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Comment by umjames | Hacker News Reader

Wouldn't u.getA() and u.getB() still be evaluated first so that their values can be passed to the debug method? Without the if statement, how is the debug method not being called?

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In the original code, the debug call is guarded by a condition:

  if (log.isDebugEnabled()) { 
    log.debug("{}", expensiveOp(), debugModeOnlyOp()); 
  }

But in the new code, the calls to expensiveOp and debugModeOnlyOp will be called every time this sequence of code is run regardless of the condition:

  log.debug("{}", expensiveOp(), debugModeOnlyOp());

which is equivalent to the old code:

  var x = expensiveOp();
  var y = debugModeOnlyOp();

  if (log.isDebugEnabled()) { 
    log.debug("{} {}", x, y); 
  }

In most languages that do not support macros, it's impossible to correctly state the dependency of expensiveOp & debugModeOnlyOp on the log.isDebugEnabled() flag using log4j's new API.

I think he's saying exactly that. His post is kinda confusing because of how he quoted the article.

u.getA() and u.getB() will still be evaluated. However I believe the conversion of A and B to strings will not be carried out if debug is disabled.

Often it is the conversion to string that is an expensive operation.

Incidentally with lots of lazy loading (Hibernate I'm looking at you...) this can lead to errors only appearing when debug is turned off - not a nice situation to be in.

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