以Swift为整数获取屏幕大小

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,需要它作为类型UInt32,而不是CGFloat.我不能用那种方式进行铸造.有任何想法吗？

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)