Re: [PATCH 1/2] x86/bitops: implement __test_bit

From: H. Peter Anvin
Date: Tue Sep 01 2015 - 19:49:28 EST

On 09/01/2015 08:03 AM, Michael S. Tsirkin wrote:
>>> Hmm - so do you take back the ack?
>> I have no strong feelings either way, it simply strikes me as misguided to
>> explicitly optimize for something that is listed as a high overhead instruction.
> [mst@robin test]$ diff a.c b.c
> 31c31
> < if (__variable_test_bit(1, &addr))
> ---
> > if (__constant_test_bit(1, &addr))
> [mst@robin test]$ gcc -Wall -O2 a.c; time ./a.out
> real 0m0.532s
> user 0m0.531s
> sys 0m0.000s
> [mst@robin test]$ gcc -Wall -O2 b.c; time ./a.out
> real 0m0.517s
> user 0m0.517s
> sys 0m0.000s
> So __constant_test_bit is faster even though it's using more
> instructions
> $ gcc -Wall -O2 a.c; -objdump -ld ./a.out

I think this is well understood. The use of bts/btc in locked
operations is sometimes justified since it reports the bit status back
out, whereas in unlocked operations bts/btc has no benefit except for
code size. bt is a read operation, and is therefore "never/always"
atomic; it cannot be locked because there is no read/write pair to lock.

So it is strictly an issue of code size versus performance.

However, your test is simply faulty:

804843f: 50 push %eax
8048440: 6a 01 push $0x1
8048442: e8 b4 ff ff ff call 80483fb <__variable_test_bit>

You're encapsulating the __variable_test_bit() version into an expensive
function call, whereas the __constant_test_bit() seems to emit code that
is quite frankly completely bonkers insane:

8048444: 8b 45 ec mov -0x14(%ebp),%eax
8048447: 83 e0 1f and $0x1f,%eax
804844a: 89 c1 mov %eax,%ecx
804844c: d3 ea shr %cl,%edx
804844e: 89 d0 mov %edx,%eax
8048450: 83 e0 01 and $0x1,%eax
8048453: 85 c0 test %eax,%eax
8048455: 0f 95 c0 setne %al
8048458: 0f b6 c0 movzbl %al,%eax
804845b: 85 c0 test %eax,%eax
804845d: 74 00 je 804845f <main+0x64>

Observe the sequence and/test/setne/movzbl/test!


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