if(debug), optimized away?

[Michael Chermside]

> I would like to use something like this when running in
> debug mode:
> 
> if(debug):
>    if(items > 9):
>      print "fatal error..." 
>    do_some_check_foo(42)
>    ...
> 
> 
> If I set debug=0, is this code optimized away as in most
> C/C++?? compilers?
> 
> The reason for asking is that I would like to have A LOT of
> debugging code and would like to avoid the performance penalty
> when not runing in debug mode.
> 
> Are there other, perhaps more Pythonish, ways of doing the same
> thing?

No, it is not optimized away.

I would recomend the following as being pythonic, but others might not 
agree, so get other opinions:


     # WARNING: Untested Code

     def itemsAreLT9AndValid(items):
         if items > 9:
             print "fatal error..."
         do_some_check_foo(42)


     ... LATER ON, in your code ...

     ... code here ...
     assert itemsAreLT9AndValid()
     ... more code here ...

This way, you've put all of your code into an assert statement. Of 
course, (in my experience) 90% of all the debug code can be expressed as 
a single boolean expression, so 90% of the time you won't need the 
helper function, and you can just write something like this:

     assert items > 9

but there are a few cases (like running multiple things or testing with 
a loop) which are difficult or impossible to express without creating 
the extra function to wrap it.


You'll probably want to enclose the definitions of ALL of the test 
functions inside an "if debug:" (one which executes just once), so that 
you don't even create the functions unless you need them. For instance, 
the following:

     ... code here ...
     def itemsAreLT9AndValid(items):
         if items > 9:
             print "fatal error..."
         do_some_check_foo(42)
     assert itemsAreLT9AndValid()
     ... more code here ...

would be a bad idea for performance reasons (we may not CALL the 
function if debug is off, but we CREATE it each time.

-- Michael Chermside