#ifdef replacement in the Swift language

Question

In C/C++/Objective C you can define a macro using compiler preprocessors. Moreover, you can include/exclude some parts of code using compiler preprocessors.

Answer 1

As of Swift 4.1, if all you need is just check whether the code is built with debug or release configuration, you may use the built-in functions:

• _isDebugAssertConfiguration() (true when optimization is set to -Onone)
• _isReleaseAssertConfiguration() (true when optimization is set to -O) (not available on Swift 3+)
• _isFastAssertConfiguration() (true when optimization is set to -Ounchecked)

e.g.

func obtain() -> AbstractThing {
    if _isDebugAssertConfiguration() {
        return DecoratedThingWithDebugInformation(Thing())
    } else {
        return Thing()
    }
}

Compared with preprocessor macros,

• ✓ You don't need to define a custom -D DEBUG flag to use it
• ~ It is actually defined in terms of optimization settings, not Xcode build configuration

• ✗ Undocumented, which means the function can be removed in any update (but it should be AppStore-safe since the optimizer will turn these into constants)

  • these once removed, but brought back to public to lack of @testable attribute, fate uncertain on future Swift.

• ✗ Using in if/else will always generate a "Will never be executed" warning.