iOS detect if user is on an iPad

There are quite a few ways to check if a device is an iPad. This is my favorite way to check whether the device is in fact an iPad:

if ( UI_USER_INTERFACE_IDIOM() == UIUserInterfaceIdiomPad )
{
    return YES; /* Device is iPad */
}

The way I use it

#define IDIOM    UI_USER_INTERFACE_IDIOM()
#define IPAD     UIUserInterfaceIdiomPad

if ( IDIOM == IPAD ) {
    /* do something specifically for iPad. */
} else {
    /* do something specifically for iPhone or iPod touch. */
}   

Other Examples

if ( [(NSString*)[UIDevice currentDevice].model hasPrefix:@"iPad"] ) {
    return YES; /* Device is iPad */
}

#define IPAD     (UI_USER_INTERFACE_IDIOM() == UIUserInterfaceIdiomPad)
if ( IPAD ) 
     return YES;

For a Swift solution, see this answer: https://stackoverflow.com/a/27517536/2057171

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In Swift you can use the following equalities to determine the kind of device on Universal apps:

UIDevice.current.userInterfaceIdiom == .phone
// or
UIDevice.current.userInterfaceIdiom == .pad

Usage would then be something like:

if UIDevice.current.userInterfaceIdiom == .pad {
    // Available Idioms - .pad, .phone, .tv, .carPlay, .unspecified
    // Implement your logic here
}

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This is part of UIDevice as of iOS 3.2, e.g.:

[UIDevice currentDevice].userInterfaceIdiom == UIUserInterfaceIdiomPad

---

You can also use this

#define IPAD UI_USER_INTERFACE_IDIOM() == UIUserInterfaceIdiomPad
...
if (IPAD) {
   // iPad
} else {
   // iPhone / iPod Touch
}

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UI_USER_INTERFACE_IDIOM() only returns iPad if the app is for iPad or Universal. If its an iPhone app running on an iPad then it won't. So you should instead check the model.

---

I found that some solution didn't work for me in the Simulator within Xcode. Instead, this works:

ObjC

NSString *deviceModel = (NSString*)[UIDevice currentDevice].model;

if ([[deviceModel substringWithRange:NSMakeRange(0, 4)] isEqualToString:@"iPad"]) {
    DebugLog(@"iPad");
} else {
    DebugLog(@"iPhone or iPod Touch");
}

Swift

if UIDevice.current.model.hasPrefix("iPad") {
    print("iPad")
} else {
    print("iPhone or iPod Touch")
}

Also in the 'Other Examples' in Xcode the device model comes back as 'iPad Simulator' so the above tweak should sort that out.

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Be Careful: If your app is targeting iPhone device only, iPad running with iphone compatible mode will return false for below statement:

#define IPAD     UI_USER_INTERFACE_IDIOM() == UIUserInterfaceIdiomPad

The right way to detect physical iPad device is:

#define IS_IPAD_DEVICE      ([(NSString *)[UIDevice currentDevice].model hasPrefix:@"iPad"])

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Many Answers are good but I use like this in swift 4

Create Constant struct App { static let isRunningOnIpad = UIDevice.current.userInterfaceIdiom == .pad ? true : false } Use like this if App.isRunningOnIpad { return load(from: .main, identifier: identifier) } else { return load(from: .ipad, identifier: identifier) }

Edit: As Suggested Cœur simply create an extension on UIDevice

extension UIDevice {
    static let isRunningOnIpad = UIDevice.current.userInterfaceIdiom == .pad ? true : false
}

---

Many ways to do that in Swift:

We check the model below (we can only do a case sensitive search here):

class func isUserUsingAnIpad() -> Bool {
    let deviceModel = UIDevice.currentDevice().model
    let result: Bool = NSString(string: deviceModel).containsString("iPad")
    return result
}

We check the model below (we can do a case sensitive/insensitive search here):

    class func isUserUsingAnIpad() -> Bool {
        let deviceModel = UIDevice.currentDevice().model
        let deviceModelNumberOfCharacters: Int = count(deviceModel)
        if deviceModel.rangeOfString("iPad",
                                     options: NSStringCompareOptions.LiteralSearch,
                                     range: Range<String.Index>(start: deviceModel.startIndex,
                                                                end: advance(deviceModel.startIndex, deviceModelNumberOfCharacters)),
                                     locale: nil) != nil {
            return true
        } else {
            return false
        }
   }

UIDevice.currentDevice().userInterfaceIdiom below only returns iPad if the app is for iPad or Universal. If it is an iPhone app being ran on an iPad then it won't. So you should instead check the model. :

    class func isUserUsingAnIpad() -> Bool {
        if UIDevice.currentDevice().userInterfaceIdiom == UIUserInterfaceIdiom.Pad {
            return true
        } else {
            return false
        }
   }

This snippet below does not compile if the class does not inherit of an UIViewController, otherwise it works just fine. Regardless UI_USER_INTERFACE_IDIOM() only returns iPad if the app is for iPad or Universal. If it is an iPhone app being ran on an iPad then it won't. So you should instead check the model. :

class func isUserUsingAnIpad() -> Bool {
    if (UI_USER_INTERFACE_IDIOM() == UIUserInterfaceIdiom.Pad) {
        return true
    } else {
        return false
    }
}

---

*

In swift 3.0

*

 if UIDevice.current.userInterfaceIdiom == .pad {
        //pad
    } else if UIDevice.current.userInterfaceIdiom == .phone {
        //phone
    } else if UIDevice.current.userInterfaceIdiom == .tv {
        //tv
    } else if UIDevice.current.userInterfaceIdiom == .carPlay {
        //CarDisplay
    } else {
        //unspecified
    }

---

In Swift 4.2 and Xcode 10

if UIDevice().userInterfaceIdiom == .phone {
    //This is iPhone
} else if UIDevice().userInterfaceIdiom == .pad { 
    //This is iPad
} else if UIDevice().userInterfaceIdiom == .tv {
    //This is Apple TV
}

If you want to detect specific device

let screenHeight = UIScreen.main.bounds.size.height
if UIDevice().userInterfaceIdiom == .phone {
    if (screenHeight >= 667) {
        print("iPhone 6 and later")
    } else if (screenHeight == 568) {
        print("SE, 5C, 5S")
    } else if(screenHeight<=480){
        print("4S")
    }
} else if UIDevice().userInterfaceIdiom == .pad { 
    //This is iPad
}

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You can check the rangeOfString to see of the word iPad exists like this.

NSString *deviceModel = (NSString*)[UIDevice currentDevice].model;

if ([deviceModel rangeOfString:@"iPad"].location != NSNotFound)  {
NSLog(@"I am an iPad");
} else {
NSLog(@"I am not an iPad");
}

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Yet another Swifty way:

//MARK: -  Device Check
let iPad = UIUserInterfaceIdiom.Pad
let iPhone = UIUserInterfaceIdiom.Phone
@available(iOS 9.0, *) /* AppleTV check is iOS9+ */
let TV = UIUserInterfaceIdiom.TV

extension UIDevice {
    static var type: UIUserInterfaceIdiom 
        { return UIDevice.currentDevice().userInterfaceIdiom }
}

Usage:

if UIDevice.type == iPhone {
    //it's an iPhone!
}

if UIDevice.type == iPad {
    //it's an iPad!
}

if UIDevice.type == TV {
    //it's an TV!
}

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Why so complicated? This is how I do it...

Swift 4:

var iPad : Bool {
    return UIDevice.current.model.contains("iPad")
}

This way you can just say if iPad {}

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I don't think any of these answers meet my need, unless I am fundamentally misunderstanding something.

I have an app (originally an iPad app) that I want to run both on an iPad and on the Mac, under Catalyst. I'm using the plist option to scale the Mac interface to match the iPad, but would like to migrate to AppKit if that is reasonable. When running on a Mac, I believe that all of the aforementioned approaches tell me that I'm on an iPad. The Catalyst fake-out is pretty thorough.

For most concerns I indeed understand that the code should pretend it's on an iPad when thus running on a Mac. One exception is that the rolling picker is not available on the Mac under Catalyst, but is on the iPad. I want to figure out whether to create a UIPickerView or to do something different, at run time. Run-time selection is crucial because I want to use a single binary to run both on the iPad and Mac in the long term, while making the best use of the supported UI standards on each.

The APIs give potentially misleading results to the casual pre-Catalyst reader. For example, [UIDevice currentDevice].model returns @"iPad" when running under Catalyst on a Mac. The user interface idiom APIs sustain the same illusion.

I found that you really need to look deeper. I start with this information:

NSString *const deviceModel = [UIDevice currentDevice].model;
NSProcessInfo *const processInfo = [[NSProcessInfo alloc] init];
const bool isIosAppOnMac = processInfo.iOSAppOnMac;  // Note: this will be "no" under Catalyst
const bool isCatalystApp = processInfo.macCatalystApp;

Then you can combine these queries with expressions like [deviceModel hasPrefix: @"iPad"] to sort out the kinds of subtleties I'm facing. For my case, I explicitly want to avoid making a UIPickerView if the indicated isCatalystApp is true, independent of "misleading" information about the interface idiom, or the illusions sustained by isIosAppOnMac and deviceModel.

Now I'm curious what happens if I move the Mac app to run over on my iPad sidecar...

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For the latest versions of iOS, simply add UITraitCollection:

extension UITraitCollection {

    var isIpad: Bool {
        return horizontalSizeClass == .regular && verticalSizeClass == .regular
    }
}

and then within UIViewController just check:

if traitCollection.isIpad { ... }

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if(UI_USER_INTERFACE_IDIOM () == UIUserInterfaceIdiom.pad)
 {
            print("This is iPad")
 }else if (UI_USER_INTERFACE_IDIOM () == UIUserInterfaceIdiom.phone)
 {
            print("This is iPhone");
  }

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