How to check if iPad is iPad Pro

Question

The new iPad pro has different dimension and resolutions. If I check based on screen width would it be correct? I havent upgraded to Xcode 7.1 nor do I have the device so I can't check it yet. Will this check work?

if([UIScreen mainScreen].bounds.size.width>1024)
    {
        // iPad is an iPad Pro
    }

You disregard device's orientation. – El Tomato Nov 20 '15 at 09:34 — El Tomato, Nov 20 '15 at 09:34

Leo · Answer 1 · 2015-11-20T09:25:17.337

14

You may use this

#define IS_IPAD (UI_USER_INTERFACE_IDIOM() == UIUserInterfaceIdiomPad)
#define SCREEN_WIDTH ([[UIScreen mainScreen] bounds].size.width)
#define SCREEN_HEIGHT ([[UIScreen mainScreen] bounds].size.height)
#define IS_IPAD_PRO_1366 (IS_IPAD && MAX(SCREEN_WIDTH,SCREEN_HEIGHT) == 1366.0)
#define IS_IPAD_PRO_1024 (IS_IPAD && MAX(SCREEN_WIDTH,SCREEN_HEIGHT) == 1024.0)

Then

 if (IS_IPAD_PRO_1366) {
    NSLog(@"It is ipad pro 1366");
  }

edited Nov 20 '15 at 09:25

answered Nov 20 '15 at 07:01

Leo

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What is an iPad Pro 1024? Are you suggesting there are 2 models? – TahoeWolverine Jan 15 '16 at 23:45
I believe, the last line should be MIN, not MAX. – FreeNickname Aug 10 '16 at 11:16

score 14 · Answer 2 · edited Nov 20 '15 at 09:26

14

+(BOOL) isIpad_1024
{

    if ([UIScreen mainScreen].bounds.size.height == 1024) {
        return  YES;
    }
    return NO;
}

+(BOOL) isIpadPro_1366
{

    if ([UIScreen mainScreen].bounds.size.height == 1366) {
        return  YES;
    }
    return NO;
}

edited Nov 20 '15 at 09:26

Leo

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answered Nov 20 '15 at 07:57

pankaj raghav

587
2
9

score 9 · Accepted Answer · answered Sep 23 '16 at 10:47

9

So far this macro seems to do the trick without any issues.

#define IS_IPAD_PRO (MAX([[UIScreen mainScreen]bounds].size.width,[[UIScreen mainScreen] bounds].size.height) > 1024)

answered Sep 23 '16 at 10:47

Francis F

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3
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score 5 · Answer 4 · edited May 23 '17 at 10:31

As stated by HAS in their answer here, add this extension in your code:

public extension UIDevice {
    var modelName: String {
        var systemInfo = utsname()
        uname(&systemInfo)
        let machineMirror = Mirror(reflecting: systemInfo.machine)
        let identifier = machineMirror.children.reduce("") { identifier, element in
            guard let value = element.value as? Int8 where value != 0 else { return identifier }
            return identifier + String(UnicodeScalar(UInt8(value)))
        }

        switch identifier {
        case "iPod5,1":                                 return "iPod Touch 5"
        case "iPod7,1":                                 return "iPod Touch 6"
        case "iPhone3,1", "iPhone3,2", "iPhone3,3":     return "iPhone 4"
        case "iPhone4,1":                               return "iPhone 4s"
        case "iPhone5,1", "iPhone5,2":                  return "iPhone 5"
        case "iPhone5,3", "iPhone5,4":                  return "iPhone 5c"
        case "iPhone6,1", "iPhone6,2":                  return "iPhone 5s"
        case "iPhone7,2":                               return "iPhone 6"
        case "iPhone7,1":                               return "iPhone 6 Plus"
        case "iPhone8,1":                               return "iPhone 6s"
        case "iPhone8,2":                               return "iPhone 6s Plus"
        case "iPad2,1", "iPad2,2", "iPad2,3", "iPad2,4":return "iPad 2"
        case "iPad3,1", "iPad3,2", "iPad3,3":           return "iPad 3"
        case "iPad3,4", "iPad3,5", "iPad3,6":           return "iPad 4"
        case "iPad4,1", "iPad4,2", "iPad4,3":           return "iPad Air"
        case "iPad5,1", "iPad5,3", "iPad5,4":           return "iPad Air 2"
        case "iPad2,5", "iPad2,6", "iPad2,7":           return "iPad Mini"
        case "iPad4,4", "iPad4,5", "iPad4,6":           return "iPad Mini 2"
        case "iPad4,7", "iPad4,8", "iPad4,9":           return "iPad Mini 3"
        case "iPad5,1", "iPad5,2":                      return "iPad Mini 4"
        case "iPad6,7", "iPad6,8":                      return "iPad Pro"
        case "i386", "x86_64":                          return "Simulator"
        default:                                        return identifier
        }
    }
}

And for check

if(UIDevice.currentDevice().modelName == "iPad Pro"){//Your code}

This code has been posted repeatedly, e.g. in http://stackoverflow.com/a/26962452/1187415. If you copy code from another source, please add a link to the original for proper attribution. — Martin R, Nov 20 '15 at 10:25

score 2 · Answer 5 · answered Jul 12 '18 at 14:29

2

SWIFT

This is the accepted answer written all swift like.

let isIpadPro:Bool = max(UIScreen.main.bounds.size.width, UIScreen.main.bounds.size.height) > 1024

answered Jul 12 '18 at 14:29

Brandon A

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Harrison Xi · Answer 6 · 2015-11-20T07:27:04.287

Try this library: https://github.com/fahrulazmi/UIDeviceHardware

Then your codes should be:

NSString *platform = [UIDeviceHardware platformString];
if ([platform isEqualToString:@"iPad6,7"] || [platform isEqualToString:@"iPad6,8"]) {
     // iPad is an iPad Pro
}

Or this more powerful library: https://github.com/InderKumarRathore/DeviceUtil

The solution cannot work for simulators. I you want to check device type of simulator, looks like you have to check the screen size.

score 0 · Answer 7 · answered Nov 20 '15 at 12:28

you can use this code:

#include <sys/types.h>
#include <sys/sysctl.h>

- (BOOL) isIpadPro{
    size_t size;
    sysctlbyname("hw.machine", NULL, &size, NULL, 0);
    char *machine = malloc(size);
    sysctlbyname("hw.machine", machine, &size, NULL, 0);
    NSString *platform = [NSString stringWithUTF8String:machine];
    free(machine);

    if ([platform isEqualToString:@"iPad6,8"])
        return YES;

    return NO;
}

score 0 · Answer 8 · answered Sep 25 '16 at 13:45

This macro works in both landscape and portrait:

#define IS_IPAD_PRO_12_INCH (([UIScreen mainScreen].bounds.size.width == 1366 && [UIScreen mainScreen].bounds.size.height == 1024) || ([UIScreen mainScreen].bounds.size.width == 1024 && [UIScreen mainScreen].bounds.size.height == 1366))

score 0 · Answer 9 · answered Jan 02 '17 at 20:04

When I tested in simulator in Xcode 8 none of these solutions worked.

The trick is to look for "nativeBounds" size hight otherwise you will keep getting 1024 as height in simulator

#define iPadPro12 (UIDevice.currentDevice.userInterfaceIdiom == UIUserInterfaceIdiomPad && UIScreen.mainScreen.nativeBounds.size.height > 1024)

if (iPadPro12)
{
  //its ipad Pro 12.9 inch screen
}

score 0 · Answer 10 · answered Jan 30 '19 at 07:00

My complete set of device detection.

#define IS_IPHONE (UI_USER_INTERFACE_IDIOM() == UIUserInterfaceIdiomPhone)
#define IS_RETINA ([[UIScreen mainScreen] scale] >= 2.0)

#define SCREEN_WIDTH ([[UIScreen mainScreen] bounds].size.width)
#define SCREEN_HEIGHT ([[UIScreen mainScreen] bounds].size.height)
#define SCREEN_MAX_LENGTH (MAX(SCREEN_WIDTH, SCREEN_HEIGHT))
#define SCREEN_MIN_LENGTH (MIN(SCREEN_WIDTH, SCREEN_HEIGHT))

#define IS_IPHONE_4_OR_LESS (IS_IPHONE && SCREEN_MAX_LENGTH < 568.0)
#define IS_IPHONE_5 (IS_IPHONE && SCREEN_MAX_LENGTH == 568.0)
#define IS_IPHONE_6 (IS_IPHONE && SCREEN_MAX_LENGTH == 667.0)
#define IS_IPHONE_6P (IS_IPHONE && SCREEN_MAX_LENGTH == 736.0)
#define IS_IPHONE_X (IS_IPHONE && SCREEN_MAX_LENGTH == 812.0)

#define IS_IPAD_PRO_97 (IS_IPAD && SCREEN_MAX_LENGTH == 1024.0)
#define IS_IPAD_PRO_105 (IS_IPAD && SCREEN_MAX_LENGTH == 1112.0)
#define IS_IPAD_PRO_129 (IS_IPAD && SCREEN_MAX_LENGTH == 1366.0)

score 0 · Answer 11 · answered Jan 24 '20 at 10:24

0

You can use Regular Expression for detecting iPad in userAgent

var isIPadPro = /Macintosh/.test(navigator.userAgent) && 'ontouchend' in document;

answered Jan 24 '20 at 10:24

Lesha Petrov

131
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score -1 · Answer 12 · answered Feb 06 '16 at 05:31

Are you guys joking with your complex answers?

if([UIScreen mainScreen].bounds.size.width >= 1024) {
    // iPad pro (or hypothetical/future huge-screened iOS device)
} else {
    // not iPad pro
}

If you just did a >= sign instead of a > sign, it will work wonderfully.

(okay, I know I shouldn't be so dismissive of your thorough, specific answers. Surely there are times that specific device matters more than screen size. But for the quick, obvious answer...!)

But this will fail on landscape. At that time for iPad, width is 1024. — Francis F, Aug 11 '16 at 09:13

score -2 · Answer 13 · edited Nov 20 '15 at 07:02

-2

Follow the below steps for checking

if([[[UIDevice currentDevice] name] isEqualToString:@"iPad Pro"])
{
  //  do your stuff
}

edited Nov 20 '15 at 07:02

user3182143

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answered Nov 20 '15 at 06:39

Divyanshu Sharma

551
2
12

2

This will not work.This is the name in settings,user can change the name – Leo Nov 20 '15 at 06:55
if this is not working then you can manually check your screen height and width like [UIScreen mainScreen].size.height == 2048 then its ipad pro – Divyanshu Sharma Nov 20 '15 at 07:20

score -2 · Answer 14 · answered Feb 23 '16 at 22:00

There is a bug for iPad Pro which make it has a wrong useragent for webview at the moment. User Agent will look likes this:

Mozilla/5.0 (iPhone; CPU iPhone OS9_1 like Mac OS X) AppleWebKit/601.1.46(KHTML, like Gecko)Mobile/13B143

I think we can use this bug for detecting iPad Pro for apps running on compatible mode.

-(BOOL)isiPadPro;
{
    UIWebView* webView = [[UIWebView alloc] initWithFrame:CGRectZero];
    NSString* userAgent = [webView stringByEvaluatingJavaScriptFromString:@"navigator.userAgent"];
    return [userAgent containsString:@"iPhone"] && ([[UIDevice currentDevice] userInterfaceIdiom] == UIUserInterfaceIdiomPad);
}

How to check if iPad is iPad Pro

14 Answers14