python equivalent of '#define func() ' or how to comment out a function call in python

Question

my python code is interlaced with lots of function calls used for (debugging|profiling|tracing etc.) for example: ``` import logging logging.root.setLevel(logging.DEBUG) logging.debug('hello') j = 0 for i in range(10): j += i logging.debug('i %d j %d' % (i,j)) print(j) logging.debug('bye') ``` i want to #define these resource consuming functions out of the code. something like the c equivalent ``` #define logging.debug(val) ``` yes, i know the logging module logging level mechanism can be used to mask out loggings below set log level. but, im asking for a general way to have the python interpreter skip functions (that take time to run even if they dont do much) one idea is to redefine the functions i want to comment out into empty functions: ``` def lazy(*args): pass logging.debug = lazy ``` the above idea still calls a function, and may create a myriad of other problems

Accepted Answer

Python does not have a preprocessor, although you could run your python source through an external preprocessor to get the same effect - e.g. `sed "/logging.debug/d"` will strip out all the debug logging commands. This is not very elegant though - you will end up needing some sort of build system to run all your modules through the preprocessor and perhaps create a new directory tree of the processed .py files before running the main script.

Alternatively if you put all your debug statements in an `if __debug__:` block they will get optimised out when python is run with the -O (optimise) flag.

As an aside, I checked the code with the dis module to ensure that it did get optimised away. I discovered that both

```
if __debug__: doStuff()
```

and

```
if 0: doStuff()
```

are optimised, but

```
if False: doStuff()
```

is not. This is because False is a regular Python object, and you can in fact do this:

```
>>> False = True
>>> if False: print "Illogical, captain"
Illogical, captain
```

Which seems to me a flaw in the language - hopefully it is fixed in Python 3.

Edit:

This is fixed in Python 3: Assigning to True or False now gives a SyntaxError. Since True and False are constants in Python 3, it means that `if False: doStuff()` is now optimised:

```
>>> def f():
...     if False: print( "illogical")
... 
>>> dis.dis(f)
  2           0 LOAD_CONST               0 (None) 
              3 RETURN_VALUE         
```

python equivalent of '#define func() ' or how to comment out a function call in python

Solution

Problem