How to Detect If I am in Release or Debug Mode

How do I detect if I am in release or debug mode?

The simplest, and best long-term solution, is to use BuildConfig.DEBUG. This is a boolean value that will be true for a debug build, false otherwise:

if (BuildConfig.DEBUG) {
  // do something for a debug build
}

There have been reports that this value is not 100% reliable from Eclipse-based builds, though I personally have not encountered a problem, so I cannot say how much of an issue it really is.

If you are using Android Studio, or if you are using Gradle from the command line, you can add your own stuff to BuildConfig or otherwise tweak the debug and release build types to help distinguish these situations at runtime.

The solution from Illegal Argument is based on the value of the android:debuggable flag in the manifest. If that is how you wish to distinguish a "debug" build from a "release" build, then by definition, that's the best solution. However, bear in mind that going forward, the debuggable flag is really an independent concept from what Gradle/Android Studio consider a "debug" build to be. Any build type can elect to set the debuggable flag to whatever value that makes sense for that developer and for that build type.

How to determine whether code is running in DEBUG / RELEASE build?

Check your project's build settings under 'Apple LLVM - Preprocessing', 'Preprocessor Macros' for debug to ensure that DEBUG is being set - do this by selecting the project and clicking on the build settings tab. Search for DEBUG and look to see if indeed DEBUG is being set.

Pay attention though. You may see DEBUG changed to another variable name such as DEBUG_MODE.

then conditionally code for DEBUG in your source files

#ifdef DEBUG

// Something to log your sensitive data here

#else

// 

#endif

How can I check if a Flutter application is running in debug?

While this works, using constants kReleaseMode or kDebugMode is preferable. See Rémi's answer below for a full explanation, which should probably be the accepted question.

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The easiest way is to use assert as it only runs in debug mode.

Here's an example from Flutter's Navigator source code:

assert(() {
  if (navigator == null && !nullOk) {
    throw new FlutterError(
      'Navigator operation requested with a context that does not include a Navigator.\n'
      'The context used to push or pop routes from the Navigator must be that of a '
      'widget that is a descendant of a Navigator widget.'
    );
  }
  return true;
}());

Note in particular the () at the end of the call - assert can only operate on a Boolean, so just passing in a function doesn't work.

How to check if a app is in debug or release

At compile time or runtime? At compile time, you can use #if DEBUG. At runtime, you can use [Conditional("DEBUG")] to indicate methods that should only be called in debug builds, but whether this will be useful depends on the kind of changes you want to make between debug and release builds.

Detect DEBUG, RELEASE and my own mode C#

MyMode is a configuration. But, in and of itself, that doesn't define any conditional compilation symbols.

You change these through the projects compilation settings¹ or by passing the -define option to csc. If you look through the Debug configuration's compilation settings, you'll find that the DEBUG conditional compilation symbol was already defined², but there's no RELEASE symbol defined in the Release configuration.

There is no requirement (as you'll find above) that there be any relation between configurations and the symbols that they define.

#if (and family) is defined to work with conditional compilation.

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¹Project -> Properties -> Build -> General.

²In some versions of Visual Studio, there's a dedicated checkbox for it rather than it being listed in the Conditional Compilation symbols, but the effect is the same. If you unload the project and examine the XML, you'll find that all constants are stored in the <DefineConstants> element.

C# if/then directives for debug vs release

DEBUG/_DEBUG should be defined in VS already.

Remove the #define DEBUG in your code. Set preprocessors in the build configuration for that specific build.

The reason it prints "Mode=Debug" is because of your #define and then skips the elif.

The right way to check is:

#if DEBUG
    Console.WriteLine("Mode=Debug"); 
#else
    Console.WriteLine("Mode=Release"); 
#endif

Don't check for RELEASE.

How to check if APK is signed or debug build ?

There are different way to check if the application is build using debug or release certificate, but the following way seems best to me.

According to the info in Android documentation Signing Your Application, debug key contain following subject distinguished name: "CN=Android Debug,O=Android,C=US". We can use this information to test if package is signed with debug key without hardcoding debug key signature into our code.

Given:

import android.content.pm.Signature;
import java.security.cert.CertificateException;
import java.security.cert.X509Certificate;

You can implement an isDebuggable method this way:

private static final X500Principal DEBUG_DN = new X500Principal("CN=Android Debug,O=Android,C=US");
private boolean isDebuggable(Context ctx)
{
    boolean debuggable = false;

    try
    {
        PackageInfo pinfo = ctx.getPackageManager().getPackageInfo(ctx.getPackageName(),PackageManager.GET_SIGNATURES);
        Signature signatures[] = pinfo.signatures;

        CertificateFactory cf = CertificateFactory.getInstance("X.509");

        for ( int i = 0; i < signatures.length;i++)
        {   
            ByteArrayInputStream stream = new ByteArrayInputStream(signatures[i].toByteArray());
            X509Certificate cert = (X509Certificate) cf.generateCertificate(stream);       
            debuggable = cert.getSubjectX500Principal().equals(DEBUG_DN);
            if (debuggable)
                break;
        }
    }
    catch (NameNotFoundException e)
    {
        //debuggable variable will remain false
    }
    catch (CertificateException e)
    {
        //debuggable variable will remain false
    }
    return debuggable;
}

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