Messages in this thread | | | Date | Tue, 8 Mar 2016 11:27:55 +0100 | From | Ingo Molnar <> | Subject | Re: [PATCH] x86/entry: Improve system call entry comments |
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* Andy Lutomirski <luto@amacapital.net> 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
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