[Python-ideas] Efficient debug logging
Kyle Lahnakoski
klahnakoski at mozilla.com
Tue Feb 14 12:39:02 EST 2017
Can you wrap the expensive functions in lambdas? And have your logger
evaluate it, only if required?
> debugLog( ‘info is %r’ % (lambda: expensiveFunction(),) )
On 2017-02-14 10:51, Barry Scott wrote:
> A common pattern I use is to have logging calls for debug and information with my applications.
> The logging calls can be separately enabled and disabled.
>
> For example:
>
> debug_log_enabled = False
> def debugLog( msg ):
> If debug_log_enabled:
> print( ‘Debug: %s’ % (msg,) )
>
> Then the caller can simple write:
>
> def main():
> debugLog( ‘Start of main’ )
>
> This is fine until the evaluation of the msg becomes expensive.
>
> debugLog( ‘info is %r’ % (expensiveFunction(),) )
>
> What would be nice is to be able to avoid evaluation the tuple of arguments if debug is
> disabled as this can be expensive. I can write this:
>
> if debug_log_enabled: debugLog( ‘info is %r’ % (expensiveFunction(),) )
>
> But that is a more code then I would like to write. And if the debug code is a performance problem cannot
> be left in the production code.
>
> I could combine the boolean and the log function by using a class to tidy up the implementation.
>
> class DebugLog:
> def __init__( self, enabled = False ):
> self.enabled = enabled
>
> def __bool__( self ):
> return self.enabled
>
> def __call__( self, msg ):
> if self.enabled: print( ‘Debug: %s’ % (msg,) )
>
> And call like this:
>
> dbg_log = DebugLog()
>
> If dbg_log: dbg_log( ‘a debug message’ )
>
> But I’d like to only write:
>
> dbg_log( ‘a debug message’ )
>
> And have the evaluation of the argument skipped unless its dbg_log is enabled.
>
> I cannot see how to do this with python as it stands.
>
> Something would have to be added to allow python to short circuit the argument tuple evaluation.
>
> Maybe python can check for a special dunder on the class that know how to do this idiom, __if_true_call__?
>
> Thoughts?
>
> Barry
>
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