Java - try-with-resources 中的 ZipOutputStream

Question

I have this Java code that creates and saves a zip file with a text file inside it:

public static void main(String[] args) {
    try (ByteArrayOutputStream baos = new ByteArrayOutputStream();
         ZipOutputStream zos = new ZipOutputStream(baos)) {
        String s = "Hello, world!";
        ZipEntry entry = new ZipEntry("text_file.txt");
        zos.putNextEntry(entry);
        zos.write(s.getBytes());
        zos.closeEntry();

        try (OutputStream os = new FileOutputStream("zip_file.zip")) {
            baos.writeTo(os);
            System.out.println("ZIP file created successfully");
        }
    } catch (IOException e) {
        System.out.println("Error: " + e);
    }
}

The problem with this code is that it creates a zip file that is corrupted, so I cannot extract or even open the content within it.

Eventually I found out that if I manually close the ZipOutputStream by putting zos.close() after zos.closeEntry() the zip file is created successfully without corruption. Honestly this does not make any sense to me since I have declared the ZipOutputStream inside a try-with-resource statement, so I would have expected it to close automatically.

So it seems that the try-with-resource statement does not actually close the stream. Am I doing something wrong?

Any help would be appreciated.

OS: Windows 10

Note: I used ByteArrayOutputStream because in the real scenario I have to create the zip file in memory, since I don't have a folder on a disk to rely on.

Answer 1

The try-with-resources does close the ZipOutputStream . It's just that it closes the zip stream after you copy its content to the FileOutputStream . How do you expect work that happens after the file has been closed to affect the content of the file?

I recommend replacing the ByteArrayOutputStream with the FileOutputStream . What need do you have to buffer?

try (OutputStream os = new FileOutputStream("zip_file.zip");
     ZipOutputStream zos = new ZipOutputStream(os)) {
  String s = "Hello, world!";
  ZipEntry entry = new ZipEntry("text_file.txt");
  zos.putNextEntry(entry);
  zos.write(s.getBytes());
  zos.closeEntry();
}
System.out.println("ZIP file created successfully");

Answer 2

It turned out that I was misplacing the ByteArrayOutputStream object: I should have not closed it by placing it inside the try-with-resource statement because, as explained here , ByteArrayOutputStream is purely memory based so the allocated memory will be released automatically by the garbage collector.

The following code produces the desired result:

public static void main(String[] args) {
    ByteArrayOutputStream baos = new ByteArrayOutputStream();

    try (ZipOutputStream zos = new ZipOutputStream(baos)) {
        String s = "Hello, world!";
        ZipEntry entry = new ZipEntry("text_file.txt");
        zos.putNextEntry(entry);
        zos.write(s.getBytes());
        zos.closeEntry();
    } catch (IOException e) {
        System.out.println("Error: " + e);
    }

    try (OutputStream os = new FileOutputStream("zip_file.zip")) {
        baos.writeTo(os);
        System.out.println("ZIP file created successfully");
    } catch (IOException e) {
        System.out.println("Error: " + e);
    }
}

Answer 3

To keep everything in same Try-catch, another approach is to finish & flush the ZipOutputStream before using the content of ByteArrayOutputStream inside it (which can be incomplete, this is why it may generate corrupted zip), for example:

 try (ByteArrayOutputStream baos = new ByteArrayOutputStream();
     ZipOutputStream zos = new ZipOutputStream(baos)) {
    //codes insert into zip
    zos.finish(); //finish writing
    zos.flush(); //flush any buffer inside
    //use baos 
 } catch (IOException e) {
    System.out.println("Error: " + e);
 }