Conditional compilation in Python

Question

How to do conditional compilation in Python ?

Is it using DEF ?

Answer 1

Python isn't compiled in the same sense as C or C++ or even Java, python files are compiled "on the fly", you can think of it as being similar to a interpreted language like Basic or Perl.₁

You can do something equivalent to conditional compile by just using an if statement. For example:

if FLAG:
    def f():
        print "Flag is set"
else:
    def f():
        print "Flag is not set"

You can do the same for the creation classes, setting of variables and pretty much everything.

The closest way to mimic IFDEF would be to use the hasattr function. E.g.:

if hasattr(aModule, 'FLAG'):
    # do stuff if FLAG is defined in the current module.

You could also use a try/except clause to catch name errors, but the idiomatic way would be to set a variable to None at the top of your script.

1. Python code is byte compiled into an intermediate form like Java, however there generally isn't a separate compilation step. The "raw" source files that end in .py are executable.

Answer 2

There is actually a way to get conditional compilation, but it's very limited.

if __debug__:
    doSomething()

The __debug__ flag is a special case. When calling python with the -O or -OO options, __debug__ will be false, and the compiler will ignore that statement. This is used primarily with asserts, which is why assertions go away if you 'really compile' your scripts with optimization.

So if your goal is to add debugging code, but prevent it from slowing down or otherwise affecting a 'release' build, this does what you want. But you cannot assign a value to __debug__, so that's about all you can use it for.