In absence of preprocessor macros, is there a way to define practical scheme specific flags at project level in Xcode project

In Swift you can still use the "#if/#else/#endif" preprocessor macros (although more constrained), as per Apple docs. Here's an example:

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

Now, you must set the "DEBUG" symbol elsewhere, though. Set it in the "Swift Compiler - Custom Flags" section, "Other Swift Flags" line. You add the DEBUG symbol with the -D DEBUG entry.

(Build Settings -> Swift Compiler - Custom Flags)

As usual, you can set a different value when in Debug or when in Release.

I tested it in real code; it doesn't seem to be recognized in a playground.

We ran into an issue with not wanting to set swift compiler flags because we didn't want to have to set them and keep them up to date for different targets etc. Also, in our mixed codebase, we didn't want to make remember to set our flags appropriately all the time for each language.

For ours, we declared a file in ObjC

PreProcessorMacros.h

extern BOOL const DEBUG_BUILD;

In the .m

PreProcessorMacros.m

#ifdef DEBUG
    BOOL const DEBUG_BUILD = YES;
#else
    BOOL const DEBUG_BUILD = NO;
#endif

Then, in your Objective-C Bridging Header

#import "PreProcessorMacros.h"

Now, use this in your Swift codebase

if DEBUG_BUILD {
    println("debug")
} else {
    println("release")
}

This is definitely a workaround, but it solved our problem so I posted it here in the hopes that it will help. It is not meant to suggest that the existing answers are invalid.

More swifty solution to Logans method. Set -D DEBUG in Other Swift Flags of Swift Compiler - Custom Flags section in build settings of your target.

Then declare following method in global scope:

#if DEBUG
let isDebugMode = true
#else
let isDebugMode = false
#endif

Now use it as

if isDebugMode {
    // Do debug stuff
}