Determine if an iOS device supports TouchID withou

Question

I'm currently developing an iOS app that enables users to log in to the app using TouchID, but firstly they must set up a password inside the app first. Problem is, to show the setup password option to enable the TouchID login, I need to detect if the iOS device supports TouchID.

Using the LAContext and canEvaluatePolicy (like the answers in here If Device Supports Touch ID), I am able to determine whether the current device supports TouchID if the user has set up passcode on their iOS device. Here is a my code snippet (I'm using Xamarin, so it's in C#):

static bool DeviceSupportsTouchID () 
{
    if (UIDevice.CurrentDevice.CheckSystemVersion(8, 0))
    {
        var context = new LAContext();
        NSError authError;
        bool touchIDSetOnDevice = context.CanEvaluatePolicy(LAPolicy.DeviceOwnerAuthenticationWithBiometrics, out authError);

        return (touchIDSetOnDevice || (LAStatus) Convert.ToInt16(authError.Code) != LAStatus.TouchIDNotAvailable);
    }

    return false;
}

If the user has not set up the device passcode, the authError will just return "PasscodeNotSet" error regardless of whether the device actually supports TouchID or not.

If the user's device supports TouchID, I want to always show the TouchID option in my app regardless of whether the user has set up passcode on their device (I will just warn the user to setup passcode on their device first). Vice versa, if the user's device doesn't support TouchID, I obviously don't want to show the TouchID option in my app.

So my question is, is there a nice way to consistently determine whether an iOS device supports TouchID regardless of whether the user has set up passcode on their device?

The only workaround I can think of is to determine the architecture of the device (which is answered in Determine if iOS device is 32- or 64-bit), as TouchID is only supported on devices with 64-bit architecture. However, I'm looking if there's any nicer way to do this.

Thanks beforehand! :)

Answer 1

In conclusion of the discussion below, for the time being it is not possible to determine whether a device actually supports TouchID or not when the user hasn't set up passcode on their device.

I have reported this TouchID flaw on the Apple bug reporter. Those who want to follow the issue can see it on Open Radar here: http://www.openradar.me/20342024

Thanks @rckoenes for the input :)

EDIT

Turns out that someone has reported a similar issue already (#18364575). Here is Apple's reply regarding the issue:

"Engineering has determined that this issue behaves as intended based on the following information:

If passcode is not set, you will not be able to detect Touch ID presence. Once the passcode is set, canEvaluatePolicy will eventually return LAErrorTouchIDNotAvailable or LAErrorTouchIdNotEnrolled and you will be able to detect Touch ID presence/state.

If users have disabled passcode on phone with Touch ID, they knew that they will not be able to use Touch ID, so the apps don't need to detect Touch ID presence or promote Touch ID based features. "

So..... the final answer from Apple is No. :(

Note: similar StackOverflow question from the person who reported this -> iOS8 check if device has Touch ID (wonder why I didn't find this question before despite my extensive searching...)

Answer 2

Here is a bit tedious way to figure out if device has physical touch id sensor.

+ (BOOL)isTouchIDExist {
if(![LAContext class]) //Since this mandotory class is not there, that means there is no physical touch id.
    return false;

//Get the current device model name
size_t size;
sysctlbyname("hw.machine", NULL, &size, NULL, 0);
char *model = malloc(size);
sysctlbyname("hw.machine", model, &size, NULL, 0);
NSString *deviceModel = [NSString stringWithCString:model encoding:NSUTF8StringEncoding];

//Devices that does not support touch id
NSArray *deviceModelsWithoutTouchID = [[NSArray alloc]
                                       initWithObjects:
                                       @"iPhone1,1", //iPhone
                                       @"iPhone1,2", //iPhone 3G
                                       @"iPhone2,1", //iPhone 3GS
                                       @"iPhone3,1", //iPhone 4
                                       @"iPhone3,2",
                                       @"iPhone3,3",
                                       @"iPhone4,1", //iPhone 4S
                                       @"iPhone5,1", //iPhone 5
                                       @"iPhone5,2",
                                       @"iPhone5,3", //iPhone 5C
                                       @"iPhone5,4",
                                       @"iPod1,1", //iPod
                                       @"iPod2,1",
                                       @"iPod3,1",
                                       @"iPod4,1",
                                       @"iPod5,1",
                                       @"iPod7,1",
                                       @"iPad1,1", //iPad
                                       @"iPad2,1", //iPad 2
                                       @"iPad2,2",
                                       @"iPad2,3",
                                       @"iPad2,4",// iPad mini 1G
                                       @"iPad2,5",
                                       @"iPad2,5",
                                       @"iPad2,7",
                                       @"iPad3,1", //iPad 3
                                       @"iPad3,2",
                                       @"iPad3,3",
                                       @"iPad3,4", //iPad 4
                                       @"iPad3,5",
                                       @"iPad3,6",
                                       @"iPad4,1", //iPad Air
                                       @"iPad4,2",
                                       @"iPad4,3",
                                       @"iPad4,4", //iPad mini 2
                                       @"iPad4,5",
                                       @"iPad4,6",
                                       @"iPad4,7",
                                       nil];

return ![deviceModelsWithoutTouchID containsObject:deviceModel];

}

Reference: https://www.theiphonewiki.com/wiki/Models https://en.wikipedia.org/wiki/IOS

Answer 3

The correct way to detect if TouchID is available:

BOOL hasTouchID = NO;
// if the LAContext class is available
if ([LAContext class]) {
    LAContext *context = [LAContext new];
    NSError *error = nil;
    hasTouchId = [context canEvaluatePolicy:LAPolicyDeviceOwnerAuthenticationWithBiometrics error:&error];
}

Sorry it is in Objective-C, you might have to translate it to C#.

You should refrain from checking the system version and just check whether or not the class or methods are available.

Answer 4

I know this is a question from last year, but this solution does not make what you need? (Swift code)

if #available(iOS 8.0, *) {
    var error: NSError?
    let hasTouchID = LAContext().canEvaluatePolicy(LAPolicy.DeviceOwnerAuthenticationWithBiometrics, error: &error)

    //Show the touch id option if the device has touch id hardware feature (even if the passcode is not set or touch id is not enrolled)
    if(hasTouchID || (error?.code != LAError.TouchIDNotAvailable.rawValue)) {
        touchIDContentView.hidden = false
    } 
}

Then, when the user presses the button to log in with touch id:

@IBAction func loginWithTouchId() {
    let context = LAContext()

    var error: NSError?
    let reasonString = "Log in with Touch ID"

    if (context.canEvaluatePolicy(LAPolicy.DeviceOwnerAuthenticationWithBiometrics, error: &error)) {
        [context.evaluatePolicy(LAPolicy.DeviceOwnerAuthenticationWithBiometrics, localizedReason: reasonString, reply: { (success: Bool, evalPolicyError: NSError?) -> Void in
            //Has touch id. Treat the success boolean
        })]
    } else { 
        //Then, if the user has touch id but is not enrolled or the passcode is not set, show a alert message
        switch error!.code{

        case LAError.TouchIDNotEnrolled.rawValue:
            //Show alert message to inform that touch id is not enrolled
            break

        case LAError.PasscodeNotSet.rawValue:
            //Show alert message to inform that passcode is not set
            break

        default:
            // The LAError.TouchIDNotAvailable case.
            // Will not catch here, because if not available, the option will not visible
        }
    }
}

Hope it helps!

Answer 5

For iOS 11+ you can use biometryType: LABiometryType of LAContext. More from Apple documentation:

/// Indicates the type of the biometry supported by the device.
///
/// @discussion  This property is set only when canEvaluatePolicy succeeds for a biometric policy.
///              The default value is LABiometryTypeNone.
@available(iOS 11.0, *)
open var biometryType: LABiometryType { get }

@available(iOS 11.0, *)
public enum LABiometryType : Int {

    /// The device does not support biometry.
    @available(iOS 11.2, *)
    case none

    /// The device does not support biometry.
    @available(iOS, introduced: 11.0, deprecated: 11.2, renamed: "LABiometryType.none")
    public static var LABiometryNone: LABiometryType { get }

    /// The device supports Touch ID.
    case touchID

    /// The device supports Face ID.
    case faceID
}

Answer 6

For Objective C
It works great on all devices without checking device version.

- (void)canAuthenticatedByTouchID{
LAContext *myContext = [[LAContext alloc] init];
NSError *authError = nil;
NSString *myLocalizedReasonString = touchIDRequestReason;

if ([myContext canEvaluatePolicy:LAPolicyDeviceOwnerAuthenticationWithBiometrics error:&authError]) {
 }else{
    switch (authError.code) {
        case kLAErrorTouchIDNotAvailable:
            [labelNotSupportTouchID setHidden:NO];
            [switchBtn setHidden:YES];
            [labelEnableTouchid setHidden:YES];
            static dispatch_once_t onceToken;
            dispatch_once(&onceToken, ^{
                [self showAlertMessage:@"EyeCheck Pro" message:@"Device does not support Touch ID Service."];
            });

            break;
    }
  }
}