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Home/ Questions/Q 8975711
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Editorial Team
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Editorial Team
Asked: June 15, 20262026-06-15T18:56:58+00:00 2026-06-15T18:56:58+00:00

Take the following example: public void Foo() { //Code… Debug.Assert(ExpensiveTest()); //Code… } What happens

  • 0

Take the following example:

public void Foo()
{
    //Code...

    Debug.Assert(ExpensiveTest());

    //Code...
}

What happens to the the Debug.Assert method when I compile in release mode? Would ExpensiveTest() still run? If not, then how does it work (since it is not a macro that can be set to evaluate to nothing)? If it does run, then doesn’t that defeat the purpose of debug assertions?

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  1. Editorial Team
    Editorial Team
    2026-06-15T18:57:00+00:00Added an answer on June 15, 2026 at 6:57 pm

    What happens to the the Debug.Assert method when I compile in release mode?

    It’s completely removed (including the call to ExpensiveTest), assuming you don’t have the DEBUG conditional compilation symbol defined in your release configuration.

    If you look at the documentation, the declaration uses [ConditionalAttribute("DEBUG")]:

    [ConditionalAttribute("DEBUG")]
    public static void Assert(
        bool condition
    )
    

    ConditionalAttribute is used for conditional compilation. See Bart de Smet’s blog post on conditional compilation for more details, along with section 17.4.2 of the C# 4 specification.

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