我正在尝试制作无尽的奔跑者，目前，我有2个瓷砖预制件要重复生成

Question

zSpawn是瓷砖产生的z位置每个图块长66个单位有2个瓷砖，它们都是预制件

    using System.Collections;
    using System.Collections.Generic;
    using UnityEngine;

    public class HurdleManager : MonoBehaviour
    {
      public GameObject[] tilePrefabs;
      public float zSpawn = 0;
      public float tileLength = 66;
        void Start()
        {
          SpawnTile(0);
          SpawnTile(1);
          SpawnTile(1);
          SpawnTile(0);

        }

        // Update is called once per frame
        void Update()
        {

        }
        public void SpawnTile(int tileIndex)
        {

        Instantiate(tilePrefabs[tileIndex],transform.forward* zSpawn, transform.rotation);
        zSpawn+=tileLength;
      }
    }

Answer 1

您是否尝试过使用递归函数/协程？您可以使用递归协程，并具有一个空的变换生成点，该生成点在每次生成对象时都会移动。

using System.Collections;
using System.Collections.Generic;
using UnityEngine;

public class HurdleManager : MonoBehaviour
{
     public GameObject[] tilePrefabs; 
     public Transform spawnPosition = 0;
     public float tileLength = 66;

     private void Start ()
     {
         StartCoroutine(SpawnTile(0))
     }

     private IEnumerator SpawnTile(int tileIndex)
     {
         Instantiate(tilePrefabs[tileIndex], spawnPosition, Quaternion.identity);
         spawnPosition.z += tileLength; // move the spawn position

         yield return new WaitForSeconds(5f); // how long do you want to wait before spawning the next tile

         StartCoroutine(SpawnTile(Random.Range(0, tilePrefabs.Length))); // spawn a random new tile prefab
     }
 }