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Message-ID: <Y5oCD0gFV+Cq1JqJ@casper.infradead.org>
Date:   Wed, 14 Dec 2022 17:04:15 +0000
From:   Matthew Wilcox <willy@...radead.org>
To:     Nico Pache <npache@...hat.com>
Cc:     linux-kernel@...r.kernel.org, linux-mm@...ck.org,
        muchun.song@...ux.dev, mike.kravetz@...cle.com,
        akpm@...ux-foundation.org, gerald.schaefer@...ux.ibm.com
Subject: Re: [RFC V2] mm: add the zero case to page[1].compound_nr in
 set_compound_order

On Tue, Dec 13, 2022 at 04:45:05PM -0700, Nico Pache wrote:
> Since commit 1378a5ee451a ("mm: store compound_nr as well as
> compound_order") the page[1].compound_nr must be explicitly set to 0 if
> calling set_compound_order(page, 0).
> 
> This can lead to bugs if the caller of set_compound_order(page, 0) forgets
> to explicitly set compound_nr=0. An example of this is commit ba9c1201beaa
> ("mm/hugetlb: clear compound_nr before freeing gigantic pages")
> 
> Collapse these calls into the set_compound_order by utilizing branchless
> bitmaths [1].
> 
> [1] https://graphics.stanford.edu/~seander/bithacks.html#ConditionalSetOrClearBitsWithoutBranching
> 
> V2: slight changes to commit log and remove extra '//' in the comments

We don't usually use // comments anywhere in the kernel other than
the SPDX header.

>  static inline void set_compound_order(struct page *page, unsigned int order)
>  {
> +	unsigned long shift = (1U << order);

Shift is a funny name for this variable.  order is the shift.  this is 'nr'.

>  	page[1].compound_order = order;
>  #ifdef CONFIG_64BIT
> -	page[1].compound_nr = 1U << order;
> +	// Branchless conditional:
> +	// order  > 0 --> compound_nr = shift
> +	// order == 0 --> compound_nr = 0
> +	page[1].compound_nr = shift ^ (-order  ^ shift) & shift;

Can the compiler see through this?  Before, the compiler sees:

	page[1].compound_order = 0;
	page[1].compound_nr = 1U << 0;
...
	page[1].compound_nr = 0;

and it can eliminate the first store.  Now the compiler sees:

	unsigned long shift = (1U << 0);
	page[1].compound_order = order;
	page[1].compound_nr = shift ^ (0  ^ shift) & shift;

Does it do the maths at compile-time, knowing that order is 0 at this
callsite and deducing that it can just store a 0?

I think it might, since shift is constant-1,

	page[1].compound_nr = 1 ^ (0 ^ 1) & 1;
->	page[1].compound_nr = 1 ^ 1 & 1;
->	page[1].compound_nr = 0 & 1;
->	page[1].compound_nr = 0;

But you should run it through the compiler and check the assembly
output for __destroy_compound_gigantic_page().

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