linux-kernel - Re: [PATCH] x86/mm: Simplify redundant overlap calculation

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Date: Tue, 23 Jan 2024 09:00:45 -0800
From: Dave Hansen <dave.hansen@...el.com>
To: David Binderman <dcb314@...mail.com>,
 Dave Hansen <dave.hansen@...ux.intel.com>,
 "linux-kernel@...r.kernel.org" <linux-kernel@...r.kernel.org>
Cc: Andy Lutomirski <luto@...nel.org>, Peter Zijlstra <peterz@...radead.org>,
 Thomas Gleixner <tglx@...utronix.de>, Ingo Molnar <mingo@...hat.com>,
 Borislav Petkov <bp@...en8.de>, "x86@...nel.org" <x86@...nel.org>
Subject: Re: [PATCH] x86/mm: Simplify redundant overlap calculation

On 1/23/24 08:54, David Binderman wrote:
>> Remove the second condition.  It is exactly the same as the first.
> I don't think the first condition is sufficient. I suspect something like
> 
>        return (r2_start <= r1_start && r1_start <= r2_end) ||
>                (r2_start <= r1_end && r1_end <= r2_end);
> 
> Given the range [r2_start .. r2_end], then if r1_start or r1_end
> are in that range, you have overlap.
> 
> Unless you know different.

First of all, I've gotten these bounds checks wrong in code more times
than I can count.  I have zero trust that I'll get them right. :)

But the compiler seems to know different at least:

int  overlaps1(unsigned long r1_start, unsigned long r1_end,
	      unsigned long r2_start, unsigned long r2_end)
{
	return  (r1_start <= r2_end && r1_end >= r2_start) ||
		(r2_start <= r1_end && r2_end >= r1_start);
}

int  overlaps2(unsigned long r1_start, unsigned long r1_end,
	      unsigned long r2_start, unsigned long r2_end)
{
	return (r1_start <= r2_end && r1_end >= r2_start);
}

Results in:

0000000000001180 <overlaps1>:
    1180:	f3 0f 1e fa          	endbr64
    1184:	48 39 cf             	cmp    %rcx,%rdi
    1187:	49 89 d0             	mov    %rdx,%r8
    118a:	0f 96 c2             	setbe  %dl
    118d:	31 c0                	xor    %eax,%eax
    118f:	4c 39 c6             	cmp    %r8,%rsi
    1192:	0f 93 c0             	setae  %al
    1195:	21 d0                	and    %edx,%eax
    1197:	c3                   	ret
    1198:	0f 1f 84 00 00 00 00 	nopl   0x0(%rax,%rax,1)
    119f:	00

00000000000011a0 <overlaps2>:
    11a0:	f3 0f 1e fa          	endbr64
    11a4:	48 39 cf             	cmp    %rcx,%rdi
    11a7:	49 89 d0             	mov    %rdx,%r8
    11aa:	0f 96 c2             	setbe  %dl
    11ad:	31 c0                	xor    %eax,%eax
    11af:	4c 39 c6             	cmp    %r8,%rsi
    11b2:	0f 93 c0             	setae  %al
    11b5:	21 d0                	and    %edx,%eax
    11b7:	c3                   	ret
    11b8:	0f 1f 84 00 00 00 00 	nopl   0x0(%rax,%rax,1)
    11bf:	00

I also wrote a quick program to throw random numbers into both versions
and see if they differ.  They never did, which they obviously can't if
they're the exact same instructions.