8

我一直在寻找这个问题的答案，但实际上找不到任何东西。今天早些时候，我问我如何通过字节数组将文件变成字符串，然后再返回，以便稍后检索。

人们告诉我，我必须只存储字节数组，以避免讨厌的编码问题。所以现在我已经开始努力了，但我现在碰壁了。

基本上，我之前使用无缓冲流将文件转换为字节数组。这在理论上很有效，但它会占用大量内存，最终会抛出堆大小异常。我应该改用缓冲流（或者我被告知），我现在遇到的问题是从 BufferedInputStream 到 byte[]。我尝试复制和使用本文档中的方法

http://docs.guava-libraries.googlecode.com/git/javadoc/index.html?com/google/common/io/package-summary.html

我将无缓冲流交换为缓冲流。唯一的问题是，我不能直接将缓冲的输出流转换为字节数组，就像使用无缓冲的流一样。

帮助？:)

import java.io.BufferedInputStream;
import java.io.BufferedOutputStream;
import java.io.ByteArrayOutputStream;
import java.io.IOException;

public final class BufferedByteStream {

    private static final int BUF_SIZE = 1024000;

    public static long copy(BufferedInputStream from, BufferedOutputStream to) throws IOException { 
        byte[] buf = new byte[BUF_SIZE];
        long total = 0;
        while(true) {
            int r = from.read(buf);
            if(r == -1) {
                break;
            }
            to.write(buf, 0, r);
            total += r;
        }
        return total;
    }

    public static byte[] toByteArray(BufferedInputStream in) throws IOException {
        BufferedOutputStream out = new BufferedOutputStream(new ByteArrayOutputStream());
        copy(in, out);
        return out. // <--- Problem is here
    }
}

编辑：

我仍然收到堆空间错误。所以我现在将发布所有代码：

main.java

import java.io.*;
import java.util.logging.Level;
import java.util.logging.Logger;
import jserver.io.BufferedByteStream;
/**
*
* @author Vipar
*/
public class main {
    public static void main(String[] args) {
    File f = new File("<doesn't matter>");
        try {
            byte[] buf;
            try (BufferedInputStream bis = new BufferedInputStream(new FileInputStream(f))) {
                buf = BufferedByteStream.toByteArray(bis);
                bis.close();
            }
            File f2 = new File("<doesn't matter>");
            try (FileOutputStream fos = new FileOutputStream(f2)) {
                fos.write(buf);
                fos.close();
            }
        } catch (FileNotFoundException ex) {
            Logger.getLogger(main.class.getName()).log(Level.SEVERE, null, ex);
        } catch (IOException ex) {
            Logger.getLogger(main.class.getName()).log(Level.SEVERE, null, ex);
        }
    }
}

BufferedByteStream.java

import java.io.BufferedInputStream;
import java.io.BufferedOutputStream;
import java.io.ByteArrayOutputStream;
import java.io.IOException;

public final class BufferedByteStream {

    private static final int BUF_SIZE = 1024000;

    public static long copy(BufferedInputStream from, BufferedOutputStream to) throws IOException { 
        byte[] buf = new byte[BUF_SIZE];
        long total = 0;
        while(true) {
            int r = from.read(buf);
            if(r == -1) {
                break;
            }
            to.write(buf, 0, r);
            total += r;
        }
        return total;
    }

    public static byte[] toByteArray(BufferedInputStream in) throws IOException {
        ByteArrayOutputStream bytesOut = new ByteArrayOutputStream();
        BufferedOutputStream out = new BufferedOutputStream(bytesOut);
        copy(in, out);
        return bytesOut.toByteArray();
    }
}
4

2 回答 2

11

Have a look at ByteArrayOutputStream: Java 7 API java.io.ByteArrayOutputStream 

bytesOut = new ByteArrayOutputStream();
byte[] bytes = bytesOut.toByteArray();

Update: If you insist on doing what you are doing you can just assign the intermediate ByteArrayOutputStream to a variable and get hold of the array that way:

ByteArrayOutputStream bytesOut = new ByteArrayOutputStream()
BufferedOutputStream out = new BufferedOutputStream(bytesOut);
copy(in, out);
return bytesOut.toByteArray();

Update 2: The real question seems to be how to copy a file without reading it all into memory first:

1) Manually:

    byte[] buff = new byte[64*1024]; //or some size, can try out different sizes for performance
    BufferedInputStream in = new BufferedInputStream(new FileInputStream("fromFile"));
    BufferedOutputStream out = new BufferedOutputStream(new FileoutputStream("toFile"));
    int n = 0;
    while ((n = in.read(buff)) >= 0) {
        out.write(buff, 0, n);
    }
    in.close();
    out.close();

2) Efficiently by the OS and no loop etc:

FileChannel from = new FileInputStream(sourceFile).getChannel();
FileChanngel to = new FileOutputStream(destFile).getChannel();
to.transferFrom(from, 0, from.size());
//or from.transferTo(0, from.size(), to);
from.close();
to.close();

3) If you have Java 7 you can simplify exception and stream closing or just copy the file with the new APIs i in java 7:

java.nio.file.Files.copy(...);

see java.nio.file.Files

于 2012-05-17T20:19:55.367 回答
-1

DataFetcher 将是完美的：

http://tus.svn.sourceforge.net/viewvc/tus/tjacobs/io/DataFetcher.java?revision=34&view=markup

如果内存不足，可以使用 clearBuffer 清除读取之间的缓冲区

您还需要 Timeout 类 - 它在同一个包中的同一个项目中。

于 2012-05-17T20:39:21.390 回答