Re: [Benchmark] Chain calls

["Alexander Gladysh" <agladysh@...>]

On Sat, Nov 15, 2008 at 3:09 PM, Alexander Gladysh <agladysh@gmail.com> wrote:
> On Sat, Nov 15, 2008 at 3:05 PM, Florian Weimer <fw@deneb.enyo.de> wrote:

>> * Alexander Gladysh:

>> > I guess that the cost of returning function is compensated by proper tail
>> > recursion in the "chain" case, and that x1.09 slowdown of the "plain_chain"
>> > case is that cost itself.

>> I think you are probably measuring GETUPVAL performance.

> I think, that does not explain the difference between "plain_chain"
> and "plain" cases. I would, however, adjust my cases to cache function
> locally and post updated results here.

Or maybe I misunderstood you. Could you please elaborate a little?

Here are the results for the cached upvalue modification (code attached as chaincallbench.lua).

lua
-------------------------------------------------------------------
                name |     rel | abs s / iter = us (1e-6 s) / iter
-------------------------------------------------------------------
         chain_upval |  1.0000 |  83.45 /  100000000 = 0.834500 us
         plain_local |  1.0115 |  84.41 /  100000000 = 0.844100 us
         plain_upval |  1.0260 |  85.62 /  100000000 = 0.856200 us
         chain_local |  1.0405 |  86.83 /  100000000 = 0.868300 us
   plain_chain_upval |  1.1081 |  92.47 /  100000000 = 0.924700 us
   plain_chain_local |  1.1395 |  95.09 /  100000000 = 0.950900 us
luajit -O
-------------------------------------------------------------------
                name |     rel | abs s / iter = us (1e-6 s) / iter
-------------------------------------------------------------------
         plain_local |  1.0000 |  10.35 /  100000000 = 0.103500 us
         plain_upval |  1.1884 |  12.30 /  100000000 = 0.123000 us
   plain_chain_local |  1.3652 |  14.13 /  100000000 = 0.141300 us
   plain_chain_upval |  1.4541 |  15.05 /  100000000 = 0.150500 us
         chain_upval |  1.5527 |  16.07 /  100000000 = 0.160700 us
         chain_local |  1.5691 |  16.24 /  100000000 = 0.162400 us

The code to run benchmark is attached as kbench.tar.bz2. (It was slightly updated from the last time I posted.) Run it as

  $ ./kbench.sh 'bench.lua chaincallbench.lua' 1e8 2>&1 | luajit -O kbenchparse.lua

Alexander.

Attachment: chaincallbench.lua
Description: Binary data

Attachment: kbench.tar.bz2
Description: BZip2 compressed data