Does the C# compiler optimize out an empty try/catch?

Question

In c#, how much time is expended in an empty try catch block?

I've heard that it is between 10-55ms, but i'm leaning to zero/one as the compiler will replace the empty try catch with a no-op placeholder.

any compiler experts that can verify this information?

I believe the reason that you're getting so many downvotes is that this probably isn't the appropriate place to ask this particular question... — Jfabs
– Jfabs, Commented Mar 13, 2014 at 19:43
You can review the generated IL/bytecode yourself. How can I view MSIL / CIL generated by C# compiler? Why is it called assembly? — Jonathan Lonowski
– Jonathan Lonowski, Commented Mar 13, 2014 at 19:46
Make sure to choose a good title, it really helps avoid negative initial reactions/votes. The hypothesis and fundamental question that has been presented is "I'm leaning to zero/one as the compiler will replace the empty try catch with a no-op placeholder.", so focus on that. — user2864740
– user2864740, Commented Mar 13, 2014 at 19:47
When you say "empty try catch block", do you mean if you literally have: try { } catch { }? — Chris Sinclair
– Chris Sinclair, Commented Mar 13, 2014 at 19:50

Chris Sinclair · Accepted Answer · 2014-03-13 20:11:23Z

If when you say "empty try catch block", you mean literally:

try
{
}
catch
{
}

Then yes, when building with compiler optimizations turned on (that is, "release" mode), it will not emit any IL instructions. For example:

private void Test()
{
    try
    {
    }
    catch
    {
    }
}

Compiles to:

IL_0000:  ret

Which is the same as if it were an empty method.

However, when optimizations are turned off (that is, "debug" mode), it emits:

IL_0000:  nop         
IL_0001:  nop         
IL_0002:  nop         
IL_0003:  leave.s     IL_000A
IL_0005:  pop         
IL_0006:  nop         
IL_0007:  nop         
IL_0008:  leave.s     IL_000A
IL_000A:  nop         
IL_000B:  ret

Whereas an empty method would be:

IL_0000:  nop         
IL_0001:  ret

Whether or not these IL instructions are stripped when the JIT compiler is executed, I am not positive. But this would still only apply when compiling without compiler optimizations as with them turned on the try/catch is stripped out during the initial compile to IL before it hits the JIT compiler.

EDIT: I just realized I may not have actually answered the question:

In release mode, zero time is expended on an empty try/catch.
In debug mode, a non-zero time is expended on an empty try/catch as it still emits the necessary IL code for debugging/breakpoint purposes and requires some JIT compilation. However, it would be a negligibly small amount of time. Definitely nowhere near the neighbourhood of "10-55 milliseconds".

kjbartel · Accepted Answer · 2015-08-06 01:44:50Z

2

If you check compiled code by proper software (like ILspy or Reflector), you can see that empty block will be deleted completley by compiler.

edited Aug 6, 2015 at 1:44

kjbartel

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answered Mar 13, 2014 at 19:43

nuclear sweet

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3 Comments

Jfabs Over a year ago

In the future, you'll want to back up your answers with sources and such

nuclear sweet Over a year ago

@Jfabs check this prntscr.com/30ielc (i hope i get this question right, because name of topic is pretty confsing)

Jfabs Over a year ago

looks like you are correct to me. Thank you for following up. I would recommend you edit your post to include that screenshot as evidence of your statement.

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