On 04/30/2015 05:13 PM, Eric Dumazet wrote:
On Thu, 2015-04-30 at 16:24 -0700, Alexander Duyck wrote:
Actually a byte operation itself is not faster. Note in the next line
we are returning the value. So what you typically end up with by doing
it that way would be 2 reads, one for the u8 and one for the u16 return
value. That is actually what I am trying to address in the second patch
in the set since we were doing a 8b test on the first byte of the
address followed by a 64b read.
The advantage with the way I wrote this is that the compiler itself
should be able to sort out how it wants to test the value while
accessing it in a 16b size. So at worst case it is a mask and compare,
followed by a return of the value. From what I have seen the compiler
seems to be smart enough on x86 anyway to just convert this into a one
byte compare on AL and then return the result in AX. I would suspect
that for bit-endian systems it would likely just perform the compare.
My compiler (4.8.2 (Ubuntu 4.8.2-19ubuntu1)) does the following :
62d: 0f b7 42 0c movzwl 0xc(%rdx),%eax
631: 0f b6 d0 movzbl %al,%edx
634: 83 fa 05 cmp $0x5,%edx
637: 7e 02 jle 63b <eth_type_trans+0x8b>
639: c9 leaveq
63a: c3 retq
Presumably this would be possible
movzwl 0xc(%rdx),%eax
cmp $0x5,%al
jle 63b <eth_type_trans+0x8b>
leaveq
retq
My compiler (5.0.1 (Red Hat 5.0.1-0.1)) does like what you have in the
"would be possible" example. What I end up with is something like this:
648: 0f b7 42 0c movzwl 0xc(%rdx),%eax
64c: 3c 05 cmp $0x5,%al
64e: 76 40 jbe 690 <eth_type_trans+0xc0>
The assembler before my patch was:
652: 0f b7 40 0c movzwl 0xc(%rax),%eax
656: 89 c2 mov %eax,%edx
658: 66 c1 c2 08 rol $0x8,%dx
65c: 0f b7 d2 movzwl %dx,%edx
65f: 81 fa ff 05 00 00 cmp $0x5ff,%edx
665: 7e 41 jle 6a8 <eth_type_trans+0xd8>
The savings isn't meant to be anything huge for the patch, maybe a cycle
or two. I suspect the before on your system is probably something
similar to what I had so we are still probably dropping at least 2
instructions.
- Alex
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