Seems slightly simpler to just make debugLog accept a callable as an alternative to a string.
debugLog(lambda:( ‘info is %s’ % expensiveFunction()) ) Op 14 feb. 2017 18:42 schreef "Kyle Lahnakoski" <klahnako...@mozilla.com>: 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 > > _______________________________________________ > Python-ideas mailing list > Python-ideas@python.org > https://mail.python.org/mailman/listinfo/python-ideas > Code of Conduct: http://python.org/psf/codeofconduct/ _______________________________________________ Python-ideas mailing list Python-ideas@python.org https://mail.python.org/mailman/listinfo/python-ideas Code of Conduct: http://python.org/psf/codeofconduct/
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