I know that java NIO have two modes , the asynchronized and synchronized mode.When I am reading the javadoc of SocketChannel.read(), I get the explaination below:
Reads a sequence of bytes from this channel into the given buffer.
An attempt is made to read up to r bytes from the channel, where r is the number of bytes remaining in the buffer, that is, dst.remaining(), at the moment this method is invoked.
Suppose that a byte sequence of length n is read, where 0 <= n <= r. This byte sequence will be transferred into the buffer so that the first byte in the sequence is at index p and the last byte is at index p + n - 1, where p is the buffer's position at the moment this method is invoked. Upon return the buffer's position will be equal to p + n; its limit will not have changed.
A read operation might not fill the buffer, and in fact it might not read any bytes at all. Whether or not it does so depends upon the nature and state of the channel. **A socket channel in non-blocking mode, for example, cannot read any more bytes than are immediately available from the socket's input buffer; similarly, a file channel cannot read any more bytes than remain in the file. It is guaranteed, however, that if a channel is in blocking mode and there is at least one byte remaining in the buffer then this method will block until at least one byte is read.**
This method may be invoked at any time. If another thread has already initiated a read operation upon this channel, however, then an invocation of this method will block until the first operation is complete.
what make me confused is the explanation of asynchronized and sychronized read .Yes , in asynchronized mode, it will immediately read what is now already in buffer and return.But in synchronized mode,it is not the same?if there is something in buffer, why is does't not read it and return immediately?wait for what?
if there is something in buffer, why is does't not read it and return immediately?
I think you understood it wrong. As it is said in docs for int read(ByteBuffer dst)
The buffer into which bytes are to be transferred.
The channel reads bytes and writes them into buffer.
It is guaranteed, however, that if a channel is in blocking mode and
there is at least one byte remaining in the buffer then this method
will block until at least one byte is read.
at least one byte remaining in the buffer means that buffer still has free space. Functions boolean hasRemaining() and int remaining() are used for checking this.
wait for what?
Wait for reading some bytes from the channel or reaching the end-of-stream.
Reading on blocking channel couldn't return 0.
Related
I have a few questions about the way InputStream.read() works. I'm trying to listen on InputStream and whenever bytes are available, I need them to be copied to a byte array and then the byte array is handed off to another method for further processing and then return to listening to the InputStream for next set of bytes. Here's how far I've gotten:
while(true){
while((i = inputstream.read(recvBuffer, 0, recvBuffer.length)) != -1){
doSomething(recvBuffer);
}
}
I read that read(byte[] b, int off, int len) method returns an int that is supposed to represent the number of bytes read from the stream. Doesn't that mean whenever bytes are available, it sets the value of i to the respective number of bytes? Once it reads x number of bytes and reaches the end of the stream, wouldn't it return a positive integer representing the number of bytes, instead of -1? Then, when would the check against -1 happen for i? I know I'm interpreting this wrong but I can't say how. Any help understanding this would be appreciated
Also, I know the max amount of bytes that a sender can push onto a stream. In this case, is it sufficient to specify the size of recvBuffer as the max amount of bytes or is it prudent to allocate a bit more than that?
doSomething(recvBuffer);
That should be
doSomething(recvBuffer, i);
The method needs to know how many bytes were actually received.
I read that read(byte[] b, int off, int len) method returns an int that is supposed to represent the number of bytes read from the stream.
Correct.
Doesn't that mean whenever bytes are available, it sets the value of
i to the respective number of bytes?
Yes.
Once it reads x number of bytes
and reaches the end of the stream, wouldn't it return a positive
integer representing the number of bytes, instead of -1?
No, it transfers the bytes and returns the count, then the next time there are no bytes, only end of stream, so it returns -1.
Then, when would the check against -1 happen for i? I know I'm interpreting this
wrong but I can't say how.
See above.
Also, I know the max amount of bytes that a sender can push onto a stream. In this case, is it sufficient to specify the size of recvBuffer as the max amount of bytes or is it prudent to allocate a bit more than that?
Most people use 4096 or 8192 bytes. There's not a lot of point in specifying a buffer larger than the path MTU in truth, which is normally < 1500, unless you are slow at reading so that the kernel socket receive buffer fills up.
Consider we have a socket connection between two device (A and B). Now if I write only 16 bytes (size doesn't matter here) to the output stream (not BufferedOutputStream) of the socket in side A 3 times or in general more than once like this:
OutputStream outputStream = socket.getOutputStream();
byte[] buffer = new byte[16];
OutputStream.write(buffer);
OutputStream.write(buffer);
OutputStream.write(buffer);
I read the data in side B using the socket input stream (not BufferedInputStream) with a buffer larger than sending buffer for example 1024:
InputStream inputStream = socket.getInputStream();
byte[] buffer = new byte[1024];
int read = inputStream.read(buffer);
Now I wonder how the data is received on side B? May it get accumulated or it exactly read the data as A sends it? In another word may the read variable get more than 16?
InputStream makes very few guarantees about how much data will be read by any invocation of the multi-byte read() methods. There is a whole class of common programming errors that revolve around misunderstandings and wrong assumptions about that. For example,
if InputStream.read(byte[]) reads fewer bytes than the provided array can hold, that does not imply that the end of the stream has been reached, or even that another read will necessarily block.
the number of bytes read by any one invocation of InputStream.read(byte[]) does not necessarily correlate to any characteristic of the byte source on which the stream draws, except that it will always read at least one byte when not at the end of the stream, and that it will not read more bytes than are actually available by the time it returns.
the number of available bytes indicated by the available() method does not reliably indicate how many bytes a subsequent read should or will read. A nonzero return value reliably indicates only that the next read will not block; a zero return value tells you nothing at all.
Subclasses may make guarantees about some of those behaviors, but most do not, and anyway you often do not know which subclass you actually have.
In order to use InputStreams properly, you generally must be prepared to perform repeated reads until you get sufficient data to process. That can mean reading until you have accumulated a specific number of bytes, or reading until a delimiter is encountered. In some cases you can handle any number of bytes from any given read; generally these are cases where you are looping anyway, and feeding everything you read to a consumer that can accept variable length chunks (many compression and encryption interfaces are like that, for example).
Per the docs:
public int read(byte[] b) throws IOException
Reads some number of bytes from the input stream and stores them into the buffer array b. The number of bytes
actually read is returned as an integer. This method blocks until
input data is available, end of file is detected, or an exception is
thrown. If the length of b is zero, then no bytes are read and 0 is
returned; otherwise, there is an attempt to read at least one byte. If
no byte is available because the stream is at the end of the file, the
value -1 is returned; otherwise, at least one byte is read and stored
into b.
The first byte read is stored into element b[0], the next one into
b[1], and so on. The number of bytes read is, at most, equal to the
length of b. Let k be the number of bytes actually read; these bytes
will be stored in elements b[0] through b[k-1], leaving elements b[k]
through b[b.length-1] unaffected.
Read(...) tells you how many bytes it put into the array and yes, you can read further; you'll get whatever was already there.
This is an exact quote from my text:
The purpose of I/O buffering is to improve system performance.
Rather than reading a byte at a time, a large number of bytes are read together
the first time the read() method is invoked.
However, when I use BufferedInputStream.read() all I can do is get a single byte. What am I doing wrong and what do I need to do?
It's not you, it is the stream that reads more than one character at a time. The BufferedInputStream keeps a buffer, and next time you call read() the next byte from that buffer is returned without any access to a physical drive (unless the buffer is empty and the next chunk of data has to be read into the buffer).
Note there are methods that read more than a byte, but these don't really have to do with the difference you explicitly asked for in your question.
The BufferedInputStream class facilitates buffering to your input streams. Rather than read one byte at a time from the network or disk, you read a larger block at a time.
You can set the buffer size to be used internally by the BufferedInputStream with the following constructor
InputStream input = new BufferedInputStream(new FileInputStream("PathOfFile"), 2 * 1024);
This example sets the buffer size to 2 KB
When the BufferedInputStream is created, an internal buffer array is created. As bytes from the stream are read or skipped, the internal buffer is refilled as necessary from the contained input stream, many bytes at a time
The javadoc says the following.
Parameters:
b - destination buffer.
off - offset at which to start storing bytes.
len - maximum number of bytes to read.
I would like to confirm my understanding of the "offset at which to start storing bytes". Does this mean that off is "the index at the destination buffer b at which to start storing bytes"? It does sound like off means it. I think the method is more usable if off is the "offset at the BufferedInputStream at which to start reading bytes", but I want to confirm. I tried looking at the source but it's hard to read.
A side question: When 1024 bytes of a stream is read, will the said 1024 bytes always be removed from the stream?
Does this mean that off is "the index at the destination buffer b at which to start storing bytes"?
It's documented: "The first byte read is stored into element b[off]".
When 1024 bytes of a stream is read, will the said 1024 bytes always be removed from the stream?
Of course, but you seem to be really asking whether 1024 bytes will always be read if you supply a buffer of 1024 bytes. Answer is no. It's documented: "there is an attempt to read at least one byte".
Yes. off is the index in b where the stream will start entering len bytes.
When 1024 bytes of a stream is read, will the said 1024 bytes always be removed from the stream?
Using an InputStream, you have no knowledge of what's going on underneath. All you know are the methods available to you and what they do (what their documentation says). Implementations can do whatever they want.
Does the read command check the size of the buffer when filling it with data or is there a chance that data is lost because buffer isn't big enough? In other words, if there are ten bytes of data available to be read, will the server continue to store the remaining 2 bytes of data until the next read.
I'm just using 8 as an example here to over dramatise the situation.
InputStream stdout;
...
while(condition)
{
...
byte[] buffer = new byte[8];
int len = stdout.read(buffer);
}
No, read() won't lose any data just because you haven't given it enough space for all the available bytes.
It's not clear what you mean by "the server" here, but the final two bytes of a 10 byte message would be available after the first read. (Or possible, the first read() would only read the first six bytes, leaving four still to read, for example.)