文章目录
引入
为什么要使用零拷贝技术？
传统写入数据需要4次拷贝，如下图：
传统IO
java
import java.io.*;
import java.net.Socket;
public class TranditionIOClient {
private static final int PORT = 8888;
private final static String FILE_NAME = "D:\\test.mp4";
// 接收缓冲区大小
private static final int BUFFER_SIZE = 1024;
public static void main(String[] args) throws Exception {
try (Socket socket = new Socket("localhost", PORT);
InputStream inputStream = new FileInputStream(FILE_NAME);
DataOutputStream dos = new DataOutputStream(socket.getOutputStream());) {
byte[] buffer = new byte[BUFFER_SIZE];
long readCount = 0;
long total = 0;
long startTime = System.currentTimeMillis();
// 读取文件：从硬盘读取到内存，发生2次copy(DMA拷贝和CPU拷贝)
while ((readCount = inputStream.read(buffer)) >= 0) {
total += readCount;
// 网络发送：从内存到网卡，发生2次copy(DMA拷贝和CPU拷贝)
dos.write(buffer, 0, (int) readCount);
}
System.out.println("TranditionIOClient发送总字节数：" + total + "，耗时：" + (System.currentTimeMillis() - startTime) + " ms");
} catch (IOException e) {
e.printStackTrace();
}
}
}内存映射mmap
java
import java.io.FileInputStream;
import java.net.InetSocketAddress;
import java.nio.ByteBuffer;
import java.nio.channels.FileChannel;
import java.nio.channels.SocketChannel;
public class MmapClient {
private static final int PORT = 8888;
private final static String FILE_NAME = "D:\\test.mp4";
public static void main(String[] args) throws Exception {
try (SocketChannel socketChannel = SocketChannel.open();
FileChannel fileChannel = new FileInputStream(FILE_NAME).getChannel()) {
socketChannel.connect(new InetSocketAddress("localhost", PORT));
socketChannel.configureBlocking(true);
long startTime = System.currentTimeMillis();
// 获取文件大小
long size = fileChannel.size();
// 内存映射整个文件，发生3次copy（DMA拷贝和CPU拷贝）
ByteBuffer buffer = fileChannel.map(FileChannel.MapMode.READ_ONLY, 0, size);
// 发送文件
while (buffer.hasRemaining()) {
socketChannel.write(buffer);
}
System.out.println("MmapClient发送总字节数：" + size + "，耗时：" + (System.currentTimeMillis() - startTime) + " ms");
} catch (Exception e) {
e.printStackTrace();
}
}
}文件描述符sendFile
java
import java.io.FileInputStream;
import java.net.InetSocketAddress;
import java.nio.channels.FileChannel;
import java.nio.channels.SocketChannel;
public class SendFileClient {
private static final int PORT = 8888;
private final static String FILE_NAME = "D:\\test.mp4";
public static void main(String[] args) throws Exception {
try (SocketChannel socketChannel = SocketChannel.open();
FileChannel fileChannel = new FileInputStream(FILE_NAME).getChannel()) {
socketChannel.connect(new InetSocketAddress("localhost", PORT));
socketChannel.configureBlocking(true);
long startTime = System.currentTimeMillis();
long position = 0;
// 8MB，与系统缓冲区大小匹配或略小以避免问题
long chunkSize = 8 * 1024 * 1024;
long size = fileChannel.size();
while (position < size) {
long bytesRemaining = size - position;
// 确保最后一次传输不会超过文件大小
long count = Math.min(bytesRemaining, chunkSize);
// transferTo⽅法⽤到了零拷⻉，底层是sendfile，发生2次copy（DMA拷贝）
long transferCount = fileChannel.transferTo(position, count, socketChannel);
if (transferCount == 0) {
// 如果一次传输没有发生，可能需要检查连接是否仍然活跃或处理其他错误情况
break;
}
position += transferCount;
}
System.out.println("SendFileClient发送总字节数：" + size + "，耗时：" + (System.currentTimeMillis() - startTime) + " ms");
} catch (Exception e) {
e.printStackTrace();
}
}
}如果发送的文件不大于8M，则可以简单写，如下：
java
import java.io.FileInputStream;
import java.net.InetSocketAddress;
import java.nio.channels.FileChannel;
import java.nio.channels.SocketChannel;
public class SendFileClient {
private static final int PORT = 8888;
private final static String FILE_NAME = "D:\\test.mp4";
public static void main(String[] args) throws Exception {
try (SocketChannel socketChannel = SocketChannel.open();
FileChannel fileChannel = new FileInputStream(FILE_NAME).getChannel()) {
socketChannel.connect(new InetSocketAddress("localhost", PORT));
socketChannel.configureBlocking(true);
long startTime = System.currentTimeMillis();
// transferTo⽅法⽤到了零拷⻉，底层是sendfile，发生2次copy（DMA拷贝）
long transferCount = fileChannel.transferTo(0, fileChannel.size(), socketChannel);
System.out.println("SendFileClient发送总字节数：" + transferCount + "，耗时：" + (System.currentTimeMillis() - startTime) + " ms");
} catch (Exception e) {
e.printStackTrace();
}
}
}测试
服务端代码如下：
java
import java.io.DataInputStream;
import java.io.IOException;
import java.net.ServerSocket;
import java.net.Socket;
public class Server {
private static final int PORT = 8888;
// 接收缓冲区大小
private static final int BUFFER_SIZE = 1024;
public static void main(String[] args) throws Exception {
try (ServerSocket ss = new ServerSocket(PORT);) {
while (true) {
try (Socket s = ss.accept();
DataInputStream dis = new DataInputStream(s.getInputStream());) {
int byteCount = 0;
byte[] bytes = new byte[BUFFER_SIZE];
while (true) {
int readCount = dis.read(bytes, 0, BUFFER_SIZE);
if (readCount == -1) {
break;
}
byteCount = byteCount + readCount;
}
System.out.println("服务端接受字节数：" + byteCount + "字节");
} catch (IOException e) {
e.printStackTrace();
}
}
} catch (IOException e) {
e.printStackTrace();
}
}
}我们都使用了test.mp4进行测试，文件大小500M，测试结果如下：
java
TranditionIOClient发送总字节数：524288000，耗时：9590 ms
MmapClient发送总字节数：524288000，耗时：1182 ms
SendFileClient发送总字节数：524288000，耗时：983 ms总结
零拷贝并不是不需要拷贝，而是指计算机执行操作时，不需要将数据从内存复制到应用程序
效率高到低：sendFile>mmap>传统IO
明明传了500M的文件，但实际读出来8M?代码如下：
java
try (SocketChannel socketChannel = SocketChannel.open();
FileChannel fileChannel = new FileInputStream(FILE_NAME).getChannel()) {
socketChannel.connect(new InetSocketAddress("localhost", PORT));
socketChannel.configureBlocking(true);
long startTime = System.currentTimeMillis();
// transferTo⽅法⽤到了零拷⻉，底层是sendfile，发生2次copy（DMA拷贝）
long transferCount = fileChannel.transferTo(0, fileChannel.size(), socketChannel);
System.out.println("SendFileClient发送总字节数：" + transferCount + "，耗时：" + (System.currentTimeMillis() - startTime) + " ms");
} catch (Exception e) {
e.printStackTrace();
}
// 输出结果：SendFileClient发送总字节数：8388608，耗时：15 ms原因：由于操作系统的默认socket缓冲区大小限制所导致的。当使用transferTo进行大文件传输时，如果文件大小超过了操作系统为socket分配的缓冲区大小，那么transferTo可能在达到这个限制后停止，因为它试图一次性将数据从文件通道转移到socket通道，但缓冲区不足以容纳整个文件内容。
解决：分批次进行文件传输