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

[Ingo Molnar]

* Andy Lutomirski <luto@xxxxxxxxxxxxxx> wrote:

> > >  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. [...]

How can 64-bit programs trigger entry_INT80_32? It's only ever set on 32-bit 
kernels:

#ifdef CONFIG_X86_32
        set_system_trap_gate(IA32_SYSCALL_VECTOR, entry_INT80_32);
        set_bit(IA32_SYSCALL_VECTOR, used_vectors);
#endif

> [...]  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.

That's true of INT $0x80, but I'm talking about the entry point: AFAICS 
entry_INT80_32 can only ever execute on 32-bit kernels.

We don't even build the entry_32.S::entry_INT80_32 entry point on 64-bit kernels:

obj-y                           := entry_$(BITS).o [...]

> 
> > 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.

So the part in parentheses conflict with your above statement :)

What I wanted to say with this:

> > > + * 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.)

... is what it says: that entry_INT80_32 is only active on 32-bit kernels, running 
32-bit programs, performing 32-bit system calls.

Programs running on 64-bit kernels can use INT $0x80 as well, but will land on 
another, different, 64-bit kernel specific entry point.

What am I missing?

Thanks,

	Ingo