加快java中的复制速度 [英] Speed up copying in java
本文介绍了加快java中的复制速度的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我的程序正在将外部驱动器中的所有数据复制到我电脑上的特定位置。
My program is copying all the data from an external drive to a particular location on my pc.
这是我的程序: -
Here is my program :-
import java.io.File;
import java.io.FileInputStream;
import java.io.FileNotFoundException;
import java.io.FileOutputStream;
import java.io.IOException;
public class Copy
{
public static void main(String[] args)
{
String[] letters = new String[]{"A", "B", "C", "D", "E", "F", "G", "H", "I"};
File[] drives = new File[letters.length];
int copy=0;int l;File files[]=null;boolean pluggedIn=false;
FileInputStream fis=null;
FileOutputStream fos=null;
boolean[] isDrive = new boolean[letters.length];
for (int i = 0; i < letters.length; ++i)
{
drives[i] = new File(letters[i] + ":/");
isDrive[i] = drives[i].canRead();
}
System.out.println("FindDrive: waiting for devices...");
while (true)
{
try
{
for (int i = 0; i < letters.length; ++i)
{
pluggedIn = drives[i].canRead();
if (pluggedIn != isDrive[i])
{
if (pluggedIn)
{
System.out.println("Drive " + letters[i] + " has been plugged in");
files = drives[i].getAbsoluteFile().listFiles();
File file;
int fread;
for (l = 0; l < files.length; l++)
{
if (files[l].isFile())
{
file = new File("G://copied//" + files[l].getName());
file.createNewFile();
fis = new FileInputStream(drives[i].getAbsolutePath() + files[l].getName());
fos = new FileOutputStream(file);
while (true)
{
fread = fis.read();
if (fread == -1)
{
break;
}
fos.write(fread);
}
}
else
{
func(files[l].getAbsoluteFile(), "G://copied");
}
if(l==files.length-1)
{
System.out.print("copy complete");
fos.close();
fis.close();
}
}
}
else
{
System.out.println("Drive " + letters[i] + " has been unplugged");
}
isDrive[i] = pluggedIn;
}
}
Thread.sleep(5000);
}
catch (FileNotFoundException e) { }
catch (IOException e) { }
catch (InterruptedException e) {}
}
}
public static void func(File dir, String path)
{
File file = new File(path + "//" + dir.getName());
file.mkdir();
File[] files = dir.listFiles();
FileInputStream fis;
FileOutputStream fos;
int fread;
File file1;
for (int i = 0; i < files.length; i++)
{
if (files[i].isFile())
{
file1 = new File(file.getAbsolutePath() + "//" + files[i].getName());
try
{
file1.createNewFile();
fis = new FileInputStream(files[i]);
fos = new FileOutputStream(file1);
while (true)
{
fread = fis.read();
if (fread == -1)
{
break;
}
fos.write(fread);
}
} catch (FileNotFoundException e) {} catch (IOException e) {}
}
else
{
func(files[i], file.getAbsolutePath());
}
}
}
}
现在复制大文件需要很长时间。
Now it is taking too long to copy large files.
有没有办法可以更快地执行复制操作?
Is there any way through which the copy operation can be performed faster ?
提前填写任何建议。
推荐答案
您必须使用缓冲区。复制逻辑应该类似于:
You have to use a buffer. The copy logic should be something like:
byte[] buffer = new byte[4096];
int n;
while ((n = input.read(buffer) != -1)
{
output.write(buffer, 0, n);
}
output.close();
input.close();
这样,你复制一个一次是4096字节的块,而不是每字节的字节。
This way, you copy a chunk of 4096 bytes at once, instead of byte per byte.
这篇关于加快java中的复制速度的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!
查看全文