Date: Sun, 5 Feb 2012 20:00:22 +0100 From: Jilles Tjoelker <jilles@stack.nl> To: Daniel Eischen <deischen@freebsd.org> Cc: Garrett Cooper <yanegomi@gmail.com>, freebsd-hackers@freebsd.org, Ivan Voras <ivoras@freebsd.org>, Xin LI <delphij@delphij.net>, Jan Mikkelsen <janm-freebsd-hackers@transactionware.com>, davidxu@freebsd.org Subject: Re: sem(4) lockup in python? Message-ID: <20120205190022.GC57421@stack.nl> In-Reply-To: <Pine.GSO.4.64.1202051330020.28531@sea.ntplx.net> References: <jejrbe$or8$1@dough.gmane.org> <201201110806.30620.jhb@freebsd.org> <CAF-QHFWFvYTPeM68Mk%2BOYVX--MNhKOJ2o1GF9ZOsBmtiC5fYFQ@mail.gmail.com> <CAGH67wRsek2-WY_ETW6QEER1r5dDXLXfDjbzpHMjtv059Y8cJw@mail.gmail.com> <5D37298B-9D68-4F0F-8AAB-E8F2DBB9D9C3@transactionware.com> <CAGH67wT3HuxPHUXeTib0qJNH%2BO5snn3Eiim1bfj8LewYoKdXdA@mail.gmail.com> <CAF-QHFVADLkduLH1AG_hSZeDtDVCC=FkqZxbxrsMY3Y3%2BsMZ8A@mail.gmail.com> <Pine.GSO.4.64.1202051330020.28531@sea.ntplx.net>
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On Sun, Feb 05, 2012 at 01:32:42PM -0500, Daniel Eischen wrote: > On Sun, 5 Feb 2012, Ivan Voras wrote: > > On 5 February 2012 11:44, Garrett Cooper <yanegomi@gmail.com> wrote: > >> 'make MAKE_JOBS_NUMBER=1' is the workground used right now.. > > David Xu suggested that it is a bug in Python - it doesn't set > > process-shared attribute when it calls sem_init(), but i've tried > > patching it (replacing the port patchfile file the one I've attached) > > and I still get the hang. > I don't understand how process shared semaphores can work. Perhaps > I'm dumb and ignorant, but a sem_id_t is an allocated struct. The > actual kernel sem_id is inside the struct. Isn't this the same > reason pthread_mutex_t and pthread_cond_t cannot be process-shared? That's how the old implementation works. It does not support process-shared semaphores although they may happen to work in some specific cases. However, in 9.0, sem_t works differently and contains the actual lock word directly, so that process-shared semaphores work. The implementation is in lib/libc/gen/sem_new.c. The pshared flag to sem_init() is not a no-op because it tells the kernel to allow for use from multiple processes. Note that the old implementation is still present as well, for compatibility with old binaries. -- Jilles Tjoelker
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