SWIFT 4/5可重用扩展，支持iPhone 12

    extension UIDevice {
    
    enum `Type` {
        case iPhone_5_5S_5C_SE1
        case iPhone_6_6S_7_8_SE2
        case iPhone_6_6S_7_8_PLUS
        case iPhone_X_XS_12mini
        case iPhone_XR_11
        case iPhone_XS_11Pro_Max
        case iPhone_12_Pro
        case iPhone_12_Pro_Max
    }
    
    var hasHomeButton: Bool {
        switch type {
        case . iPhone_X_XS_12mini, . iPhone_XR_11, .iPhone_XS_11Pro_Max, . iPhone_XS_11Pro_Max, .iPhone_12_Pro, .iPhone_12_Pro_Max:
            return false
        default:
            return true
        }
    }
    
    var type: Type {
        if UI_USER_INTERFACE_IDIOM() == .phone {
        switch UIScreen.main.nativeBounds.height {
        case 1136:
            return .iPhone_5_5S_5C_SE1
        case 1334:
            return .iPhone_6_6S_7_8_SE2
        case 1920, 2208:
            return .iPhone_6_6S_7_8_PLUS
        case 2436:
            return .iPhone_X_XS_12mini
        case 2532:
            return .iPhone_12_Pro
        case 2688:
            return .iPhone_XS_11Pro_Max
        case 2778:
            return .iPhone_12_Pro_Max
        case 1792:
            return .iPhone_XR_11
        default:
            assertionFailure("Unknown phone device detected!")
            return .iPhone_6_6S_7_8_SE2
        }
    } else {
        assertionFailure("Unknown idiom device detected!")
        return .iPhone_6_6S_7_8_SE2
    }
   }
}

#define IS_IPHONE (UI_USER_INTERFACE_IDIOM() == UIUserInterfaceIdiomPhone)
#define IS_IPHONE_X (IS_IPHONE && [[UIScreen mainScreen] bounds].size.height == 812.0f)

在看完所有的答案后，我最后是这么做的:

解决方案(兼容Swift 4.1)

extension UIDevice {
    static var isIphoneX: Bool {
        var modelIdentifier = ""
        if isSimulator {
            modelIdentifier = ProcessInfo.processInfo.environment["SIMULATOR_MODEL_IDENTIFIER"] ?? ""
        } else {
            var size = 0
            sysctlbyname("hw.machine", nil, &size, nil, 0)
            var machine = [CChar](repeating: 0, count: size)
            sysctlbyname("hw.machine", &machine, &size, nil, 0)
            modelIdentifier = String(cString: machine)
        }

        return modelIdentifier == "iPhone10,3" || modelIdentifier == "iPhone10,6"
    }

    static var isSimulator: Bool {
        return TARGET_OS_SIMULATOR != 0
    }
}

Use

if UIDevice.isIphoneX {
    // is iPhoneX
} else {
    // is not iPhoneX
}

Note

Pre Swift 4.1你可以检查应用程序是否在模拟器上运行，就像这样:

TARGET_OS_SIMULATOR != 0

从Swift 4.1开始，你可以使用目标环境平台条件检查应用程序是否在模拟器上运行:

#if targetEnvironment(simulator)
    return true
#else
    return false
#endif

(旧的方法仍然有效，但这个新方法更经得起未来的考验)

我详细阐述了你对其他人的回答，并在UIDevice上进行了快速扩展。我喜欢快速枚举和“一切有序”&原子化。我已经创建了解决方案，工作在设备和模拟器。

优点: -界面简单，使用方法如UIDevice.current.isIPhoneX - UIDeviceModelType enum让你能够轻松地扩展模型特定的特性和你想在你的应用中使用的常量，例如拐角半径

劣势: -这是特定于型号的解决方案，而不是特定于分辨率的解决方案-例如，如果苹果将生产另一种具有相同规格的型号，这将无法正常工作，你需要添加另一种型号才能使其工作=>你需要更新你的应用程序。

extension UIDevice {

    enum UIDeviceModelType : Equatable {

        ///iPhoneX
        case iPhoneX

        ///Other models
        case other(model: String)

        static func type(from model: String) -> UIDeviceModelType {
            switch model {
            case "iPhone10,3", "iPhone10,6":
                return .iPhoneX
            default:
                return .other(model: model)
            }
        }

        static func ==(lhs: UIDeviceModelType, rhs: UIDeviceModelType) -> Bool {
            switch (lhs, rhs) {
            case (.iPhoneX, .iPhoneX):
                return true
            case (.other(let modelOne), .other(let modelTwo)):
                return modelOne == modelTwo
            default:
                return false
            }
        }
    }

    var simulatorModel: String? {
        guard TARGET_OS_SIMULATOR != 0 else {
            return nil
        }

        return ProcessInfo.processInfo.environment["SIMULATOR_MODEL_IDENTIFIER"]
    }

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

        return model
    }

    var modelType: UIDeviceModelType {
        let model = self.simulatorModel ?? self.hardwareModel
        return UIDeviceModelType.type(from: model)
    }

    var isIPhoneX: Bool {
        return modelType == .iPhoneX
    }
}

根据你的问题，答案是否定的。没有直接的方法。欲了解更多信息，您可以在这里获得信息:

如何在iOS上获得设备制作和模型?

and

如何通过swift编程检查iPhone 4和iPhone 5的屏幕大小

iPhone X的高度是2436像素

从设备屏幕尺寸和分辨率:

从设备屏幕尺寸和方向:

Swift 3及以后:

if UIDevice().userInterfaceIdiom == .phone {
    switch UIScreen.main.nativeBounds.height {
        case 1136:
            print("iPhone 5 or 5S or 5C")
        
        case 1334:
            print("iPhone 6/6S/7/8")
        
        case 1920, 2208:
            print("iPhone 6+/6S+/7+/8+")
        
        case 2436:
            print("iPhone X/XS/11 Pro")
        
        case 2688:
            print("iPhone XS Max/11 Pro Max")
        
        case 1792:
            print("iPhone XR/ 11 ")
        
        default:
            print("Unknown")
        }
    }

objective - c:

if([[UIDevice currentDevice] userInterfaceIdiom] == UIUserInterfaceIdiomPhone) {
        switch ((int)[[UIScreen mainScreen] nativeBounds].size.height) {
            case 1136:
                printf("iPhone 5 or 5S or 5C");
                    break;

            case 1334:
                printf("iPhone 6/6S/7/8");
                break;

            case 1920:
            case 2208:
                printf("iPhone 6+/6S+/7+/8+");
                break;

           case 2436:
                printf("iPhone X/XS/11 Pro");
                 break;

            case 2688:
                printf("iPhone XS Max/11 Pro Max");
                 break;

            case 1792:
                printf("iPhone XR/ 11 ");
                 break;

            default:
                printf("Unknown");
                break;
        }
    }

Xamarin.iOS:

if (UIDevice.CurrentDevice.UserInterfaceIdiom == UIUserInterfaceIdiom.Phone) {
    if ((UIScreen.MainScreen.Bounds.Height * UIScreen.MainScreen.Scale) == 1136) {
        Console.WriteLine("iPhone 5 or 5S or 5C");
    } else if ((UIScreen.MainScreen.Bounds.Height * UIScreen.MainScreen.Scale) == 1334) {
        Console.WriteLine("iPhone 6/6S/7/8");
    } else if ((UIScreen.MainScreen.Bounds.Height * UIScreen.MainScreen.Scale) == 1920 || (UIScreen.MainScreen.Bounds.Height * UIScreen.MainScreen.Scale) == 2208) {
        Console.WriteLine("iPhone 6+/6S+/7+/8+");
    } else if ((UIScreen.MainScreen.Bounds.Height * UIScreen.MainScreen.Scale) == 2436) {
        Console.WriteLine("iPhone X, XS, 11 Pro");
    } else if ((UIScreen.MainScreen.Bounds.Height * UIScreen.MainScreen.Scale) == 2688) {
        Console.WriteLine("iPhone XS Max, 11 Pro Max");
    } else if ((UIScreen.MainScreen.Bounds.Height * UIScreen.MainScreen.Scale) == 1792) {
        Console.WriteLine("iPhone XR, 11");
    } else {
        Console.WriteLine("Unknown");
    }
}

根据你的问题如下:

或者使用screenSize。高度为float 812.0f而不是int 812。

if (UI_USER_INTERFACE_IDIOM() == UIUserInterfaceIdiomPhone) {
    CGSize screenSize = [[UIScreen mainScreen] bounds].size;
        // 812.0 on iPhone X, XS
        // 896.0 on iPhone XS Max, XR.

    if (screenSize.height >= 812.0f)
        NSLog(@"iPhone X");
    }

有关更多信息，您可以参考以下页面的iOS人机界面指南:

适应性和布局-视觉设计- iOS -人机界面指南

迅速:

用topNotch检测:

如果有人考虑使用notch来检测iPhoneX，请注意，在“横向”上，它对所有iphone都是一样的。

var hasTopNotch: Bool {
    if #available(iOS 13.0,  *) {
        return UIApplication.shared.windows.filter {$0.isKeyWindow}.first?.safeAreaInsets.top ?? 0 > 20
    }else{
     return UIApplication.shared.delegate?.window??.safeAreaInsets.top ?? 0 > 20
    }

    return false
}

objective - c:

- (BOOL)hasTopNotch {
   if (@available(iOS 13.0, *)) {
       return [self keyWindow].safeAreaInsets.top > 20.0;
   }else{
       return [[[UIApplication sharedApplication] delegate] window].safeAreaInsets.top > 20.0;
   }
   return  NO;
}

- (UIWindow*)keyWindow {
    UIWindow        *foundWindow = nil;
    NSArray         *windows = [[UIApplication sharedApplication]windows];
    for (UIWindow   *window in windows) {
        if (window.isKeyWindow) {
            foundWindow = window;
            break;
        }
    }
    return foundWindow;
}

更新:

不要使用userInterfaceIdiom属性来识别设备类型，正如userInterfaceIdiom的文档解释的那样:

对于通用应用程序，您可以使用此属性为特定类型的设备定制应用程序的行为。例如，iPhone和iPad设备具有不同的屏幕尺寸，因此您可能希望根据当前设备的类型创建不同的视图和控件。

也就是说，这个属性只是用来识别正在运行的应用程序的视图样式。然而，iPhone应用程序(不是通用的)可以通过app store安装在iPad设备上，在这种情况下，userInterfaceIdiom也将返回UIUserInterfaceIdiomPhone。

正确的方法是通过uname获取机器名。详情请查阅以下资料:

如何在iOS上获得设备制作和模型?

struct ScreenSize {
    static let width = UIScreen.main.bounds.size.width
    static let height = UIScreen.main.bounds.size.height
    static let maxLength = max(ScreenSize.width, ScreenSize.height)
    static let minLength = min(ScreenSize.width, ScreenSize.height)
    static let frame = CGRect(x: 0, y: 0, width: ScreenSize.width, height: ScreenSize.height)
}

struct DeviceType {
    static let iPhone4orLess = UIDevice.current.userInterfaceIdiom == .phone && ScreenSize.maxLength < 568.0
    static let iPhone5orSE = UIDevice.current.userInterfaceIdiom == .phone && ScreenSize.maxLength == 568.0
    static let iPhone678 = UIDevice.current.userInterfaceIdiom == .phone && ScreenSize.maxLength == 667.0
    static let iPhone678p = UIDevice.current.userInterfaceIdiom == .phone && ScreenSize.maxLength == 736.0
    static let iPhoneX = UIDevice.current.userInterfaceIdiom == .phone && ScreenSize.maxLength == 812.0

    static let IS_IPAD              = UIDevice.current.userInterfaceIdiom == .pad && ScreenSize.maxLength == 1024.0
    static let IS_IPAD_PRO          = UIDevice.current.userInterfaceIdiom == .pad && ScreenSize.maxLength == 1366.0
}