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Ben, Right. But before Log.Log is called arguments of methods are copied on the stack. That means, also the current content of errno is copied. And "current" means in that case before the call to Log.Log is performed (errno is transferred by value - not by reference). -Peter Benjamin LaHaise wrote: > On Tue, Sep 20, 2005 at 05:20:03PM +0200, Peter Duellings wrote: > >>Hi Ben, >> >>if Log.Log would modify errno the Log.Log debug output should >>not be affected since the value of errno - from my understanding - >>is copied on the stack *before* Log.Log is called. >>Or did I forget something? > > > errno does not reside on the stack. > > -ben - To unsubscribe from this list: send the line "unsubscribe linux-kernel" in the body of a message to majordomo@vger.kernel.org More majordomo info at http://vger.kernel.org/majordomo-info.html Please read the FAQ at http://www.tux.org/lkml/ | ||||||||||||
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