#if check (preprocessor macro) to differentiate between iPhone and iPad

Question

Is there a build preprocessor macro I can check, with #if or #ifdef to determine if my current Xcode project is being built for iPhone or iPad?

EDIT

As several answers have pointed out, often apps are universal, and the same binary can run on both devices. Conditional behavior between these very similar devices should be solved at runtime rather than compile time.

score 25 · Answer 1 · answered Sep 13 '10 at 00:08

25

Some ideas in the comment section of this blog

http://greensopinion.blogspot.com/2010/04/from-iphone-to-ipad-creating-universal.html

Mostly using

UI_USER_INTERFACE_IDIOM()

Such as:

#ifdef UI_USER_INTERFACE_IDIOM()
  #define IS_IPAD() (UI_USER_INTERFACE_IDIOM() == UIUserInterfaceIdiomPad)
#else
  #define IS_IPAD() (false)
#endif

answered Sep 13 '10 at 00:08

Lou Franco

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3

this is actually suggested in apple lecture from wwdc. – Vojto Sep 13 '10 at 01:06
Why bother with the macro? Nobody supports iOS lower than 3.2, so the advantage of the `UI_USER_INTERFACE_IDIOM` is none. – Tricertops May 09 '13 at 14:28
This question and answer is from 2010. – Lou Franco May 09 '13 at 18:07
@LouFranco Sorry, next time I will check the date. – Tricertops May 10 '13 at 11:19

score 9 · Answer 2 · edited May 23 '17 at 10:29

9

NSString *deviceType = [UIDevice currentDevice].model;

if([deviceType isEqualToString:@"iPhone"]) {
    //iPhone
}
else if([deviceType isEqualToString:@"iPod touch"]) {
    //iPod Touch
}
else {
    //iPad
}

You cannot, as far as I am concerned, use #if or #ifdef to do this but, it is supported because Obj-C is a strict superset of C.

Related: Determine device (iPhone, iPod Touch) with iPhone SDK

edited May 23 '17 at 10:29

Community

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1

answered Sep 12 '10 at 23:58

Thats what I was afraid of since they essential run the same applications. – Justin Meiners Sep 13 '10 at 00:03
Thanks for the help though. I think I can think of a configuration that I can setup that creates a #define based on this. – Justin Meiners Sep 13 '10 at 00:05
@Alexander Rafferty: Oops sorry. Yeah, C is what I meant to say. – Sep 13 '10 at 00:06
@Alexander Ha I know what he means. – Justin Meiners Sep 13 '10 at 00:06
4

You can't do a compile-time check for a "universal binary" because it's being *built for both*. If you build separate iPad and iPhone apps, then there are many ways, including defining your own compiler macros for the different targets. – tc. Sep 13 '10 at 00:43
2

This isn't safe! Tomorrow Apple change the string and you App stop working correctly. The userInterfaceIdiom property from UIDevice is the safest way. – SEQOY Development Team Aug 12 '11 at 15:30
Thanks for all the comments guys! I know understand the dangers of conditional building vs a runtime check. – Justin Meiners Apr 09 '12 at 18:03
1

Not safe. You shouldn't compare strings but use the UIDevice API instead. – DZenBot Jan 02 '13 at 04:01

score 8 · Accepted Answer · answered May 09 '13 at 14:43

There is no way to determine whether your app is built for iPhone or iPad. Preprocessor #if directives are resolved during build. Once your app is built and flagged as Universal, it has to run correctly on both devices. During building nobody knows where it will be installed later and one build can be installed on both.

However you may want to do one of these:

Detect device model during runtime.

To do this, use [[UIDevice currentDevice] model] and compare to iPhone, iPod touch or iPad strings. This will return you correct device even when running in compatibility mode on iPad (for iPhone-only apps). This can be usefull for usage analytics.

Detect user interface idiom during runtime.

This is what everyone checks for, when providing different content for iPhone and iPad. Use [[UIDevice currentDevice] userInterfaceIdiom] and compare to UIUserInterfaceIdiomPhone or UIUserInterfaceIdiomPad. You may want to make convenience methods like this:

@implementation UIDevice (UserInterfaceIdiom)

- (BOOL)iPhone {
    return (self.userInterfaceIdiom == UIUserInterfaceIdiomPhone);
}
+ (BOOL)iPhone {
    return [[UIDevice currentDevice] iPhone];
}

- (BOOL)iPad {
    return (self.userInterfaceIdiom == UIUserInterfaceIdiomPad);
}
+ (BOOL)iPad {
    return [[UIDevice currentDevice] iPad];
}

@end

Then you can use:

if ([[UIDevice currentDevice] iPhone]) { }
// or
if ([UIDevice iPhone]) { }
// or
if (UIDevice.iPhone) { }

I asked this over 2 years ago, I now understand when runtime vs compile time checks should be used. Thanks for the Answer! Im sure it will help others out. — Justin Meiners, May 09 '13 at 18:36

score 0 · Answer 4 · answered Jun 09 '14 at 11:26

0

Updation for swift:

Couldn't use preprocessor. Make global function as

func IS_IPAD() -> Bool { 
( return (UIDevice.respondsToSelector(Selector("userInterfaceIdiom"))) && (UIDevice.currentDevice().userInterfaceIdiom == UIUserInterfaceIdiom.Pad) )}

answered Jun 09 '14 at 11:26

Mani

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As documented in the UI_USER_INTERFACE_IDIOM() macro, you only need to check for the selector if you're deploying for an iOS version less than 3.2. Using Swift precludes that. ;) – Christopher Rogers Aug 08 '14 at 02:42

score 0 · Answer 5 · answered Sep 10 '15 at 09:04

I use the following code for AppleTV 4 because UIUserInterfaceIdiomUnspecified, doesn't seem to work, nor can I find any other enums:

#ifdef TARGET_OS_IOS
CGSize result = [[UIScreen mainScreen] bounds].size;
    if(result.width == 1920) {
        NSLog(@"tvOS");
    }
#endif

I used to use this for iPad and such before the dark times, before the empire- OB1, hah good night. But you can use this similar technique for other screen sizes you know of.

#if check (preprocessor macro) to differentiate between iPhone and iPad

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