Re: Is _PAGE_PROTNONE set only for user mappings?

[Dave Hansen]

On 5/12/22 22:33, Hyeonggon Yoo wrote:
> Thanks Mel, and IIUC nor does do_kern_addr_fault() in arch/x86/mm/fault.c
> expect _PAGE_PROTNONE instead of _PAGE_GLOBAL. I want to make it clear
> in the code that _PAGE_PROTNONE is only used for user mappings.
> 
> How does below look?
> 
> diff --git a/arch/x86/include/asm/pgtable_types.h b/arch/x86/include/asm/pgtable_types.h
> index 40497a9020c6..f8d02b91a90c 100644
> --- a/arch/x86/include/asm/pgtable_types.h
> +++ b/arch/x86/include/asm/pgtable_types.h
> @@ -35,7 +35,10 @@
>  #define _PAGE_BIT_DEVMAP	_PAGE_BIT_SOFTW4
>  
>  /* If _PAGE_BIT_PRESENT is clear, we use these: */
> -/* - if the user mapped it with PROT_NONE; pte_present gives true */
> +/*
> + * if the user mapped it with PROT_NONE; pte_present gives true
> + * this is only used for user mappings (with _PAGE_BIT_USER)
> + */
>  #define _PAGE_BIT_PROTNONE	_PAGE_BIT_GLOBAL
>  
>  #define _PAGE_PRESENT	(_AT(pteval_t, 1) << _PAGE_BIT_PRESENT)
> @@ -115,7 +118,8 @@
>  #define _PAGE_DEVMAP	(_AT(pteval_t, 0))
>  #endif
>  
> -#define _PAGE_PROTNONE	(_AT(pteval_t, 1) << _PAGE_BIT_PROTNONE)
> +#define _PAGE_PROTNONE	((_AT(pteval_t, 1) << _PAGE_BIT_USER) | \
> +			 (_AT(pteval_t, 1) << _PAGE_BIT_PROTNONE))
>  
>  /*
>   * Set of bits not changed in pte_modify.  The pte's

I don't like the idea of _PAGE_BIT_USER being so implicit.  It is
something kernel users should know explicitly that they are messing with.

I was thinking of something more along the lines of taking the
set_memory.c code and ensuring that it never sets (or even observes)
_PAGE_BIT_GLOBAL on a _PAGE_USER mapping.  There was also a question of
if set_memory.c is ever used on userspace mappings.  It would be good to
validate whether it's possible in-tree today and if not, enforce that
_PAGE_USER PTEs should never even be observed with set_memory.c.

The arch/x86/mm/dump_pagetables.c code is also a reasonable place to put
assumptions about the page tables since it walks *everything* when asked.