in/out with -release

[Kai Meyer]

On 03/05/2011 11:14 AM, Lars T. Kyllingstad wrote:
> 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

All very welcome responses. Thanks for your time :) Got lots of reading 
to do.

-Kai Meyer