在Java中将二进制文件读入ArrayList [英] Reading a binary file into ArrayList in Java

问题描述

我想在Java中读取二进制文件。我有153（1.bin，2.bin ... 153.bin）bin文件。我必须阅读。我以为我必须使用ArrayList缓冲。但是我做不到。我该怎么做？

经过一番研究，我发现在这个标题中（

  StringBuilder sb = new StringBuilder（）; 
 String fileName =/path/10.bin; 
 byte [] buffer; 
 
 try {
 
 buffer = Files.readAllBytes（Paths.get（fileName））; 
 
（byte b：buffer）{
 
 sb.append（String.format（％02X，b））; 
} 
 System.out.println（sb.toString（））; 
 $ b $ catch（IOException e）{
 // TODO自动生成的catch块
 e.printStackTrace（）; 
 
 
 
 $ b $ p 
 $ b $是我的问题的真正的方式还是我必须使用Arraylist缓冲？如果我使用单字节数组进行缓冲，是否必须清除其他二进制文件的缓冲区。
 
 
 
编辑：153单位表示153文件（1.bin，2。 bin ... 153.bin）
 
解决方案
我假定一个单元是一个字节。这是正确的吗？
 
 
 
 ArrayList不适合字节缓冲。它是一个实现了List接口的数组的包装类（通过这个接口，大部分的权力都被定义了）。         http://docs.oracle.com/javase/7/docs/api/java/util/ArrayList。如果你只是想从文件中读取字节，你可以使用FileInputStream。
 
 
  p>  http://docs.oracle.com/javase/tutorial/essential /io/bytestreams.html  
 
 
 
以下是一个简单的例子，它从包含hello world！的文件中读取输入。
 
 
  import java.io.FileInputStream; 
 import java.io.FileNotFoundException; 
 import java.io.IOException; 
 $ b $ public class Input {
 public static void main（String [] args）throws FileNotFoundException，IOException {
 try（FileInputStream in = new FileInputStream（test.txt）） {
 int c; （（c = in.read（））！= -1）
 
 System.out.print（（byte）c +）; 
 
 
 
 
 
 
以字节为单位的输出：104 101 108 108 111 32 119 111 114 108 100 33 13 10 
I want to read binary file in java. I have 153(1.bin, 2.bin...153.bin) bin file. I must read that. I thought that I must use ArrayList for buffering. But I could not do that. How can I do this ? 

After some research, I found that way in this title(Reading a binary input stream into a single byte array in Java). There is code like below in this title. 
    StringBuilder sb = new StringBuilder();
    String fileName  = "/path/10.bin";
    byte[] buffer;

    try {

        buffer = Files.readAllBytes(Paths.get(fileName));

        for(byte b : buffer){

            sb.append(String.format("%02X ", b));
        }
        System.out.println(sb.toString());

    } catch (IOException e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
    }
Is it true way for my question or do I must use Arraylist for buffering ? If I use single byte array for buffering, do I must clear the buffer for the other binary files.

Edit : 153 unit means 153 file(1.bin,2.bin ... 153.bin)
解决方案
I'm assuming a unit is one byte.  Is this correct?

An ArrayList is not appropriate for byte buffering.  It is a wrapper class for an array that implements the List interface (by which most of it's power is defined.)

http://docs.oracle.com/javase/7/docs/api/java/util/ArrayList.html

If you simply want to read bytes in from a file, you could use FileInputStream.

http://docs.oracle.com/javase/tutorial/essential/io/bytestreams.html

Here's a simple example that reads input from a file containing "hello world!"
import java.io.FileInputStream;
import java.io.FileNotFoundException;
import java.io.IOException;

public class Input {
public static void main(String[] args) throws FileNotFoundException, IOException {
    try (FileInputStream in = new FileInputStream("test.txt")) {
        int c;
        while ( (c = in.read()) != -1 )
            System.out.print((byte)c + " ");
    }
}
}
Output in bytes: 104 101 108 108 111 32 119 111 114 108 100 33 13 10

                        
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