Checking for debug and some other condition in Python

In Python sometimes I want to do something like (1)

if __debug__ and verbose: print "whatever"

If Python is run with -O, then I'd like for that whole piece of code to disappear, as it would if I just had (2)

if __debug__: print "whatever"

or even (3)

if __debug__:
    if verbose: print foo

However, that doesn't seem to happen (see below). Is there a way I can get the run-time efficiency of #3 with compact code more like #1?

Here's how I tested that I'm not getting the efficient code I want:

#!/usr/bin/python2.7

from dis import dis
import sys

cmds = ["""
def func ():
    if __debug__ and 1+1: sys.stdout.write('spam')""",   """
def func():
    if __debug__: sys.stdout.write('ham')""",     """
def func():
    __debug__ and sys.stdout.write('eggs')"""]

print "__debug__ is", __debug__, "\n\n\n"

for cmd in cmds:
    print "*"*80, "\nSource of {}\n\ncompiles to:".format(cmd)
    exec(cmd)
    dis(func)
    print "\n"*4

Running this gives

__debug__ is False 



******************************************************************************** 
Source of 
def func ():
    if __debug__ and 1+1: sys.stdout.write('spam')

compiles to:
  3           0 LOAD_GLOBAL              0 (__debug__)
              3 POP_JUMP_IF_FALSE       31
              6 LOAD_CONST               3 (2)
              9 POP_JUMP_IF_FALSE       31
             12 LOAD_GLOBAL              1 (sys)
             15 LOAD_ATTR                2 (stdout)
             18 LOAD_ATTR                3 (write)
             21 LOAD_CONST               2 ('spam')
             24 CALL_FUNCTION            1
             27 POP_TOP             
             28 JUMP_FORWARD             0 (to 31)
        >>   31 LOAD_CONST               0 (None)
             34 RETURN_VALUE        





******************************************************************************** 
Source of 
def func():
    if __debug__: sys.stdout.write('ham')

compiles to:
  3           0 LOAD_CONST               0 (None)
              3 RETURN_VALUE        





******************************************************************************** 
Source of 
def func():
    __debug__ and sys.stdout.write('eggs')

compiles to:
  3           0 LOAD_GLOBAL              0 (__debug__)
              3 JUMP_IF_FALSE_OR_POP    21
              6 LOAD_GLOBAL              1 (sys)
              9 LOAD_ATTR                2 (stdout)
             12 LOAD_ATTR                3 (write)
             15 LOAD_CONST               1 ('eggs')
             18 CALL_FUNCTION            1
        >>   21 POP_TOP             
             22 LOAD_CONST               0 (None)
             25 RETURN_VALUE

Solution

No, you can't. Python's compiler is not nearly smart enough to detect in what cases it could remove the code block and if statement.

Python would have to do a whole lot of logic inference otherwise. Compare:

if __debug__ or verbose:

with

if __debug__ and verbose:

for example. Python would have to detect the difference between these two expressions at compile time; one can be optimised away, the other cannot.

Note that the difference in runtime between code with and without if __debug__ statements is truly minute, everything else being equal. A small constant value test and jump is not anything to fuss about, really.

Checking for __debug__ and some other condition in Python

Checking for debug and some other condition in Python