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
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