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
