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

Question

Before swift I would define a set of schemes for alpha, beta, and distribution builds. Each of these schemes would have a set of macros that were defined to gate certain behaviors at the project level. The simplest example is the DEBUG=1 macro that is defined by default for all Xcode projects in the default scheme for the Run build. One could query #ifdef DEBUG ... and make decisions in the code accordingly, even compiling out non-necessary code.

It seems that this type of configurational gating is not as easy using swift, as macros are not supported. Can someone suggest a comparable approach, I don't care if the code is compiled out, per se. I would like to gate features based on build scheme, though.

Answer 1

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.

Answer 2

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.

Answer 3

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
}

Answer 4

For me, set the debug item of "Active Compilation Condition" to "DEBUG" worked.

Then using DEBGU key work in #IF DEBUG works in debug mode and #ELSE in release mode:

1. Select your target,
2. In Build Setting tab search for "Active Compilation Condition",
3. Set the value of its "Debug" item to "YourKeyWord",

4. Use simply as follow:

   #if DEBUG
       print("You'r running in DEBUG mode!")
   #else
       print("You'r running in RELEASE mode!")
   #endif
   

Answer 5

I'm working in a mixed language code base where the obj-c code uses a macro to send debug messages to the console (and that macro relies on our debug preprocessor flag). I wanted to be able to call that same macro in the swift code...

1. I created a class method on one of my obj-c classes that is a wrapper around that macro.
2. I added that obj-c header to our bridge header file.
3. Now my swift code calls that class method as a "proxy" to the obj-c macro.

It's mildly annoying that I can't just call the macro straight up in the swift code, but at least now I only have one place in the project to worry about turning my debug flag on/off.