Looks resonable.
/ Peter Bortas
Previous text:
2003-02-05 18:29: Subject: synchronized
Even in that case the declaration for the variable ought have some sort of flag that it has a mutex, i.e. that the mutex still is declared in one way or the other. E.g:
synchronized mapping foo; mapping bar; ... void f() { synchronized (foo) {...} // Ok. synchronized (bar) {...} // Error: bar doesn't have a mutex. }
There are two reasons for this requirement: As I've said earlier it avoids silently getting mutexes on the wrong things when one happens to mixup variables. Another reason is that it allows more efficient implementation since the space for a mutex can be allocated at compile time. (Note that the proposed solution with a weak mapping is prone to cause gc overhead since the locked things can't be refcount garbed.)
/ Martin Stjernholm, Roxen IS