Date: Wed, 3 Oct 2012 15:12:37 -0400 From: John Baldwin <jhb@freebsd.org> To: freebsd-arch@freebsd.org Cc: Andriy Gapon <avg@freebsd.org> Subject: Re: x86 boot code build Message-ID: <201210031512.37718.jhb@freebsd.org> In-Reply-To: <506C385C.3020400@FreeBSD.org> References: <506C385C.3020400@FreeBSD.org>
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On Wednesday, October 03, 2012 9:06:36 am Andriy Gapon wrote: > > Currently we produce "slightly" different binaries for x86 boot code depending > whether MACHINE_CPUARCH is i386 or amd64. I think that there is no good reason > for this, since in both cases we use exactly the same code and target the same > classes of machines. In other words, the binaries should be interchangeable[*]. > > The difference boils down to using -march=i386 on amd64 while i386 uses default > compiler flags, which are equivalent to -march=i486 -mtune=generic. > If my analysis is correct, the only thing affected by the flags in the boot code > is use of leave instruction when -Os is _not_ specified. > For -march=i386 our gcc prefers using leave. For -march=i486 it thinks that > movs+pops are faster than leave and so prefers to not use it. If -Os is > specified, then leave is always used because it results in smaller machine code. > > So, as it is now, on amd64 we produce slightly smaller boot binaries where size > doesn't matter. Where size really matters (-Os) we produce identical binaries. > > If we decide that it makes sense to converge i386 and amd64 boot build options, > which should we pick? > > [*] It is the current state of matter, but it is not necessary that it will always > be the same. Go for smaller binaries. -- John Baldwin
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