Detect the specific iPhone/iPod touch model

Question

Possible Duplicate:
Determine device (iPhone, iPod Touch) with iOS

I am making a game that utilizes the peer-to-peer bluetooth capabilities of the iPhone (and probably the iPod touch 2nd generation). However, to stop the users from trying to play a multiplayer on an iPod 1st gen and iPhone 2G I need to check for the specific device model.

[[UIDevice currentDevice] model] will only tell me if the device is an "iPhone" or an "iPod touch". Is there a way to check for the specific device model, like: "iPhone 3GS", "iPod touch 1st generation" or something.

EDIT:

There is a category to UIDevice (I think it's created by Erica Sadun, I don't take credit for it) that uses the following code to get the specific device model. You can find the whole category here along with other useful stuff: https://github.com/erica/uidevice-extension

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

@implementation UIDevice (Hardware)

/*
 Platforms
 iPhone1,1 -> iPhone 1G
 iPhone1,2 -> iPhone 3G 
 iPod1,1   -> iPod touch 1G 
 iPod2,1   -> iPod touch 2G 
*/

- (NSString *) platform
{
  size_t size;
  sysctlbyname("hw.machine", NULL, &size, NULL, 0);
  char *machine = malloc(size);
    sysctlbyname("hw.machine", machine, &size, NULL, 0);
    NSString *platform = [NSString stringWithCString:machine encoding:NSUTF8StringEncoding];
  free(machine);
  return platform;
}

This works and apps using this have been lately approved in the AppStore.

VERY IMPORTANT: Apple recommends NOT to use this code. This is because in some scenarios this can be detected incorrectly. Check WWDC 2011 Session 123 at 3'30 and you'll see them showing this exact code and telling not to use. — Thiago Peres, Nov 25 '11 at 01:41
I saw that video. Using that kind of code is not recommended for checking for iPad (as we have UI_USER_INTERFACE_IDIOM for it now), but nothing was said about using it to determine the concrete hardware model. — zubko, Jan 20 '12 at 14:58
Yea its not that in "some scenarios this can be detected incorrectly", it is always detected correctly. You just have to future proof your code and not have some dangling "else" condition that defaults to a specific version. — Cbas, Aug 17 '16 at 03:53

score 154 · Answer 1 · edited May 04 '16 at 12:44

154

You can get the device model number using uname from sys/utsname.h. For example:

#import <sys/utsname.h>

NSString*
machineName()
{
    struct utsname systemInfo;
    uname(&systemInfo);

    return [NSString stringWithCString:systemInfo.machine
                              encoding:NSUTF8StringEncoding];
}

The result should be:

@"i386"      on the simulator
@"iPod1,1"   on iPod Touch
@"iPod2,1"   on iPod Touch Second Generation
@"iPod3,1"   on iPod Touch Third Generation
@"iPod4,1"   on iPod Touch Fourth Generation
@"iPhone1,1" on iPhone
@"iPhone1,2" on iPhone 3G
@"iPhone2,1" on iPhone 3GS
@"iPad1,1"   on iPad
@"iPad2,1"   on iPad 2
@"iPad3,1"   on iPad 3 (aka new iPad)
@"iPhone3,1" on iPhone 4
@"iPhone4,1" on iPhone 4S
@"iPhone5,1" on iPhone 5
@"iPhone5,2" on iPhone 5

edited May 04 '16 at 12:44

JAL

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answered Jul 10 '09 at 11:00

Will Harris

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Thanks for the reply. That did it! Just a correction: iPhone 3GS is "iPhone2,1". Cheers – Dimitris Jul 10 '09 at 12:34
Thanks. I've updated the answer with the result for iPhone 3GS. – Will Harris Jul 11 '09 at 12:14
What is the systemInfo for iPhone 2 gen? Is it the same as iPhone 3G? – Chilly Zhong Mar 02 '10 at 02:37
@iPhoney The original 2G iPhone is @"iPhone1,1". – Will Harris Mar 02 '10 at 17:26
Anyone know what all of the new devices spit out? iPad? iPod Touch 2010? – jtalarico Sep 08 '10 at 15:49
@jtalarico I've added the string for iPad. I don't have access to an iPod Touch 2010. – Will Harris Sep 09 '10 at 13:22
I've added iPad 2, iPhone 4S, and some newer iPods Touch. I don't have these devices to hand, so the strings are based on hearsay rather than direct evidence, but they come from a reliable-sounding source. – Will Harris Oct 27 '11 at 14:14
whether apple will allow to use this code?? – Vasu Ashok May 14 '12 at 12:40
11

@"x86_64" for simulator. – João Nunes Jan 04 '13 at 10:42
1

I'm not sure if it was me or what, but this worked for me in development when I tested on 4,4s,and 5, but it failed hard in production and didn't seem to work at all which caused lots of problems. – Dave Chenell May 07 '13 at 02:24
@"Dave Chenell", isn't it that in production [NSString stringWithCString] uses wrong encoding? Because it may change by environment and therefore return nil. – Skrew Nov 19 '13 at 11:22
6 plus: iPhone7,1 6: iPhone7,2 5S: iPhone6,2 – static0886 Jan 28 '16 at 15:27
This coding is crashing when building in Xcode 7.3 on iPhone 4-devices exclusively. – Sti Apr 21 '16 at 08:52
@RenishDadhaniya your edit completely changes this answer. I have rolled it back. If you have a new answer in Swift, you should add it as your own answer, not change an existing answer. – JAL May 04 '16 at 12:45
@JAL, Yes i will take care.I will upload as a new answer because older answer in not working in all iOS7+ versions in swift. Thank You! – Renish Dadhaniya May 05 '16 at 04:16

djromero · Accepted Answer · 2013-01-22T21:31:20.673

66

Most complete UIDevice (Hardware) category probably is http://github.com/erica/uidevice-extension/ (by Erica Sadun):

[[UIDevice currentDevice] platformType]   // ex: UIDevice4GiPhone
[[UIDevice currentDevice] platformString] // ex: @"iPhone 4G"

edited Jan 22 '13 at 21:31

answered Jul 29 '10 at 17:32

djromero

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It is not working today. I can't find platformType and platformString in [UIDevice currentDevice]. – Yi Jiang Aug 26 '13 at 01:32
@RobertYiJiang did you add the classes from the Github project linked in the answer? – rckoenes Dec 21 '15 at 11:01

Rodrigo · Answer 3 · 2013-10-22T16:30:50.113

How about this code, if new version was released, you will identifier with the last know device

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

- (NSString *)getModel {
    size_t size;
    sysctlbyname("hw.machine", NULL, &size, NULL, 0);
    char *model = malloc(size);
    sysctlbyname("hw.machine", model, &size, NULL, 0);
    NSString *sDeviceModel = [NSString stringWithCString:model encoding:NSUTF8StringEncoding];
    free(model);                              
    if ([sDeviceModel isEqual:@"i386"])      return @"Simulator";  //iPhone Simulator
    if ([sDeviceModel isEqual:@"iPhone1,1"]) return @"iPhone1G";   //iPhone 1G
    if ([sDeviceModel isEqual:@"iPhone1,2"]) return @"iPhone3G";   //iPhone 3G
    if ([sDeviceModel isEqual:@"iPhone2,1"]) return @"iPhone3GS";  //iPhone 3GS
    if ([sDeviceModel isEqual:@"iPhone3,1"]) return @"iPhone4 AT&T";  //iPhone 4 - AT&T
    if ([sDeviceModel isEqual:@"iPhone3,2"]) return @"iPhone4 Other";  //iPhone 4 - Other carrier
    if ([sDeviceModel isEqual:@"iPhone3,3"]) return @"iPhone4";    //iPhone 4 - Other carrier
    if ([sDeviceModel isEqual:@"iPhone4,1"]) return @"iPhone4S";   //iPhone 4S
    if ([sDeviceModel isEqual:@"iPhone5,1"]) return @"iPhone5";    //iPhone 5 (GSM)
    if ([sDeviceModel isEqual:@"iPod1,1"])   return @"iPod1stGen"; //iPod Touch 1G
    if ([sDeviceModel isEqual:@"iPod2,1"])   return @"iPod2ndGen"; //iPod Touch 2G
    if ([sDeviceModel isEqual:@"iPod3,1"])   return @"iPod3rdGen"; //iPod Touch 3G
    if ([sDeviceModel isEqual:@"iPod4,1"])   return @"iPod4thGen"; //iPod Touch 4G
    if ([sDeviceModel isEqual:@"iPad1,1"])   return @"iPadWiFi";   //iPad Wifi
    if ([sDeviceModel isEqual:@"iPad1,2"])   return @"iPad3G";     //iPad 3G
    if ([sDeviceModel isEqual:@"iPad2,1"])   return @"iPad2";      //iPad 2 (WiFi)
    if ([sDeviceModel isEqual:@"iPad2,2"])   return @"iPad2";      //iPad 2 (GSM)
    if ([sDeviceModel isEqual:@"iPad2,3"])   return @"iPad2";      //iPad 2 (CDMA)

    NSString *aux = [[sDeviceModel componentsSeparatedByString:@","] objectAtIndex:0];

//If a newer version exist
    if ([aux rangeOfString:@"iPhone"].location!=NSNotFound) {
        int version = [[aux stringByReplacingOccurrencesOfString:@"iPhone" withString:@""] intValue];
        if (version == 3) return @"iPhone4"
        if (version >= 4) return @"iPhone4s";

    }
    if ([aux rangeOfString:@"iPod"].location!=NSNotFound) {
        int version = [[aux stringByReplacingOccurrencesOfString:@"iPod" withString:@""] intValue];
        if (version >=4) return @"iPod4thGen";
    }
    if ([aux rangeOfString:@"iPad"].location!=NSNotFound) {
        int version = [[aux stringByReplacingOccurrencesOfString:@"iPad" withString:@""] intValue];
        if (version ==1) return @"iPad3G";
        if (version >=2) return @"iPad2";
    }
    //If none was found, send the original string
    return sDeviceModel;
}

It still works! you need to add `#include #include ` – Yoga Oct 21 '13 at 05:56 — Yoga, Oct 21 '13 at 05:56
not working an getting output as x86_64 – Sep 15 '16 at 09:55 — , Sep 15 '16 at 09:55

score 4 · Answer 4 · answered Dec 04 '10 at 13:01

4

BOOL hasHighResScreen = NO;
if ([UIScreen instancesRespondToSelector:@selector(scale)]) {
    CGFloat scale = [[UIScreen mainScreen] scale];
    if (scale > 1.0) {
        hasHighResScreen = YES;
    }
}

answered Dec 04 '10 at 13:01

gotnull

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That doesnt return the exact model. Only is it's got an "HD" or "SD" screen. – Dimitris Dec 04 '10 at 20:52
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This was perfect for me. All I needed to know if the device has high res or not – Sam B Jan 30 '13 at 05:37

Sten · Answer 5 · 2012-04-03T14:12:34.753

4

iPhone 4 is iPhone3,1 and iPhone3,2
iPhone 4S is iPhone4,1
iPad 2 is iPad2,1 iPad2,2 and iPad2,3 depending on version (GSM etc)
iPad 3 is iPad3,1 iPad3,2 and iPad3,3 depending on version (GSM etc)

See Iphone secrets (scroll down to "internal product codes")

Another good source is: everyiphone.com

edited Apr 03 '12 at 14:12

answered Jun 20 '11 at 09:14

Sten

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score -11 · Answer 6 · edited Dec 22 '14 at 14:01

-11

NSString* valueDevice = [[UIDevice currentDevice] model];

and then check if the string is equal to whatever device you are looking for like :

if(value==@"iPod1,1" ) 
{}

and you should be good to go

edited Dec 22 '14 at 14:01

Matt

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answered Feb 17 '11 at 12:08

Ajith

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Would be nice if people mentioned why this has -11 on it but no comment saying why. – Brad Moore Mar 25 '16 at 04:42
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The code `[[UIDevice currentDevice] model]` just returns "iPhone" or "iPad" not a model number. – Chris Stillwell Apr 14 '16 at 16:21

Detect the specific iPhone/iPod touch model

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