ios - 无法将函数转换为方形颜色

Question

我想我已经了解了 Swift 编程语言的基础知识，所以现在我一直在尝试更进一步。今天早些时候，我读到一个关于如何生成随机颜色的问题，所以我尝试使用该函数在我的空白方 block (UIView) 中着色。

import UIKit

func generateColors() -> (main: UIColor, diff: UIColor) {

    let redMain: CGFloat = CGFloat(arc4random_uniform(104))
    let greenMain: CGFloat = redMain + 105.0
    let blueMain: CGFloat = redMain + 83.0

    let redColorMain: CGFloat = redMain / 255.0
    let greenColorMain: CGFloat = greenMain / 255.0
    let blueColorMain: CGFloat = blueMain / 255.0
    let main = UIColor(red: redColorMain, green: greenColorMain, blue: blueColorMain, alpha: 0.8)

    let redDiff: CGFloat = CGFloat(arc4random_uniform(104))
    let greenDiff: CGFloat = redDiff + 105.0
    let blueDiff: CGFloat = redDiff + 83.0

    let redColorDiff: CGFloat = redDiff / 255.0
    let greenColorDiff: CGFloat = greenDiff / 255.0
    let blueColorDiff: CGFloat = blueDiff / 255.0
    let diff = UIColor(red: redColorDiff, green: greenColorDiff, blue: blueColorDiff, alpha: 0.8)

    return (main, diff)
}

var square: UIView = UIView(frame: CGRect(x: 0, y: 0, width: 150, height: 150))
square.backgroundColor = generateColors()

我试了很多东西，我觉得最后好像是函数里的参数。我是否必须以特定方式调用 main 或 diff，以便尝试显而易见的事情。

Answer 1

我不了解 Swift 但 a cursory look at UIView docs建议 backgroundColor 只是一种颜色，而不是一对，除非有一些我不知道的语言。所以不是一对尝试:

return main;

或者:

return diff;

然后将您的函数返回类型更改为单一颜色，无论其语法是什么(我假设是 func generateColors() -> UIColor {)。

希望对您有所帮助...