在Objective-C中,我们可以使用宏知道应用程序是为设备还是模拟器构建的:
#if TARGET_IPHONE_SIMULATOR
// Simulator
#else
// Device
#endif
这些是编译时宏,在运行时不可用。
我如何在Swift中实现同样的目标?
在Objective-C中,我们可以使用宏知道应用程序是为设备还是模拟器构建的:
#if TARGET_IPHONE_SIMULATOR
// Simulator
#else
// Device
#endif
这些是编译时宏,在运行时不可用。
我如何在Swift中实现同样的目标?
当前回答
来自Xcode 9.3
#if targetEnvironment(simulator)
Swift supports a new platform condition targetEnvironment with a single valid argument simulator. Conditional compilation of the form '#if targetEnvironment(simulator)' can now be used to detect when the build target is a simulator. The Swift compiler will attempt to detect, warn, and suggest the use of targetEnvironment(simulator) when evaluating platform conditions that appear to be testing for simulator environments indirectly, via the existing os() and arch() platform conditions. (SE-0190)
iOS 9+:
extension UIDevice {
static var isSimulator: Bool {
return NSProcessInfo.processInfo().environment["SIMULATOR_DEVICE_NAME"] != nil
}
}
斯威夫特3:
extension UIDevice {
static var isSimulator: Bool {
return ProcessInfo.processInfo.environment["SIMULATOR_DEVICE_NAME"] != nil
}
}
iOS 9之前:
extension UIDevice {
static var isSimulator: Bool {
return UIDevice.currentDevice().model == "iPhone Simulator"
}
}
objective - c:
@interface UIDevice (Additions)
- (BOOL)isSimulator;
@end
@implementation UIDevice (Additions)
- (BOOL)isSimulator {
if([[NSProcessInfo processInfo] isOperatingSystemAtLeastVersion:(NSOperatingSystemVersion){9, 0, 0}]) {
return [NSProcessInfo processInfo].environment[@"SIMULATOR_DEVICE_NAME"] != nil;
} else {
return [[self model] isEqualToString:@"iPhone Simulator"];
}
}
@end
其他回答
来自Xcode 9.3
#if targetEnvironment(simulator)
Swift supports a new platform condition targetEnvironment with a single valid argument simulator. Conditional compilation of the form '#if targetEnvironment(simulator)' can now be used to detect when the build target is a simulator. The Swift compiler will attempt to detect, warn, and suggest the use of targetEnvironment(simulator) when evaluating platform conditions that appear to be testing for simulator environments indirectly, via the existing os() and arch() platform conditions. (SE-0190)
iOS 9+:
extension UIDevice {
static var isSimulator: Bool {
return NSProcessInfo.processInfo().environment["SIMULATOR_DEVICE_NAME"] != nil
}
}
斯威夫特3:
extension UIDevice {
static var isSimulator: Bool {
return ProcessInfo.processInfo.environment["SIMULATOR_DEVICE_NAME"] != nil
}
}
iOS 9之前:
extension UIDevice {
static var isSimulator: Bool {
return UIDevice.currentDevice().model == "iPhone Simulator"
}
}
objective - c:
@interface UIDevice (Additions)
- (BOOL)isSimulator;
@end
@implementation UIDevice (Additions)
- (BOOL)isSimulator {
if([[NSProcessInfo processInfo] isOperatingSystemAtLeastVersion:(NSOperatingSystemVersion){9, 0, 0}]) {
return [NSProcessInfo processInfo].environment[@"SIMULATOR_DEVICE_NAME"] != nil;
} else {
return [[self model] isEqualToString:@"iPhone Simulator"];
}
}
@end
斯威夫特4:
目前,我更喜欢使用ProcessInfo类来知道设备是否是模拟器以及正在使用的设备类型:
if let simModelCode = ProcessInfo().environment["SIMULATOR_MODEL_IDENTIFIER"] {
print("yes is a simulator :\(simModelCode)")
}
但是,正如您所知道的,simModelCode并不是一个容易理解的代码,因此,如果您需要,您可以尝试查看另一个so答案,以确定当前的iPhone/设备模型,并获得一个更适合人类阅读的字符串。
Xcode 11, Swift 5
#if !targetEnvironment(macCatalyst)
#if targetEnvironment(simulator)
true
#else
false
#endif
#endif
让我在这里澄清一些事情:
TARGET_OS_SIMULATOR在很多情况下在Swift代码中没有设置;由于桥接头,您可能会意外地导入它,但这是脆弱的,不受支持。这在框架中甚至是不可能的。这就是为什么有些人对这在Swift中是否有效感到困惑。 我强烈建议不要使用架构来代替模拟器。
执行动态检查:
检查ProcessInfo.processInfo。environment["SIMULATOR_DEVICE_NAME"] != nil完全没问题。
您还可以通过检查SIMULATOR_MODEL_IDENTIFIER来获得正在模拟的底层模型,它将返回像iPhone10,3这样的字符串。
执行静态检查:
Xcode 9.2及更早版本:定义你自己的Swift编译标志(如其他答案所示)。
Xcode 9.3+使用新的targetenvirenvironment条件:
#if targetEnvironment(simulator)
// for sim only
#else
// for device
#endif
达尔文在这里描述了一切。TargetConditionals: https://github.com/apple/swift-corelibs-foundation/blob/master/CoreFoundation/Base.subproj/SwiftRuntime/TargetConditionals.h
TARGET_OS_SIMULATOR—生成的代码将在模拟器下运行