Re: [PATCH] x86/entry: Improve system call entry comments

[Ingo Molnar]

* Andy Lutomirski <luto@xxxxxxxxxx> 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