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From "Uwe Schindler" <>
Subject RE: 答复:答复:mmap confusion in lucene
Date Tue, 15 Jul 2014 09:29:03 GMT
Yes, the JVM is removing the get() call, because it knows that it has no side-effect: the position()
pointer is not used afterwards and the result of the get() call is also not used. It is partly
mapped because the optimization only starts to kick in after 10,000 method calls (the default
threshold in the JVM).


Uwe Schindler
H.-H.-Meier-Allee 63, D-28213 Bremen

> -----Original Message-----
> From: wangzhijiang999 []
> Sent: Tuesday, July 15, 2014 11:10 AM
> To: java-user
> Subject: 答复:答复:mmap confusion in lucene
> Hi 308181687,
>          I also tested in this way. If print every byte, the OS cache will consume
> the size of file at last,about 800M.
> for (int j = 0; j < len; ++j){        System.out.println(buff.get());}
> If just call buff.get() in loop, the OS cache will consume only 8M at last.
> for (int j = 0; j < len; ++j){        byte b=buff.get();}
> The buff.get() means reading the byte at this buffer's current position, and
> then increments the position. But actually if you do not use the value from
> buff.get(), FS  will not read the disk. And I monitored the disk read and cache
> condition by dstat -md command to confirm that the disk read will not
> increase for the second test.
> As you said, the jvm is so smart that if you do not use the data  , it will not
> read from disk. As my previous understanding, as long as you use get
> method to fetch data, it should read from disk no matter whether
> you actually use the data or not. I will continue researching on it to find the
> real reason.
> ------------------------------------------------------------------发件人:308181687
> <>发送时间:2014年7月15日(星期二) 13:04收件人
> :java-user <>主 题:Re:答复:mmap
> confusion in luceneHi, Zhiiang It seems that the jvm is smart enough to
> ignore the unused code. Try the following code:RandomAccessFile raf = new
> RandomAccessFile(new File("/root/xx.txt"), "r");FileChannel rafc =
> raf.getChannel();ByteBuffer buff =
>, 0, rafc.size());int
> len=buff.limit();byte b = 0;for (int i = 0; i < len; i++){b + = buff.get();}The java
> process will consume the expected 800M share memory. But if change the
> line of " b + = buff.get()" to "b = buff.get()", the java process will not
> consume so much share memory, i guess that the jvm is smart enough to
> directly skip to the the last pos of the bytebuffer .Thanks & Best Regards!‍-----
> ------------- Original ------------------From: "
> wan";<>;Date: Tue, Jul 15, 2014 10:44 AMTo:
> "java-user"<>; Subject: 答复:mmap
> confusion in luceneHi Uwe,Thank you for always help. For my first testing I
> am clear of it, it is becuase the OS cache the whole file because of copying
> data to java heap and it does not free the page, then I see 800M used by
> cache in the end.But for my last two testings, the OS has freed all the
> previous cached pages, so I see the cache used only 4M in the end.Maybe I
> am not very clear of the internal kernel mechanism. As I understand, the
> kernel will swap out the page when the memory resource is limited or the
> cached page is not used for long time. The first condition is not satisfied in my
> testing, because the OS still has 30G memory available for use. For the
> second condition, although the bytes are copied to java heap in first test, but
> when the program ends to quit, the OS still reserve the cache. In the last
> test, the OS released the page even in the running process of program.
> Would you give me some further explaination for this? I am very
> appreciated.Zhiiang Wang--------------------------------------------------------------
> ----发件人:Uwe Schindler <>发送时间:2014年7月14
> 日(星期一) 18:13收件人:java-user <>;
> wangzhijiang999 <>主 题:RE: mmap
> confusion in luceneThis is very easy to explain:In the first part you copy the
> whole memory mapped stuff into a on-heap byte array. You allocate this
> byte array in total and you then do a copy (actually this is a standard libc copy
> call) of the whole file. To do this copy, the underlying OS will need to swap in
> the whole file, because it "sees" that you want to read the whole file anyway
> (because of the size of they copy operation).The other example reads the
> stuff byte by byte in a Java for-loop. The operating system has no idea how
> to optimize that, so whenever you cross page boundaries it will swap in
> another buffer. Because of internal kernel-page-garbage collection, the
> pages swapped in are freed much faster. This is OS specific.In general
> copying a random access file to java heap with mmap is just the wrong use
> case. Lucene never does this! The idea behind mmap is to *not copy* the
> data and work on the mmapped region directly (using random access). The
> OS cache logic will then use statistics about which pages were actually used
> and keep them longer in FS cache than those used one time and then no
> longer used for very long time.Uwe-----Uwe SchindlerH.-H.-Meier-Allee 63,
> D-28213 Bremenhttp://www.thetaphi.deeMail:> -----
> Original Message-----> From: wangzhijiang999
> []> Sent: Monday, July 14, 2014 11:58
> AM> To: java-user> Subject: mmap confusion in lucene> > Hi everybody, I
> found a problem confused me when I tested the mmap> feature in lucene. I
> tested to read a file size of 800M by mmap method like> below:> >
> RandomAccessFile raf = new RandomAccessFile(new File(path), "r");>
> FileChannel rafc = raf.getChannel();ByteBuffer buff =>
>, 0, rafc.size());> int
> len=buff.limit(); byte[] b = new byte[len]; for (int i = 0; i < len;> i++){ b[i]
> buff.get(); }> After the program finished, the linux cache will be consumed
> about 800M.> > > RandomAccessFile raf = new RandomAccessFile(new
> File(path), "r");> FileChannel rafc = raf.getChannel();ByteBuffer buff =>
>, 0, rafc.size());> int
> len=buff.limit(); for (int i = 0; i < len; i++){ Byte b= buff.get(); }> But in
> way, the linux cache will be consumed just 4M.> > > RandomAccessFile raf =
> new RandomAccessFile(new File(path), "r");> FileChannel rafc =
> raf.getChannel();ByteBuffer buff =>
>, 0, rafc.size());> int
> len=buff.limit(); byte[] b = new byte[len]; for (int i = 0; i < len;> i++){ b[i]
> buff.get();> b[i]=0; }> In this way, the linux cache will be also consumed 4M.>
> > The whole content of the file should be read for above three tests, but
> for> the last two testings, the linux system only cached 4M .> Would
> somebody give me the explaination about this? Thanks in advane.> > Zhijiang
> Wang> ---------------------------------------------------------------------To
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