Date: Sat, 03 Nov 2012 14:12:11 -0600 From: Ian Lepore <freebsd@damnhippie.dyndns.org> To: d@delphij.net Cc: freebsd-hackers@freebsd.org Subject: Re: watchdogd, jemalloc, and mlockall Message-ID: <1351973531.1120.118.camel@revolution.hippie.lan> In-Reply-To: <50957793.8060709@delphij.net> References: <1351967919.1120.102.camel@revolution.hippie.lan> <50957793.8060709@delphij.net>
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On Sat, 2012-11-03 at 12:59 -0700, Xin Li wrote: > -----BEGIN PGP SIGNED MESSAGE----- > Hash: SHA256 > > On 11/3/12 11:38 AM, Ian Lepore wrote: > > In an attempt to un-hijack the thread about memory usage increase > > between 6.4 and 9.x, I'm starting a new thread here related to my > > recent discovery that watchdogd uses a lot more memory since it > > began using mlockall(2). > > > > I tried statically linking watchdogd and it made a small difference > > in RSS, presumably because it doesn't wire down all of libc and > > libm. > > Speaking for this, the last time I brought this up, someone (can't > remember, I think it was phk@) argued that the shared library would > use only one copy of memory, while statically linked ones would be > duplicated and thus use more memory. I haven't yet tried to prove or > challenge that, though. That sounds right to me... if 3 or 4 daemons were to eventually be statically linked because of mlockall(), then each of them would have its own private copy of strlen(), and malloc(), and so on; we'd be back to the bad old days before shared libs came along. Each program would contain its own copy of only the routines from the library that it uses, not the entire library in each program. On the other hand, if even one daemon linked with shared libc uses mlockall(), then all of libc gets wired. As I understand it, only one physical copy of libc would exist in memory, still shared by almost all running apps. The entire contents of the library would continuously occupy physical memory, even the parts that no apps are using. It's hard to know how to weigh the various tradeoffs. I suspect there's no one correct answer. -- Ian
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