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

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.

OUTDATED FOR SWIFT 4.1. Use #if targetEnvironment(simulator) instead. Source

To detect simulator in Swift you can use build configuration:

• Define this configuration -D IOS_SIMULATOR in Swift Compiler - Custom Flags > Other Swift Flags
• Select Any iOS Simulator SDK in this drop down

Now you could use this statement to detect simulator:

#if IOS_SIMULATOR
    print("It's an iOS Simulator")
#else
    print("It's a device")
#endif

Also you could extend UIDevice class:

extension UIDevice {
    var isSimulator: Bool {
        #if IOS_SIMULATOR
            return true
        #else
            return false
        #endif
    }
}
// Example of usage: UIDevice.current.isSimulator

Updated Info as of February 20, 2018

It looks like @russbishop has an authoritative answer that renders this answer "incorrect" - even though it appeared to work for a long time.

Detect if app is being built for device or simulator in Swift

Previous Answer

Based on @WZW's answer and @Pang's comments, I created a simple utility struct. This solution avoids warning produced by @WZW's answer.

import Foundation

struct Platform {

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

}

Example usage:

if Platform.isSimulator {
    print("Running on Simulator")
}