you want to create a new integer array size 3, it would look like this:

INT [] newArray = INT新[3] {1，2，3};

共发现和可读性，对吧？

Totally find and readable... Right?

编程语言标准似乎决定了第一指数 0 。

The standard with programming languages seem to dictate that the "first" index is 0.

根据这种逻辑，如果我感兴趣的是创建一个数组的大小 3 ，我确实应该写这样的：

Using that logic, if I am interested in creating an array the size 3, I should really be writing this:

INT [] newArray = INT新[2] {1，2，3};

等一下.. VS抛出一个错误，说数组初始化长度为2，预计。

Wait a minute.. VS is throwing an error, saying an array initialize of length 2 is expected.

所以这是在通过一个数组，数组大小声明循环的第一个索引不一致？前者采用 0 -th基于索引，第二一个 1 -th指数。

So there's an inconsistency with the first index in looping through an array and the array-size declaration? The former uses a 0-th based index, and the second a 1-th index.

这不是破坏游戏/改变以任何形式或方式，但我真的好奇，为什么这里有一个差异，还是地狱，如果这甚至是一个问题了（就像我说的，这不是游戏打破任何方式，但我很好奇，为什么它做这种方式）。

It's not game-breaking/changing in any form or way, but I'm genuinely curious why there's a discrepancy here, or hell, if this is even an issue at all (like I say, it's not game-breaking in any way, but I'm curious as to why it's done this way).

我可以在此刻想到的原因 1 -th基于索引将用于：

I can at the moment think of reasons why 1-th based index would be used:

在一个for循环，你会使用＆LT; newArray.Length ，而不是＆LT; newArray.Length - 1 或＆LT; newArray.Length 。

In a for-loop you would use < newArray.Length as opposed to < newArray.Length - 1 or < newArray.Length.

与工作列表 S代表一段时间，然后回来尺寸的需求将要申报阵列抓住我有点措手不及。

Working with Lists for awhile and then coming back to size-needs-to-be-declared-arrays caught me a bit off-guard.

推荐答案

我想你混淆与长度的索引器。你想握住你的阵列，通过

I think you are confusing the indexer with the length. You want to hold three elements (or variables) in your array, denoted by

...new int[3]...

和在大括号中的元素是的值的不是指数。指数仍基于0。长篇是这样的：

and the elements in the curly brackets are the values, not the index. Index is still 0-based. Longform would look like this:

int[] newArray = new int[3];
        newArray[0] = 1;
        newArray[1] = 2;
        newArray[2] = 3;

所以，你可以看到你的从零开始的索引，是因为它关系到你的int值[]。

So in that, you can see where your zero-based index is as it relates to the values of your int[].

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