Question

为了在屏幕中随机定位元素但在边界内，我需要快速分辨率/屏幕尺寸。我已经知道如何使用CGGraphics获得它。不幸的是，我需要计算一下，需要将其作为UInt32类型，而不是CGFloat。我不能用那种方式进行铸造。有什么想法吗？

let screenSize: CGRect = UIScreen.mainScreen().bounds
let screenWidth = screenSize.width
//What I want to do
let screenHeight = screenSize.height-arc4random_uniform(screenSize.height)

Answer 1

如果您想要元素的随机X和Y位置，您可以执行以下操作：

let screenSize = UIScreen.main.bounds
let randomXPos = CGFloat(arc4random_uniform(UInt32(screenSize.width)))
let randomYPos = CGFloat(arc4random_uniform(UInt32(screenSize.height)))

Swift 4.2

let screenSize = UIScreen.main.bounds
let randomXPos = CGFloat.random(in: 0..<screenSize.width)
let randomYPos = CGFloat.random(in: 0..<screenSize.height)

Answer 2

要在Int中直接获取值，请使用此代码

var bounds: CGRect = UIScreen.mainScreen().bounds
    var w:Int  = Int(self.bounds.size.width)
    var h:Int  = Int(self.bounds.size.height)

您所要做的就是将返回值从CGFloat转换为Integer

使用以下代码获取屏幕高度宽度的值，然后将CGFloat转换为Int

var bounds: CGRect = UIScreen.mainScreen().bounds
var width:CGFloat = bounds.size.width
var height:CGFloat = bounds.size.height

Answer 3

你试试这个

let screenSize: CGRect = UIScreen.mainScreen().bounds
let screenWidth = screenSize.width
let screenHeight = screenSize.height
let screenWidth = screenSize.width * 0.75

Answer 4

我认为这就是你所需要的：

let screenHeight = screenSize.height - CGFloat(arc4random_uniform(UInt32(screenSize.height)))

以Swift为整数获取屏幕大小

4 个答案:

Swift 4.2