As I understand in our case when:
a) semaphore is initially open;
b) we set SEM_UNDO for all semop's;
c) all semop's are symmetrical;
we're safe from this bug, because semadj will be always be either 1 or 0 and it'll always get back to zero when lock is released.
Hello, Philip! Thanks for joining us here.
As I understand in our case when:
a) semaphore is initially open;
b) we set SEM_UNDO for all semop's;
c) all semop's are symmetrical;
we're safe from this bug, because semadj will be always be either 1 or 0 and it'll always get back to zero when lock is released.