I have a
static const bool debug = false;
variable in my class. Debug output goes like this, trivially:
if(debug) cerr << "debug message";
Is it guaranteed that this if/then clause is always optimized away if the
debug variable is false?
Thanks!
Markus
11  3498 
"Markus Dehmann" <ma************ @gmail.com> wrote... I have a
static const bool debug = false;
variable in my class. Debug output goes like this, trivially:
if(debug) cerr << "debug message";
Is it guaranteed that this if/then clause is always optimized away if the
debug variable is false?
Nope. Nothing is guaranteed when it comes to optimisation. Unless
you remove it yourself somehow, it may still be there.
The more generically guaranteed approach is to use a macro that would
expand to nothing if NDEBUG is defined:
#ifndef NDEBUG
#define PRINTDEBUGMESSA GE(a) cerr << (a)
#else
#define PRINTDEBUGMESSA GE(a)
#endif
Now, when you use
PRINTDEBUGMESSA GE("debug message");
in your code, it will be an empty statement if NDEBUG is defined (usually
so in release builds).
Similar approach can be taken to output more than one item at a time.
V
Nope.
For instance, if you compile debugging information into your
executable, many compilers will not optimize because a lot of
optimizations (e.g. inlining) screw with the debugger.
If you want to be 100% positive it's not there, I'm pretty sure you
have to use the preprocessor. I would do something like this:
#ifdef DEBUG
#define DEBUG_PRINT(str ing) cerr << (string)
#elseif
#else
#define DEBUG_PRINT(str ing)
#endif
(Probably there are a couple subtle errors there. I don't know whether
to put the ; in the macro or not either.)
Then you just do DEBUG_PRINT("de bug message");
That said, if you compile with optimizations the compiler almost
certainly will do it.
You could always look at the assembly output and see too be sure.
(There are things you could do to increase the chance nothing would be
there, but I don't think it's worth it.)
most commercial grade compilers will optimize it away. But the standard
doesn't guarantee it
Check the assembly generated and see it actually happens for your
compiler. Also most compilers dont perform optimizations by default for
a DEBUG build
Raj
Wow, I swear I didn't see your post before making mine...
(Also I noted an extranious #elseif in mine; get rid of that. That's me
being really tired.)
Markus Dehmann <ma************ @gmail.com> wrote in news:3a0ipdF66m igkU1
@individual.net : I have a
static const bool debug = false;
variable in my class. Debug output goes like this, trivially:
if(debug) cerr << "debug message";
Is it guaranteed that this if/then clause is always optimized away if the
debug variable is false?
No. Quality of Implementation issue. Depends on your compiler. ra******@hotmai l.com wrote: most commercial grade compilers will optimize it away. But the standard
doesn't guarantee it
On the other hand, a well-formed program can't tell whether that code
has been optimized away, because it has no observable behavior.
--
Pete Becker
Dinkumware, Ltd. ( http://www.dinkumware.com)
Pete Becker <pe********@acm .org> wrote in message news:<pP******* *************@g iganews.com>... ra******@hotmai l.com wrote: most commercial grade compilers will optimize it away. But the standard
doesn't guarantee it
On the other hand, a well-formed program can't tell whether that code
has been optimized away, because it has no observable behavior.
When will I be able to observe my program's behavior ?
What is a well-formed program ? may i ask you, a member of acm ????
Thank you
Just that wrote:
When will I be able to observe my program's behavior ?
What is a well-formed program ?
A well-formed program is one that conforms to the requirements of the
C++ Standard. Observable behavior is printed output, etc. If that code
is never executed, the output you see won't depend on whether the code
is left in the executable or removed.
--
Pete Becker
Dinkumware, Ltd. ( http://www.dinkumware.com)
Just that wrote: Pete Becker <pe********@acm .org> wrote in message
news:<pP******* *************@g iganews.com>... ra******@hotmai l.com wrote: > most commercial grade compilers will optimize it away. But the standard
> doesn't guarantee it
>
On the other hand, a well-formed program can't tell whether that code
has been optimized away, because it has no observable behavior.
When will I be able to observe my program's behavior ?
What is a well-formed program ? may i ask you, a member of acm ????
A certain slowdown might be measurable, though. Execution speed is a form of
"observable behavior," I'd say.
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