Justin Duke

Checking for Debug mode in Swift

There are often a bunch of times where you want to write Swift code that only executes when you’re in Debug mode. Some obvious examples:

  • Hiding ads when you generate screenshots
  • Logging and printing to console
  • Using a local developer/staging endpoint instead of the production one.

Still, this can kind of be a pain to manage. The Objective-C mainstay (which still works in Swift) is this very clunky macro:

#if DEBUG
    let a = 2
#else
    let a = 3
#endif

This, to me, is super gross: it’s imperative, clashes with other Swift syntax, and is hard to parse or do clever things with.

However, I stumbled on a little known (and, in my opinion, better) solution that checks the build configuration of your project:

_isDebugAssertConfiguration() -> Bool

Yup, that’s all you need to call: it’s a globally available function.

The method invocation on its own is a little clunky, so I bind it to a global struct that I use:

struct Utilities {
  // other stuff, too.
  static let isDebug = _isDebugAssertConfiguration()
}

Which means I can now use it in guards and lets and all sorts of things!

For instance, to bail out of functions early if I’m in debug mode:

// The first section only contains the ad..
// We hide the ad in debug mode so we can generate nicer screenshots.
override func tableView(_ tableView: UITableView,
                        numberOfRowsInSection section: Int) -> Int {
    let sectionIsAdSection = section == 0
    let numberOfCellsInAdSection = Constants.isDebug ? 0 : 1
    return sectionIsAdSection ?
        numberOfCellsInAdSection :
        matchingAnswers[section - 1].count
}

Some obvious caveats apply with this approach (it’s a flaky function that might get removed at some point, and it’s not publicly documented) but it’s worked great for me and I recommend trying it out rather than scattering your codebase with clunky #if DEBUG invocations.

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