java io 的问题！
请教各位大侠，java io 操作从文件读取数据那一（几）个类效率较高？
在线等>>>
谢谢！

------解决方案--------------------

java.io.FileInputStream


------解决方案--------------------
java.io.BufferedReader
------解决方案--------------------
BufferedInputStream 应该效率高些。 最基本的道理，减少disk access的次数嘛。
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用nio, 效率高:

import java.io.FileInputStream;
import java.io.FileOutputStream;
import java.io.IOException;
import java.nio.ByteBuffer;
import java.nio.MappedByteBuffer;
import java.nio.channels.FileChannel;



/** *//**
 * @author maoyeye
 *
 * TODO To change the template for this generated type comment go to
 * Window - Preferences - Java - Code Style - Code Templates
 */
public class FileUtil ...{

public static ByteBuffer readFile(String filename) throws Exception
{
FileChannel fiChannel = new FileInputStream(filename).getChannel();
MappedByteBuffer mBuf;
mBuf = fiChannel.map(FileChannel.MapMode.READ_ONLY, 0, fiChannel.size());
fiChannel.close();
fiChannel = null;

return mBuf;

}


public static void saveFile(ByteBuffer src, String filename) throws Exception
{
FileChannel foChannel = new FileOutputStream(filename).getChannel();
foChannel.write(src);
foChannel.close();
foChannel = null; 
}

public static void saveFile(FileChannel fiChannel, String filename) throws IOException
{
MappedByteBuffer mBuf;
mBuf = fiChannel.map(FileChannel.MapMode.READ_ONLY, 0, fiChannel.size());

FileChannel foChannel = new FileOutputStream(filename).getChannel();
foChannel.write(mBuf);

fiChannel.close();
foChannel.close(); 

fiChannel = null;
foChannel = null; 
}


public static void main(String[] args) throws Exception 
{
final String suffix = ".txt";
final String filename = "bufferTemp";
final String srcFile = filename + suffix ;
final String backupFile = filename + "-" + System.currentTimeMillis() + suffix;
ByteBuffer byteBuffer = FileUtil.readFile(srcFile);
FileUtil.saveFile(byteBuffer, backupFile);
byteBuffer = null;

}
}
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一般用Buffered*的类效率要高些。
------解决方案--------------------
一般用Buffered*的类效率要高些。

同意