Note that looping with primitive int is faster than with long, and native
array functions accepts/returns int instead of long for their index and
length.

It is very hard to eliminate boxing without dropping to java. In you
example, calling bit-xor does 2 autoboxing (and 1 long to int cast as
aset-byte takes int index), as there is no byte variant of
clojure.lang.Numbers/xor.

Also you cannot detect it without advanced profiler or bytecode analysis.

Best,
Jozef


On Thu, Mar 13, 2014 at 1:34 PM, Alex Miller <a...@puredanger.com> wrote:

> Agreed with all the comments on this so far. I would also say that dotimes
> is slower than loop for stuff like this so I would also make that change.
>
> (defn inplace-xor [^bytes a ^bytes b ^bytes out]
>   (let [len (alength a)]
>     (loop [i 0]
>       (when (< i len)
>         (aset-byte out i (bit-xor (aget a i) (aget b i)))
>         (recur (inc i))))))
>
> In this scenario loop/recur will use a primitive long for i.  And then I
> would look at the bytecode to verify no boxing is occurring.
>
> On Thursday, March 13, 2014 3:36:14 AM UTC-5, Michael Gardner wrote:
>>
>> Might be slow because of the polymorphic nature of nth. If you replace
>> nth with aget (and turn on *warn-on-reflection*, which is a good idea when
>> performance-tuning), you'll get reflection warnings because Clojure doesn't
>> know what Java method to use since it doesn't know what type of objects a
>> and b are. Once you silence those with type hints like in Walter's example,
>> you get roughly a 2x speedup (in my tests).
>>
>> BTW, I'd write it like this:
>>
>> (defn inplace-xor [^bytes a ^bytes b ^bytes out]
>>     (dotimes [i (alength a)]
>>         (aset-byte out i
>>             (bit-xor (aget a i) (aget b i)))))
>>
>
>
>>
>> On Mar 13, 2014, at 00:26 , Ignacio Corderi <icor...@soe.ucsc.edu>
>> wrote:
>>
>> > Hey guys, here is a huge performance problem I'm trying to figure out:
>> >
>> > ;; Say you have 2 data arrays sitting out there of 1 MB
>> >
>> > (def one-mb (byte-array (* 1024 1024)))
>> > (def another-mb (byte-array (* 1024 1024)))
>> >
>> > ;; and another one that should have the byte-by-byte XOR of the
>> previous two
>> >
>> > (def out-mb (byte-array (* 1024 1024)))
>> >
>> > ;; question is... how do you code this guy, so that it doesn't take
>> forever
>> >
>> > (defn inplace-xor [a b out]
>> >   (def ln (count a))
>> >   (loop [x 0]
>> >     (if (< x ln)
>> >       (do
>> >         (aset-byte out x (bit-xor (nth a x) (nth b x)))
>> >         (recur (+ x 1))
>> >         ))))
>> >
>> > ;; checking the time
>> >
>> > (time (inplace-xor one-mb another-mb out-mb))
>> >
>> > ;; takes about ~400ms which is.... well... A LOT
>> >
>> > ;; I'm happy to receive a solution that involves calling some java
>> library...
>> >
>> > Thanks in advance!
>> > -Ignacio
>> >
>> >
>> >
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