#ifdef replacement in the Swift language

问题:

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.

#ifdef DEBUG
    // Debug-only code
#endif

Is there a similar solution in Swift?

回答1:

Yes you can do it.

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.

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

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

You can read my original post here.

---

IMPORTANT NOTE: -DDEBUG=1 doesn\'t work. Only -D DEBUG works. Seems compiler is ignoring a flag with a specific value.

回答2:

As stated in Apple Docs

The Swift compiler does not include a preprocessor. Instead, it takes advantage of compile-time attributes, build configurations, and language features to accomplish the same functionality. For this reason, preprocessor directives are not imported in Swift.

I\'ve managed to achieve what I wanted by using custom Build Configurations:

1. Go to your project / select your target / Build Settings / search for Custom Flags
2. For your chosen target set your custom flag using -D prefix (without white spaces), for both Debug and Release
3. Do above steps for every target you have

Here\'s how you check for target:

#if BANANA
    print(\"We have a banana\")
#elseif MELONA
    print(\"Melona\")
#else
    print(\"Kiwi\")
#endif

Tested using Swift 2.2

回答3:

In many situations, you don\'t really need conditional compilation; you just need conditional behavior that you can switch on and off. For that, you can use an environment variable. This has the huge advantage that you don\'t actually have to recompile.

You can set the environment variable, and easily switch it on or off, in the scheme editor:

You can retrieve the environment variable with NSProcessInfo:

    let dic = NSProcessInfo.processInfo().environment
    if dic[\"TRIPLE\"] != nil {
        // ... do secret stuff here ...
    }

Here\'s a real-life example. My app runs only on the device, because it uses the music library, which doesn\'t exist on the Simulator. How, then, to take screen shots on the Simulator for devices I don\'t own? Without those screen shots, I can\'t submit to the AppStore.

I need fake data and a different way of processing it. I have two environment variables: one which, when switched on, tells the app to generate the fake data from the real data while running on my device; the other which, when switched on, uses the fake data (not the missing music library) while running on the Simulator. Switching each of those special modes on / off is easy thanks to environment variable checkboxes in the Scheme editor. And the bonus is that I can\'t accidentally use them in my App Store build, because archiving has no environment variables.

回答4:

A major change of ifdef replacement came up with Xcode 8. i.e use of Active Compilation Conditions.

Refer to Building and Linking in Xcode 8 Release note.

New build settings

New setting: SWIFT_ACTIVE_COMPILATION_CONDITIONS

“Active Compilation Conditions” is a new build setting for passing conditional compilation flags to the Swift compiler.

Previously, we had to declare your conditional compilation flags under OTHER_SWIFT_FLAGS, remembering to prepend “-D” to the setting. For example, to conditionally compile with a MYFLAG value:

#if MYFLAG1
    // stuff 1
#elseif MYFLAG2
    // stuff 2
#else
    // stuff 3
#endif

The value to add to the setting -DMYFLAG

Now we only need to pass the value MYFLAG to the new setting. Time to move all those conditional compilation values!

Please refer to below link for more Swift Build Settings feature in Xcode 8: http://www.miqu.me/blog/2016/07/31/xcode-8-new-build-settings-and-analyzer-improvements/

回答5:

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.

回答6:

Xcode 8 and above

Use Active Compilation Conditions setting in Build settings / Swift compiler - Custom flags.

• This is the new build setting for passing conditional compilation flags to the Swift compiler.
• Simple add flags like this: ALPHA, BETA etc.

Then check it with compilation conditions like this:

#if ALPHA
    //
#elseif BETA
    //
#else
    //
#endif

Tip: You can also use #if !ALPHA etc.

回答7:

There is no Swift preprocessor. (For one thing, arbitrary code substitution breaks type- and memory-safety.)

Swift does include build-time configuration options, though, so you can conditionally include code for certain platforms or build styles or in response to flags you define with -D compiler args. Unlike with C, though, a conditionally compiled section of your code must be syntactically complete. There\'s a section about this in Using Swift With Cocoa and Objective-C.

For example:

#if os(iOS)
    let color = UIColor.redColor()
#else
    let color = NSColor.redColor()
#endif

回答8:

My two cents for Xcode 8:

a) A custom flag using the -D prefix works fine, but...

b) Simpler use:

In Xcode 8 there is a new section: \"Active Compilation Conditions\", already with two rows, for debug and release.

Simply add your define WITHOUT -D.

回答9:

isDebug Constant Based on Active Compilation Conditions

Another, perhaps simpler, solution that still results in a boolean that you can pass into functions without peppering #if conditionals throughout your codebase is to define DEBUG as one of your project build target\'s Active Compilation Conditions and include the following (I define it as a global constant):

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

---

isDebug Constant Based on Compiler Optimization Settings

This concept builds on kennytm\'s answer

The main advantage when comparing against kennytm\'s, is that this does not rely on private or undocumented methods.

In Swift 4:

let isDebug: Bool = {
    var isDebug = false
    // function with a side effect and Bool return value that we can pass into assert()
    func set(debug: Bool) -> Bool {
        isDebug = debug
        return isDebug
    }
    // assert:
    // \"Condition is only evaluated in playgrounds and -Onone builds.\"
    // so isDebug is never changed to true in Release builds
    assert(set(debug: true))
    return isDebug
}()

Compared with preprocessor macros and kennytm\'s answer,

• ✓ 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

• ✓ Documented, which means the function will follow normal API release/deprecation patterns.

• ✓ Using in if/else will not generate a \"Will never be executed\" warning.

回答10:

After setting DEBUG=1 in your GCC_PREPROCESSOR_DEFINITIONS Build Settings I prefer using a function to make this calls:

func executeInProduction(_ block: () -> Void)
{
    #if !DEBUG
        block()
    #endif
}

And then just enclose in this function any block that I want omitted in Debug builds:

executeInProduction {
    Fabric.with([Crashlytics.self]) // Compiler checks this line even in Debug
}

The advantage when compared to:

#if !DEBUG
    Fabric.with([Crashlytics.self]) // This is not checked, may not compile in non-Debug builds
#endif

Is that the compiler checks the syntax of my code, so I am sure that its syntax is correct and builds.

回答11:

In Swift projects created with Xcode Version 9.4.1, Swift 4.1

#if DEBUG
#endif

works by default because in the Preprocessor Macros DEBUG=1 has already been set by Xcode.

So you can use #if DEBUG \"out of box\".

By the way, how to use the condition compilation blocks in general is written in Apple\'s book The Swift Programming Language 4.1 (the section Compiler Control Statements) and how to write the compile flags and what is counterpart of the C macros in Swift is written in another Apple\'s book Using Swift with Cocoa and Objective C (in the section Preprocessor Directives)

Hope in future Apple will write the more detailed contents and the indexes for their books.

回答12:

XCODE 9 AND ABOVE

#if DEVELOP
    //
#elseif PRODCTN
    //
#else
    //
#endif

回答13:

![In Xcode 8 & above go to build setting -> search for custom flags ]1

In code

 #if Live
    print(\"Live\")
    #else
    print(\"debug\")
    #endif

回答14:

This builds on Jon Willis\'s answer that relies upon assert, which only gets executed in Debug compilations:

func Log(_ str: String) { 
    assert(DebugLog(str)) 
}
func DebugLog(_ str: String) -> Bool { 
    print(str) 
    return true
}

My use case is for logging print statements. Here is a benchmark for Release version on iPhone X:

let iterations = 100_000_000
let time1 = CFAbsoluteTimeGetCurrent()
for i in 0 ..< iterations {
    Log (\"⧉ unarchiveArray:\\(fileName) memoryTime:\\(memoryTime) count:\\(array.count)\")
}
var time2 = CFAbsoluteTimeGetCurrent()
print (\"Log: \\(time2-time1)\" )

prints:

Log: 0.0

Looks like Swift 4 completely eliminates the function call.

回答15:

func inDebugBuilds(_ code: () -> Void) {
    assert({ code(); return true }())
}

Source