in/out with -release

[Lars T. Kyllingstad]

On Sat, 05 Mar 2011 18:12:30 +0000, Lars T. Kyllingstad wrote:

> On Sat, 05 Mar 2011 10:15:48 -0700, user wrote:
> 
>> On 03/04/2011 09:22 PM, Jonathan M Davis wrote:
>>> On Friday 04 March 2011 20:14:32 Kai Meyer wrote:
>>>> I have an 'enforce' function call in an 'in' block for a function.
>>>> When I compile with "-release -O -inline", the in/out blocks appear
>>>> to be skipped. It's a simple verification for a dynamic array to not
>>>> have a length of 0. In debug mode, the test condition hits the
>>>> enforce in the 'in' block, but in release mode it does not. In both
>>>> release and debug mode, the same exact enforce function works
>>>> properly.
>>>>
>>>> So am I to understand that -release will skip in/out blocks entirely?
>>>
>>> Of course. It uses asserts. asserts are disabled in -release. Asserts
>>> are for debugging, testing, and verifying code when developing, not
>>> for code which is released. So, you get the benefit of the test when
>>> you don't have -release and the benefit of speed when you do have
>>> -release. If an assertion fails, your code logic is invalid. It's for
>>> validating your code, not user input or whatnot.
>>>
>>> enforce, on the other hand, is not a language primitive. It's not
>>> intended for testing or debugging. It's intended to be used in
>>> production code to throw an exception when its condition fails. If an
>>> enforce fails, that generally means that you had bad input somewhere
>>> or that an operation failed or whatnot. It's not intended for testing
>>> the logic of your code like assert is intended to do. It's simply a
>>> shorthand way to throw an exception when your program runs into a
>>> problem.
>>>
>>> - Jonathan M Davis
>> 
>> I don't think I understand your response entirely. I understand that
>> asserts are disabled in -release mode. I understand that enforce is a
>> function that comes with std.exception, and the code isn't hard to
>> follow.
>> 
>> What I'm confused about is the in block, and why it is skipped in
>> -release mode. You say "It uses asserts." I didn't put an assert in my
>> in block, I put an enforce. So I'm guessing that you are indicating
>> that the in block is treated like an assert, and is disabled with the
>> -release flag.
>> 
>> But I think after reading your post you've helped clarify that what I'm
>> checking (that you can't pop an empty stack) based on user input is
>> something I should be checking with an enforce inside the function, and
>> not an assert or enforce inside the in block.
>> 
>> I still think I would like it if you could be a little more explicit
>> about the in/out blocks. Are they always disabled entirely (skipped)
>> with -release, or just certain things?
>> 
>> Thanks for your help!
>> 
>> -Kai Meyer
> 
> That's right.  in, out and invariant blocks are not included in release
> mode.
> 
> -Lars

It's documented here, by the way:
http://www.digitalmars.com/d/2.0/dmd-linux.html#switches

(Scroll down to -release.)

-Lars