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Home/ Questions/Q 1031445
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Editorial Team
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Editorial Team
Asked: May 16, 20262026-05-16T13:55:51+00:00 2026-05-16T13:55:51+00:00

I don’t need a lesson in switching from recursive to non-recursive means, I just

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I don’t need a lesson in switching from recursive to non-recursive means, I just want to know why we can’t deal with this type of exception. Regardless, I’m using recursive functions on very large lists.

I have written code to attempt to catch StackOverFlowExceptions:

try { recursiveFxn(100000); }
catch(Exception){}
private void recursiveFxn(int countdown)
{
if (countdown > 0)
    recursiveFxn(countdown - 1);
else
    throw new Exception("lol.  Forced exception.");
}

But still I get program crashes (in both NUnit and a webpage I’m running). Why isn’t the exception caught?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-16T13:55:52+00:00Added an answer on May 16, 2026 at 1:55 pm

    Since .NET Framework 2.0, StackOverflowException cannot be caught. This is because it is considered a bad practice. Quoting the MSDN documentation:

    Starting with the .NET Framework
    version 2.0, a StackOverflowException
    object cannot be caught by a try-catch
    block and the corresponding process is
    terminated by default. Consequently,
    users are advised to write their code
    to detect and prevent a stack
    overflow. For example, if your
    application depends on recursion, use
    a counter or a state condition to
    terminate the recursive loop.

    Now, the only way to catch a StackOverflowException is when it was thrown by user code, as explained in a blog by Jared Parsons. Other than that, by hosting the CLR, you can handle (but not catch) a StackOverflowException and devise a way to let the execution of your program continue.

    Note that because the stack is unwound when an exception occurs, in pre-2.0 versions of .Net the stack would actually be much shorter when the StackOverflowException is handled, making it possible to do so without generating another StackOverflowException.

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