I have a case where I need to peek ahead in the stream for the existence of a certain regular expression and then read data from the stream.
mark and reset allow me to do this but I am facing an issue where mark becomes invalid if the readAheadLimit goes beyond the size of the current buffer.
For example: I have a BufferedReader with buffer size of 1k.
Lets say I am at position 1000 (mark=1000) in the buffer and I need to check for the regex in the next 100 chars (readAheadLimit=100).
So while reading, the moment I cross the current buffer size (1024), a new buffer is allocated and the mark becomes invalid (not able to reset) and the data is streamed into the new buffer in a normal way.
I think this is the intended behavior but is there a way to get around this?
Appreciate your help.
regards
the moment I cross the current buffer size (1024), a new buffer is allocated
No it isn't. The existing buffer is cleared and readied for another use.
and the mark becomes invalid (not able to reset)
No it doesn't, unless you've gone beyond the read ahead limit.
You don't seem to have read the API. You call mark() with an argument that says how far ahead you want to go before calling reset(), in this case 100 bytes, and the API is required to allow you to do exactly that. So when you get up to 100 characters ahead, call reset(), and you are back where you were when you called mark(). How that happens internally isn't your problem, but it is certainly required to happen.
And how did you get a BufferedReader with a 1k buffer? The default is 4096.
There are at least two options:
Set default cache size much more than 1k:
new BufferedReader(originalReader, 1024 * 1024) // e.g. 1Mb
Apply custom buffering to increase cache size as soon as limit was exceeded. In case if you are working with huge amount of data - custom buffering can store data it in database or file.
Related
For example I have a file whose content is：
abcdefg
then i use the following code to read 'defg'.
ByteBuffer bb = ByteBuffer.allocate(4);
int read = channel.read(bb, 3);
assert(read == 4);
Because there's adequate data in the file so can I suppose so? Can I assume that the method returns a number less than limit of the given buffer only when there aren't enough bytes in the file?
Can I assume that the method returns a number less than limit of the given buffer only when there aren't enough bytes in the file?
The Javadoc says:
a read might not fill the buffer
and gives some examples, and
returns the number of bytes read, possibly zero, or -1 if the channel has reached end-of-stream.
This is NOT sufficient to allow you to make that assumption.
In practice, you are likely to always get a full buffer when reading from a file, modulo the end of file scenario. And that makes sense from an OS implementation perspective, given the overheads of making a system call.
But, I can also imagine situations where returning a half empty buffer might make sense. For example, when reading from a locally-mounted remote file system over a slow network link, there is some advantage in returning a partially filled buffer so that the application can start processing the data. Some future OS may implement the read system call to do that in this scenario. If assume that you will always get a full buffer, you may get a surprise when your application is run on the (hypothetical) new platform.
Another issue is that there are some kinds of stream where you will definitely get partially filled buffers. Socket streams, pipes and console streams are obvious examples. If you code your application assuming file stream behavior, you could get a nasty surprise when someone runs it against another kind of stream ... and fails.
No, in general you cannot assume that the number of bytes read will be equal to the number of bytes requested, even if there are bytes left to be read in the file.
If you are reading from a local file, chances are that the number of bytes requested will actually be read, but this is by no means guaranteed (and won't likely be the case if you're reading a file over the network).
See the documentation for the ReadableByteChannel.read(ByteBuffer) method (which applies for FileChannel.read(ByteBuffer) as well). Assuming that the channel is in blocking mode, the only guarantee is that at least one byte will be read.
I was first curious about how the buffer size of 'BufferedInputStream' class is decided for each program. I found out in STACKOVERFLOW that the default buffer size is 8KB.
I wonder if there is profound meaning in the default buffer size of
'BufferInputStream' class, which is 8KB in size.
When I searched java docs for information about the 'BufferedInputStream' class,
I found out there are two forms of constructors for the class.
One constructor form offers users the ability to change the buffer size.
BufferedInputStream(InputStream in, int size)
Could the buffer size of this class be critical in deciding performance of some programs? I'm curious if anyone uses the above form of the constructor to change the buffer size to fit/optimize his/her program.
Is there any profound meaning to the default buffer size of 8KB?
Thank you for reading.
Could the buffer size of this class be critical in deciding performance of some programs? I'm curious if anyone uses the above form of the constructor to change the buffer size to fit/optimize his/her program.
Probably not. Changing from a buffer size of 1 to 2 will about double your performance (by reducing system calls). Changing from 2 to 4 will double it again. Changing from 4 to 8, again. You get the idea. At some point this ceases being true, as the performance ceases being dominated by system calls and starts being dominated by transfer sizes. 8k is a good place to stop. Use more if you like but you won't notice much difference.
Is there any profound meaning to the default buffer size of 8KB?
There isn't. It is 8k in size. By default. That's the meaning. You can change it via a constructor. Nothing more to it.
When I construct a new BufferedReader it is providing me a buffer of 8192 characters. What is the logic/reason behind this?
8192 = 2 to the power of 13
Traditionally, memory managers and paging files in the operating system work on pages that are sized in powers of 2. This allows very efficient multiply/divide operations to be performed with left/right shift operations. When working with a buffer, the worst case scenario is to have a buffer with size 1 byte longer than the page size (that would result in an extra page swap with very low benefit). So the default buffer sizes will also tend to be implemented in factors of two.
I'd assume (but have not checked) that the JVM looks for buffers like this and attempts to align them on page boundaries.
Why does this matter? Page misses are quite expensive. If you are doing a ton of IO, it's better to avoid the case where the page backing the buffer gets swapped out to disk (kind of defeats the purpose of the buffer). That said, for most applications, this is a micro-optimization, and for the vast majority of cases, the default is fine.
For reference, Windows and Linux both currently use a 4KB memory page size. So the default buffer on BufferedReader will consume exactly 2 pages.
As the BufferedReader Javadoc says
The buffer size may be specified, or the default size may be used. The default is large enough for most purposes.
The default was chosen as being "large enough" (which I would interpret as "good enough").
8192, as you said, is 2^13. The exact reason for this number being the default is hard to come by, but I'd venture to say it's based on the combination of normal use scenarios and data efficiency. You can specify a buffer size of whatever you want, though, using a different object constructor.
BufferedReader(Reader in, int sz)
Creates a buffering character-input stream that uses an input buffer of the specified size.
https://docs.oracle.com/javase/7/docs/api/java/io/BufferedReader.html
and BufferedReader default buffer size? will provide further insight.
There is a JDK ticket https://bugs.openjdk.org/browse/JDK-4953311 that states
Most OSes that we support uses a buffer size of 8192 (8K) bytes for their IO buffering, and this is also the buffer size used by Microsoft VM on Win32. We should change the default buffer size in these two classes to 8K.
8192 is 2^13 and also reveals much information regarding the RIGHT v WRONG encoded in all we do. If one takes away or adds to the author intent, he modifies and therefore corrupts the entire thing. Try to add or take away from something perfect... good luck!
I have made an application in android that lets the user compress and decompress files and I used the package java.util.zip. Everything is okay. the speed, files are totally compressed and decompressed together with the directories. The only problem is that the application is not able to compress/decompress large files (greater than 1gb).
I believe the problem is the size of my buffer. Other codes that I've seen, the value of their buffer is 1024 or 2048 or 8192 but my value of my buffer is base on the size of the chosen file (just to make it flexible). But once the user chose a large file (with a size of >8 digits), that's were the error comes out. I searched over the net and also here in this site but I can't find an answer. my problem is similar to this:
To Compress a big file in a ZIP with Java
Thanks for the future help! :)
EDIT:
Thanks for the comments and answers. It really helped a lot. I thought BUFFER in compressing/decompressing in java means the size of file so in my program, I made the buffer size flexible (buffer size = file size). Will someone please explain how buffer works so I can understand why is it okay that BUFFER has a fixed value. Also for me to figure it out why others people is telling that it is much better if the buffer size is 8k or else. Thanks a lot! :)
If you size the buffer to the size of the file, then it means that you will have OutOfMemoryError whenever the file size is too big for memory available.
Use a normal buffer size and let it do it's work - buffering the data in a streaming fashion, one chunk at a time, rather than all in one go.
For explanation, see for example the documentation of BufferedOutputStream:
The class implements a buffered output stream. By setting up such an
output stream, an application can write bytes to the underlying output
stream without necessarily causing a call to the underlying system for
each byte written.
So using a buffer is more efficient than non-buffered writing.
And from the write method:
Ordinarily this method stores bytes from the given array into this
stream's buffer, flushing the buffer to the underlying output stream
as needed. If the requested length is at least as large as this
stream's buffer, however, then this method will flush the buffer and
write the bytes directly to the underlying output stream.
Each write causes the in-memory buffer to fill up, until the buffer is full. When the buffer is full, it is flushed and cleared. If you use a very large buffer, you will cause a large amount of data to be stored in memory before flushing. If your buffer is the same size as the input file, then you are saying you need to read the whole content into memory before flushing it. Using the default buffer size is usually just fine. There will be more physical writes (flushes); you avoid exploding memory.
By allowing you to specify a specific buffer size, the API is letting you choose the right balance between memory consumption and i/o to suit your application. If you tune your application for performance, you might end up tweaking buffer size. But the default size will be reasonable for many situations.
It sounds like it would help to simply set a maximum size for the buffer, something like:
//After calculating the buffer size bufSize:
bufSize = Math.min(bufSize, MAXSIZE);
For example I have a file whose content is：
abcdefg
then i use the following code to read 'defg'.
ByteBuffer bb = ByteBuffer.allocate(4);
int read = channel.read(bb, 3);
assert(read == 4);
Because there's adequate data in the file so can I suppose so? Can I assume that the method returns a number less than limit of the given buffer only when there aren't enough bytes in the file?
Can I assume that the method returns a number less than limit of the given buffer only when there aren't enough bytes in the file?
The Javadoc says:
a read might not fill the buffer
and gives some examples, and
returns the number of bytes read, possibly zero, or -1 if the channel has reached end-of-stream.
This is NOT sufficient to allow you to make that assumption.
In practice, you are likely to always get a full buffer when reading from a file, modulo the end of file scenario. And that makes sense from an OS implementation perspective, given the overheads of making a system call.
But, I can also imagine situations where returning a half empty buffer might make sense. For example, when reading from a locally-mounted remote file system over a slow network link, there is some advantage in returning a partially filled buffer so that the application can start processing the data. Some future OS may implement the read system call to do that in this scenario. If assume that you will always get a full buffer, you may get a surprise when your application is run on the (hypothetical) new platform.
Another issue is that there are some kinds of stream where you will definitely get partially filled buffers. Socket streams, pipes and console streams are obvious examples. If you code your application assuming file stream behavior, you could get a nasty surprise when someone runs it against another kind of stream ... and fails.
No, in general you cannot assume that the number of bytes read will be equal to the number of bytes requested, even if there are bytes left to be read in the file.
If you are reading from a local file, chances are that the number of bytes requested will actually be read, but this is by no means guaranteed (and won't likely be the case if you're reading a file over the network).
See the documentation for the ReadableByteChannel.read(ByteBuffer) method (which applies for FileChannel.read(ByteBuffer) as well). Assuming that the channel is in blocking mode, the only guarantee is that at least one byte will be read.