On Mar 7, 2016 12:22 AM, "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?
>

64-bit programs can and sometimes do trigger this entry point.  It
does a 32-bit syscall regardless of the caller's bitness, but it
returns back to the caller's original context, whatever it was.

> 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.)

I like the part in parentheses.

--Andy

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