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c# - 为什么我们在初始化 Vector3 时使用“new”关键字，而不是整数？

问题描述

为什么如果我要初始化一个字符串或一个整数，我可以简单地这样做：

int number = 1;
string word = "word";

但是，如果我想初始化 Vector3，那么我必须这样做：

Vector3 coordinate = new Vector3(1f,1f,1f);

上下文是我正在创建一个非常大的数组，并且每次都必须写出新的 Vector3... 并且它变得非常乏味：

Vector3[,] coordsList = new Vector3[11, 11]
{
    {new Vector3(-1.32f,1.32f), new Vector3(-1.08f,1.32f), new Vector3(-0.84f,1.32f), new Vector3(-0.6f,1.32f), new Vector3(-0.36f,1.32f), new Vector3(-0.12f,1.32f), new Vector3(0.12f,1.32f), new Vector3(0.36f,1.32f), new Vector3(0.6f,1.32f), new Vector3(0.84f,1.32f), new Vector3(1.08f,1.32f) },
    {new Vector3(-1.32f,1.08f), new Vector3(-1.08f,1.08f), new Vector3(-0.84f,1.08f), new Vector3(-0.6f,1.08f), new Vector3(-0.36f,1.08f), new Vector3(-0.12f,1.08f), new Vector3(0.12f,1.08f), new Vector3(0.36f,1.08f), new Vector3(0.6f,1.08f), new Vector3(0.84f,1.08f), new Vector3(1.08f,1.08f) },
    {new Vector3(-1.32f,0.84f), new Vector3(-1.08f,0.84f), new Vector3(-0.84f,0.84f), new Vector3(-0.6f,0.84f), new Vector3(-0.36f,0.84f), new Vector3(-0.12f,0.84f), new Vector3(0.12f,0.84f), new Vector3(0.36f,1.32f), new Vector3(0.6f,0.84f), new Vector3(0.84f,0.84f), new Vector3(1.08f,0.84f) },
    {new Vector3(-1.32f,0.60f), new Vector3(-1.08f,0.60f), new Vector3(-0.84f,0.60f), new Vector3(-0.6f,0.60f), new Vector3(-0.36f,0.60f), new Vector3(-0.12f,0.60f), new Vector3(0.12f,0.60f), new Vector3(0.36f,0.60f), new Vector3(0.6f,0.60f), new Vector3(0.84f,0.60f), new Vector3(1.08f,0.60f) },
    {new Vector3(-1.32f,0.36f), new Vector3(-1.08f,0.36f), new Vector3(-0.84f,0.36f), new Vector3(-0.6f,0.36f), new Vector3(-0.36f,0.36f), new Vector3(-0.12f,0.36f), new Vector3(0.12f,0.36f), new Vector3(0.36f,0.36f), new Vector3(0.6f,0.36f), new Vector3(0.84f,0.36f), new Vector3(1.08f,0.36f) },
    {new Vector3(-1.32f,0.12f), new Vector3(-1.08f,0.12f), new Vector3(-0.84f,0.12f), new Vector3(-0.6f,0.12f), new Vector3(-0.36f,0.12f), new Vector3(-0.12f,0.12f), new Vector3(0.12f,0.12f), new Vector3(0.36f,0.12f), new Vector3(0.6f,0.12f), new Vector3(0.84f,0.12f), new Vector3(1.08f,0.12f) },
    {new Vector3(-1.32f,-0.12f), new Vector3(-1.08f,-0.12f), new Vector3(-0.84f,-0.12f), new Vector3(-0.6f,-0.12f), new Vector3(-0.36f,-0.12f), new Vector3(-0.12f,-0.12f), new Vector3(0.12f,1.32f), new Vector3(0.36f,-0.12f), new Vector3(0.6f,-0.12f), new Vector3(0.84f,-0.12f), new Vector3(1.08f,-0.12f) },
    {new Vector3(-1.32f,-0.36f), new Vector3(-1.08f,-0.36f), new Vector3(-0.84f,-0.36f), new Vector3(-0.6f,-0.36f), new Vector3(-0.36f,-0.36f), new Vector3(-0.12f,-0.36f), new Vector3(0.12f,-0.36f), new Vector3(0.36f,-0.36f), new Vector3(0.6f,-0.36f), new Vector3(0.84f,-0.36f), new Vector3(1.08f,-0.36f) },
    {new Vector3(-1.32f,-0.6f), new Vector3(-1.08f,-0.6f), new Vector3(-0.84f,-0.6f), new Vector3(-0.6f,-0.6f), new Vector3(-0.36f,-0.6f), new Vector3(-0.12f,-0.6f), new Vector3(0.12f,-0.6f), new Vector3(0.36f,-0.6f), new Vector3(0.6f,-0.6f), new Vector3(0.84f,-0.6f), new Vector3(1.08f,-0.6f) },
    {new Vector3(-1.32f,-0.84f), new Vector3(-1.08f,-0.84f), new Vector3(-0.84f,-0.84f), new Vector3(-0.6f,-0.84f), new Vector3(-0.36f,-0.84f), new Vector3(-0.12f,-0.84f), new Vector3(0.12f,-0.84f), new Vector3(0.36f,-0.84f), new Vector3(0.6f,-0.84f), new Vector3(0.84f,-0.84f), new Vector3(1.08f,-0.84f) },
    {new Vector3(-1.32f,-1.08f), new Vector3(-1.08f,-1.08f), new Vector3(-0.84f,-1.08f), new Vector3(-0.6f,-1.08f), new Vector3(-0.36f,-1.08f), new Vector3(-0.12f,-1.08f), new Vector3(0.12f,-1.08f), new Vector3(0.36f,-1.08f), new Vector3(0.6f,-1.08f), new Vector3(0.84f,-1.08f), new Vector3(1.08f,-1.08f) },
};

但是，如果它是一个整数数组，它会容易得多吗？Vector3 需要这个new关键字，而整数或字符串不需要。

标签： c#unity3d

解决方案

Vector3[][] list = new (float x, float y)[][]
{
    new[] { (1.5f, 3.4f), (1.5f, 6.8f) },
    new[] { (1.5f, 3.4f), (1.5f, 6.8f) },
    new[] { (1.5f, 3.4f), (1.5f, 6.8f) },
    new[] { (1.5f, 3.4f), (1.5f, 6.8f) },
}.Select(r => r.Select(v => new Vector3(v.x, v.y)).ToArray()).ToArray();

您可以使用Value Tuples（现在在最新版本的 Unity 2018.3 中支持）以一种可读且不那么乏味的方式构建您的数据 - 然后您可以使用Select将它们转换为 Vector3s。

我认为大多数评论都充分涵盖了您关于此处新关键字的方式和原因的问题。

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