Hi, 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 = rafc.map(FileChannel.MapMode.READ_ONLY, 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: "[email protected] wan";<[email protected]>;
Date: Tue, Jul 15, 2014 10:44 AM
To: "java-user"<[email protected]>;
Subject: 答复：mmap confusion in lucene
Hi 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 <[email protected]>发送时间：2014年7月14日(星期一) 18:13收件人：java-user
<[email protected]>; wangzhijiang999 <[email protected]>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: [email protected]> -----Original
Message-----> From: wangzhijiang999 [mailto:[email protected]]> 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 =>
rafc.map(FileChannel.MapMode.READ_ONLY, 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 =>
rafc.map(FileChannel.MapMode.READ_ONLY, 0, rafc.size());> int len=buff.limit();
for (int i = 0; i < len; i++){ Byte b= buff.get(); }> But in this way, the
linux cache will be consumed just 4M.> > > RandomAccessFile raf = new
RandomAccessFile(new File(path), "r");> FileChannel rafc =
raf.getChannel();ByteBuffer buff => rafc.map(FileChannel.MapMode.READ_ONLY, 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
unsubscribe, e-mail: [email protected] additional
commands, e-mail: [email protected]
