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异步和同步模式下SocketChannel.read()的区别?

[英]The difference of SocketChannel.read() in async and sync mode?

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: 我知道Java NIO有两种模式,异步模式和同步模式。当我阅读SocketChannel.read()的javadoc时,得到以下解释:

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? 是让我感到困惑的是异步读取和同步读取的解释。是的,在异步模式下,它将立即读取缓冲区中已经存在的内容并返回。但是在同步模式下,它是不一样的吗?如果缓冲区中有东西,为什么不阅读并立即返回?等待什么呢?

if there is something in buffer, why is does't not read it and return immediately? 如果缓冲区中有东西,为什么不读取它并立即返回?

I think you understood it wrong. 我认为您理解错了。 As it is said in docs for int read(ByteBuffer dst) 正如在文档中所说的int read(ByteBuffer dst)

The buffer into which bytes are to be transferred. 要将字节传送到的缓冲区。

The channel reads bytes and writes them into buffer. 通道读取字节并将其写入缓冲区。

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. 但是,可以确保,如果通道处于阻塞模式并且缓冲区中至少有一个字节剩余,则此方法将阻塞直到读取至少一个字节。

at least one byte remaining in the buffer means that buffer still has free space. at least one byte remaining in the buffer意味着缓冲区仍具有可用空间。 Functions boolean hasRemaining() and int remaining() are used for checking this. 布尔值hasRemaining()和int missing()函数用于检查此内容。

wait for what? 等什么?

Wait for reading some bytes from the channel or reaching the end-of-stream. 等待从通道读取一些字节或到达流的末尾。 Reading on blocking channel couldn't return 0. 在阻塞通道上读取无法返回0。

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