SKSpriteNode 的循环位置

Posted 2023-03-16

【中文标题】SKSpriteNode 的循环位置

【英文标题】：looping position for SKSpriteNode

【发布时间】：2018-04-17 18:55:51

【问题描述】：

我认为我对 while 循环有相当深入的了解。我希望 left_spinner 从右侧的屏幕外移到左侧的屏幕外。

然后移回右侧的屏幕外并返回左侧的屏幕外。代码方面我没有发现问题。

import SpriteKit
import GameplayKit

class GameScene: SKScene 

    var left_spinner = SKSpriteNode()

    override func didMove(to view: SKView) 

        left_spinner = self.childNode(withName: "left_spinner") as! SKSpriteNode
        left_spinner.position = CGPoint(x: 675, y: 0)

        left_spinner.run(SKAction.move(to: CGPoint(x: -675, y: 0), duration: 3.0))

        goalLoop()
    

    func goalLoop()
        left_spinner.run(SKAction.move(to: CGPoint(x: -675, y: 0), duration: 3.0))

        while (left_spinner.position == CGPoint(x: -675, y: 0))
            left_spinner.run(SKAction.move(to: CGPoint(x: 675, y: 0), duration: 0.0))
            left_spinner.run(SKAction.move(to: CGPoint(x: -675, y: 0), duration: 3.0))
        
    

    override func update(_ currentTime: TimeInterval) 
        // Called before each frame is rendered
    

【参考方案1】：

您不需要任何同步循环，SKAction API 为您提供了对操作进行排序的方法。只需将您的整个代码替换为：

override func didMove(to view: SKView) 

    left_spinner = self.childNode(withName: "left_spinner") as! SKSpriteNode
    left_spinner.position = CGPoint(x: 675, y: 0)

    // Prepare base actions
    let moveLeftAction = SKAction.move(to: CGPoint(x: -675, y: 0), duration: 3.0)
    let moveRightAction = SKAction.move(to: CGPoint(x: 675, y: 0), duration: 3.0)

    // Prepare sequencing
    let loopCount = 10
    let leftRightAction = SKAction.sequence([moveLeftAction, moveRightAction])
    let pingPongAction = SKAction.repeat(leftRightAction, count: loopCount)

    // Run final action
    left_spinner.run(pingPongAction)