How Can I Correctly Use Conditional Compilation in C# to Distinguish Between Debug and Release Builds?

C# Conditional Compilation for Debug vs. Release Builds

In Visual Studio, when configuring a solution with a single project, the Configuration setting determines the build mode, typically "debug" or "release." However, when using conditional compilation directives, such as #if/#then/#elif, it's crucial to ensure that the correct conditions are being evaluated.

In the provided code snippet, the developer aims to differentiate between debug and release modes to set variable defaults. However, the code outputs "Mode=Debug" regardless of the Configuration setting.

The confusion arises from the predefined preprocessor symbols in Visual Studio. DEBUG/_DEBUG is automatically defined for debug builds, and RELEASE for release builds. This means that defining #define DEBUG in the code overrides the default VS behavior and forces the #if DEBUG condition to always be true.

The correct approach is to remove the #define DEBUG statement and instead rely on the preprocessor symbols defined by VS. The modified code should look like this:

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

By using DEBUG as the condition, the code checks for the preprocessor symbol that is defined based on the Configuration setting, accurately distinguishing between debug and release modes.

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