* H. Peter Anvin <h...@zytor.com> wrote:
> On March 7, 2016 12:22:28 AM PST, Ingo Molnar <mi...@kernel.org> wrote:
> >
> >* Andy Lutomirski <l...@kernel.org> wrote:
> >
> >> Ingo suggested that the comments should explain when the various
> >> entries are used. This adds these explanations and improves other
> >> parts of the comments.
> >
> >Thanks for doing this, this is really useful!
> >
> >One very small detail I noticed:
> >
> >> +/*
> >> + * 32-bit legacy system call entry.
> >> + *
> >> + * 32-bit x86 Linux system calls traditionally used the INT $0x80
> >> + * instruction. INT $0x80 lands here.
> >> + *
> >> + * This entry point can be used by 32-bit and 64-bit programs to
> >perform
> >> + * 32-bit system calls. Instances of INT $0x80 can be found inline
> >in
> >> + * various programs and libraries. It is also used by the vDSO's
> >> + * __kernel_vsyscall fallback for hardware that doesn't support a
> >faster
> >> + * entry method. Restarted 32-bit system calls also fall back to
> >INT
> >> + * $0x80 regardless of what instruction was originally used to do
> >the
> >> + * system call.
> >> + *
> >> + * This is considered a slow path. It is not used by modern libc
> >> + * implementations on modern hardware except during process startup.
> >> + *
> >> + * Arguments:
> >> + * eax system call number
> >> + * ebx arg1
> >> + * ecx arg2
> >> + * edx arg3
> >> + * esi arg4
> >> + * edi arg5
> >> + * ebp arg6
> >> + */
> >> ENTRY(entry_INT80_32)
> >
> >entry_INT80_32() is only used on pure 32-bit kernels, 64-bit kernels
> >use
> >entry_INT80_compat(). So the above text should not talk about 64-bit
> >programs, as
> >they can never trigger this specific entry point, right?
> >
> >So I'd change the explanation to something like:
> >
> >> + * This entry point is active on 32-bit kernels and can thus be used
> >by 32-bit
> >> + * programs to perform 32-bit system calls. (Programs running on
> >64-bit
> >> + * kernels executing INT $0x80 will land on another entry point:
> >> + * entry_INT80_compat. The ABI is identical.)
> >
> >Agreed?
> >
> >Thanks,
> >
> > Ingo
>
> Sadly I believe Android still uses int $0x80 in the upstream version.
I don't see how that fact conflicts with my statement: on 64-bit kernels INT
$0x80
will (of course) work, but will land on another entry point:
entry_INT80_compat(),
not entry_INT80_32().
On 32-bit kernels the INT $0x80 entry point is entry_INT80_32().
Thanks,
Ingo
