Detect current device with UI_USER_INTERFACE_IDIOM() in Swift

Question

What is the equivalent of UI_USER_INTERFACE_IDIOM() in Swift to detect between iPhone and iPad?

I get an Use of unresolved identifier error when compiling in Swift.

Solution 1

When working with Swift, you can use the enum UIUserInterfaceIdiom, defined as:

enum UIUserInterfaceIdiom : Int {
    case unspecified
    
    case phone // iPhone and iPod touch style UI
    case pad   // iPad style UI (also includes macOS Catalyst)
}

So you can use it as:

UIDevice.current.userInterfaceIdiom == .pad
UIDevice.current.userInterfaceIdiom == .phone
UIDevice.current.userInterfaceIdiom == .unspecified

Or with a Switch statement:

    switch UIDevice.current.userInterfaceIdiom {
    case .phone:
        // It's an iPhone
    case .pad:
        // It's an iPad (or macOS Catalyst)

     @unknown default:
        // Uh, oh! What could it be?
    }

UI_USER_INTERFACE_IDIOM() is an Objective-C macro, which is defined as:

#define UI_USER_INTERFACE_IDIOM() \ ([[UIDevice currentDevice] respondsToSelector:@selector(userInterfaceIdiom)] ? \ [[UIDevice currentDevice] userInterfaceIdiom] : \ UIUserInterfaceIdiomPhone)

Also, note that even when working with Objective-C, the UI_USER_INTERFACE_IDIOM() macro is only required when targeting iOS 3.2 and below. When deploying to iOS 3.2 and up, you can use [UIDevice userInterfaceIdiom] directly.

Solution 2

You should use this GBDeviceInfo framework or ...

Apple defines this:

public enum UIUserInterfaceIdiom : Int {

    case unspecified

    case phone // iPhone and iPod touch style UI

    case pad // iPad style UI

    @available(iOS 9.0, *)
    case tv // Apple TV style UI

    @available(iOS 9.0, *)
    case carPlay // CarPlay style UI
}

so for the strict definition of the device can be used this code

struct ScreenSize
{
    static let SCREEN_WIDTH         = UIScreen.main.bounds.size.width
    static let SCREEN_HEIGHT        = UIScreen.main.bounds.size.height
    static let SCREEN_MAX_LENGTH    = max(ScreenSize.SCREEN_WIDTH, ScreenSize.SCREEN_HEIGHT)
    static let SCREEN_MIN_LENGTH    = min(ScreenSize.SCREEN_WIDTH, ScreenSize.SCREEN_HEIGHT)
}

struct DeviceType
{
    static let IS_IPHONE_4_OR_LESS  = UIDevice.current.userInterfaceIdiom == .phone && ScreenSize.SCREEN_MAX_LENGTH < 568.0
    static let IS_IPHONE_5          = UIDevice.current.userInterfaceIdiom == .phone && ScreenSize.SCREEN_MAX_LENGTH == 568.0
    static let IS_IPHONE_6_7          = UIDevice.current.userInterfaceIdiom == .phone && ScreenSize.SCREEN_MAX_LENGTH == 667.0
    static let IS_IPHONE_6P_7P         = UIDevice.current.userInterfaceIdiom == .phone && ScreenSize.SCREEN_MAX_LENGTH == 736.0
    static let IS_IPAD              = UIDevice.current.userInterfaceIdiom == .pad && ScreenSize.SCREEN_MAX_LENGTH == 1024.0
    static let IS_IPAD_PRO          = UIDevice.current.userInterfaceIdiom == .pad && ScreenSize.SCREEN_MAX_LENGTH == 1366.0
}

how to use

if DeviceType.IS_IPHONE_6P_7P {
    print("IS_IPHONE_6P_7P")
}

to detect iOS version

struct Version{
    static let SYS_VERSION_FLOAT = (UIDevice.current.systemVersion as NSString).floatValue
    static let iOS7 = (Version.SYS_VERSION_FLOAT < 8.0 && Version.SYS_VERSION_FLOAT >= 7.0)
    static let iOS8 = (Version.SYS_VERSION_FLOAT >= 8.0 && Version.SYS_VERSION_FLOAT < 9.0)
    static let iOS9 = (Version.SYS_VERSION_FLOAT >= 9.0 && Version.SYS_VERSION_FLOAT < 10.0)
}

how to use

if Version.iOS8 {
    print("iOS8")
}

Solution 3

if/else case:

 if UIDevice.current.userInterfaceIdiom == .pad {
     // iPad
 } else {
     // not iPad (iPhone, mac, tv, carPlay, unspecified)
 }

Solution 4

Swift 2.0 & iOS 9 & Xcode 7.1

// 1. request an UITraitCollection instance
let deviceIdiom = UIScreen.mainScreen().traitCollection.userInterfaceIdiom

// 2. check the idiom
switch (deviceIdiom) {

case .Pad:
    print("iPad style UI")
case .Phone:
    print("iPhone and iPod touch style UI")
case .TV: 
    print("tvOS style UI")
default:
    print("Unspecified UI idiom")

}

Swift 3.0 and Swift 4.0

// 1. request an UITraitCollection instance
let deviceIdiom = UIScreen.main.traitCollection.userInterfaceIdiom

// 2. check the idiom
switch (deviceIdiom) {

case .pad:
    print("iPad style UI")
case .phone:
    print("iPhone and iPod touch style UI")
case .tv: 
    print("tvOS style UI")
default:
    print("Unspecified UI idiom")
}

Use UITraitCollection. The iOS trait environment is exposed though the traitCollection property of the UITraitEnvironment protocol. This protocol is adopted by the following classes:

• UIScreen
• UIWindow
• UIViewController
• UIPresentationController
• UIView

Solution 5

I do in that way:

UIDevice.current.model

It shows the name of the device.

To check if is iPad or iPhone:

if ( UIDevice.current.model.range(of: "iPad") != nil){
    print("I AM IPAD")
} else {
    print("I AM IPHONE")
}

View more solutions

Comments

[Mihai Fratu]

Never mind. I got it working with if UIDevice.currentDevice().userInterfaceIdiom == .Pad

[Zmey]

As Tony mentioned in one of the answers below, UI_USER_INTERFACE_IDIOM in Swift apps crashes when the app is deployed via TestFlight. Strangely, it works when the app is uploaded directly to device from X-Code. I've also hit this bug.

[Peacemoon]

@Zmey Yes, my app has also got rejected because UI_USER_INTERFACE_IDIOM crashes in review, very strange

[SoftDesigner]

I like struct ScreenSize/DeviceType approach since it works in Simulator

[Praveen]

Exactly correct. It does crash and caused me a lot of headache trying to figure out the issue.

[Computerspezl]

Thanks a lot. I struggled over this bug when a new version of my app was released and crashed from the App Store, but never before, when it was installed over XCode. Hopefully I'll get an expedited review for the fix.

[Luca Davanzo]

definitely best solution, for me at least. Check for userInterfaceIdiom has a problem: if your app is for iPhone only but you launch app on iPad, userInterfaceIdiom is == .Phone

[tschoffelen]

The Swift compiler kept crashing whenever I used UI_USER_INTERFACE_IDIOM() in my code, without any error message. Very strange.

[YannSteph]

Thanks @user3378170 for iOS 9

[Ashoor]

approved answer should go to this neat answer

[davew]

fwiw, Apple's doc now states "If your app runs in iOS 3.2 and later, use userInterfaceIdiom instead."

[webcpu]

If you use UIDevice.currentDevice() in swift, Xcode 7 will report an error, "LLVM ERROR: Broken module found, compilation aborted!". There is a work-around, you can add the function below to your objective-c bridging header file and use currentDevice() instead. static UIDevice* currentDevice() { return [UIDevice currentDevice]; }

[webcpu]

public enum UIUserInterfaceIdiom : Int { case Unspecified @available(iOS 3.2, *) case Phone // iPhone and iPod touch style UI @available(iOS 3.2, *) case Pad // iPad style UI @available(iOS 9.0, *) case TV // Apple TV style UI } Check out the definition of UIUserInterfaceIdiom.If it's not Pad, it could be Phone, TV, Unspecified.

[NoodleOfDeath]

Is it recommended to use this type of switch case in production when making a universal app inside local scope? Or should you make completely separate class implementations based on the device interface?

[swiftBoy]

Huge love from INDIA, appreciate you efforts, Thanks you so much for sharing and making better Stackoverflow ;)

[Fattie]

Definitely the best solution. Good one.

[Vaibhav Jhaveri]

What's the updated code? In context to DEVICE_TYPE for iPhone 7 and 7P

[Steve]

nice, though testing with simulator just returns simulator. is there a way around this?

[Sea Coast of Tibet]

In Swift 3 UIDevice.currentDevice().userInterfaceIdiom becomes UIDevice.current.userInterfaceIdiom

[Randall Wang]

If your app is iPhone only, this method will not work correctly, you will always get .phone, check Ricardo's answer.

[Rahul Iyer]

What is the equivalent way to do this with SwiftUI ? Since the above requires UIKit ? stackoverflow.com/questions/61113923/…

[Ethan Allen]

Thank you for the Obj-C answer.

[Sayalee Pote]

More Obj-C aligned writing - [[UIDevice currentDevice] userInterfaceIdiom]

[Eric Aya]

This is not new in Swift 5, and has already been posted, here for example.