Swift: Determine iOS Screen size

I would like to use Swift code to properly position items in my app for no matter what the screen size is. For example, if I want a button to be 75% of the screen wide, I could do something like `(screenWidth * .75)` to be the width of the button. I have found that this could be determined in Objective-C by doing

CGFloat screenWidth = screenSize.width;
CGFloat screenHeight = screenSize.height;

Unfortunately, I am unsure of how to convert this to Swift. Does anyone have an idea?

Thanks!

**In Swift 5.0**

let screenSize: CGRect = UIScreen.main.bounds


Pay attention that `UIScreen.main` will be deprecated in future version of iOS. So we can use `view.window.windowScene.screen`.

**Swift 4.0**

// Screen width.
public var screenWidth: CGFloat {
return UIScreen.main.bounds.width
}

// Screen height.
public var screenHeight: CGFloat {
return UIScreen.main.bounds.height
}

**In Swift 3.0**

let screenSize = UIScreen.main.bounds
let screenWidth = screenSize.width
let screenHeight = screenSize.height

**In older swift:**
Do something like this:

let screenSize: CGRect = UIScreen.mainScreen().bounds

then you can access the width and height like this:

let screenWidth = screenSize.width
let screenHeight = screenSize.height

if you want 75% of your screen's width you can go:

let screenWidth = screenSize.width * 0.75