供参考，我已经使用UI_USER_INTERFACE_IDIOM()为我的应用程序写在Swift。应用程序可以很好地使用XCode 6.3.1编译，没有任何警告，在模拟器(任何选择的设备)和我所有的真实设备(iPhone, iPad)上运行良好，iOS版本从7.1到8.3。

然而，这款应用在苹果评测者的设备上崩溃了(并被拒绝)。我花了几天时间才发现问题，并重新上传了几次到iTunes Connect。

现在我使用UIDevice.currentDevice()。取而代之的是userInterfaceIdiom，我的应用程序可以从这样的崩溃中幸存下来。

尝试添加这样的扩展:

    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 "iPhone9,1", "iPhone9,3":                  return "iPhone 7"
        case "iPhone9,2", "iPhone9,4":                  return "iPhone 7 Plus"
        case "iPhone8,4":                               return "iPhone SE"
        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 "iPad5,3", "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 "iPad6,3", "iPad6,4", "iPad6,7", "iPad6,8":return "iPad Pro"
        case "AppleTV5,3":                              return "Apple TV"
        case "i386", "x86_64":                          return "Simulator"
        default:                                        return identifier
        }
    }

}

下面是你如何使用它:

let modelName = UIDevice.currentDevice().modelName

编辑 对于模拟器，您可以在这里尝试解决方案

Swift 2.0 & iOS 7+ / iOS 8+ / iOS 9+

public class Helper {
    public class var isIpad:Bool {
        if #available(iOS 8.0, *) {
            return UIScreen.mainScreen().traitCollection.userInterfaceIdiom == .Pad
        } else {
            return UIDevice.currentDevice().userInterfaceIdiom == .Pad
        }
    }
    public class var isIphone:Bool {
        if #available(iOS 8.0, *) {
            return UIScreen.mainScreen().traitCollection.userInterfaceIdiom == .Phone
        } else {
            return UIDevice.currentDevice().userInterfaceIdiom == .Phone
        }
    }
}

使用:

if Helper.isIpad {

}

OR

guard Helper.isIpad else {
    return
} 

由于@user3378170

你可以在Swift 5上使用新的方式:

switch traitCollection.userInterfaceIdiom {
        
    case .unspecified:
        // do something
    case .phone:
        // do something
    case .pad:
        // do something
    case .tv:
        // do something
    case .carPlay:
        // do something
    case .mac:
        // do something
    @unknown default:
        // do something
}

斯威夫特3.0:

let userInterface = UIDevice.current.userInterfaceIdiom

if(userInterface == .pad){
    //iPads
}else if(userInterface == .phone){
    //iPhone
}else if(userInterface == .carPlay){
    //CarPlay
}else if(userInterface == .tv){
    //AppleTV
}

Swift 2.0 & iOS 9 & Xcode 7.1

// 1. request an UITraitCollection instance
let deviceIdiom = UIScreen.mainScreen().traitCollection.userInterfaceIdiom

// 2. check the idiom
switch (deviceIdiom) {

case .Pad:
    print("iPad style UI")
case .Phone:
    print("iPhone and iPod touch style UI")
case .TV: 
    print("tvOS style UI")
default:
    print("Unspecified UI idiom")

}

Swift 3.0和Swift 4.0

// 1. request an UITraitCollection instance
let deviceIdiom = UIScreen.main.traitCollection.userInterfaceIdiom

// 2. check the idiom
switch (deviceIdiom) {

case .pad:
    print("iPad style UI")
case .phone:
    print("iPhone and iPod touch style UI")
case .tv: 
    print("tvOS style UI")
default:
    print("Unspecified UI idiom")
}

使用UITraitCollection。 iOS trait环境是通过UITraitEnvironment协议的traitCollection属性公开的。以下类采用此协议:

UIScreen ui窗口 ui UIPresentationController UIView