How to detect if app is being built for device or simulator in Swift

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Update 30/01/19

While this answer may work, the recommended solution for a static check (as clarified by several Apple engineers) is to define a custom compiler flag targeting iOS Simulators. For detailed instructions on how to do to it, see @mbelsky’s answer.

Original answer

If you need a static check (e.g. not a runtime if/else) you can’t detect the simulator directly, but you can detect iOS on a desktop architecture like follows

#if (arch(i386) || arch(x86_64)) && os(iOS)
    ...
#endif

After Swift 4.1 version

Latest use, now directly for all in one condition for all types of simulators need to apply only one condition –

#if targetEnvironment(simulator)
  // your simulator code
#else
  // your real device code
#endif

For more clarification, you can check Swift proposal SE-0190

For older version –

Clearly, this is false on a device, but it returns true for the iOS Simulator, as specified in the documentation:

The arch(i386) build configuration returns true when the code is compiled for the 32–bit iOS simulator.

If you are developing for a simulator other than iOS, you can simply vary the os parameter: e.g.

Detect the watchOS simulator

#if (arch(i386) || arch(x86_64)) && os(watchOS)
...
#endif

Detect the tvOS simulator

#if (arch(i386) || arch(x86_64)) && os(tvOS)
...
#endif

Or, even, detect any simulator

#if (arch(i386) || arch(x86_64)) && (os(iOS) || os(watchOS) || os(tvOS))
...
#endif

If you instead are ok with a runtime check, you can inspect the TARGET_OS_SIMULATOR variable (or TARGET_IPHONE_SIMULATOR in iOS 8 and below), which is truthy on a simulator.

Please notice that this is different and slightly more limited than using a preprocessor flag. For instance you won’t be able to use it in place where a if/else is syntactically invalid (e.g. outside of functions scopes).

Say, for example, that you want to have different imports on the device and on the simulator. This is impossible with a dynamic check, whereas it’s trivial with a static check.

#if (arch(i386) || arch(x86_64)) && os(iOS)
  import Foo
#else
  import Bar
#endif

Also, since the flag is replaced with a 0 or a 1 by the swift preprocessor, if you directly use it in a if/else expression the compiler will raise a warning about unreachable code.

In order to work around this warning, see one of the other answers.

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